Gilbert Strang; Edwin “Jed” Herman

5.7 Change of Variables in Multiple Integrals

Learning Objectives

5.7.1 Determine the image of a region under a given transformation of variables.
5.7.2 Compute the Jacobian of a given transformation.
5.7.3 Evaluate a double integral using a change of variables.
5.7.4 Evaluate a triple integral using a change of variables.

Recall from Substitution Rule the method of integration by substitution. When evaluating an integral such as $\int_{2}^{3} x {(x^{2} - 4)}^{5} d x,$ we substitute $u = g (x) = x^{2} - 4 .$ Then $d u = 2 x d x$ or $x d x = \frac{1}{2} d u$ and the limits change to $u = g (2) = 2^{2} - 4 = 0$ and $u = g (3) = 9 - 4 = 5 .$ Thus the integral becomes $\int_{0}^{5} \frac{1}{2} u^{5} d u$ and this integral is much simpler to evaluate. In other words, when solving integration problems, we make appropriate substitutions to obtain an integral that becomes much simpler than the original integral.

We also used this idea when we transformed double integrals in rectangular coordinates to polar coordinates and transformed triple integrals in rectangular coordinates to cylindrical or spherical coordinates to make the computations simpler. More generally,

\int_{a}^{b} f (x) d x = \int_{c}^{d} f (g (u)) g' (u) d u,

Where $x = g (u), d x = g' (u) d u,$ and $u = c$ and $u = d$ satisfy $c = g^{- 1} (a)$ and $d = g^{- 1} (b) .$

A similar result occurs in double integrals when we substitute $x = h (r, θ) = r cos θ,$ $y = g (r, θ) = r sin θ,$ and $d A = d x d y = r d r d θ .$ Then we get

\iint_{R} f (x, y) d A = \iint_{S} f (r cos θ, r sin θ) r d r d θ

where the domain $R$ is replaced by the domain $S$ in polar coordinates. Generally, the function that we use to change the variables to make the integration simpler is called a transformation or mapping.

Planar Transformations

A planar transformation $T$ is a function that transforms a region $G$ in one plane into a region $R$ in another plane by a change of variables. Both $G$ and $R$ are subsets of $ℝ^{2} .$ For example, Figure 5.71 shows a region $G$ in the $u v -plane$ transformed into a region $R$ in the $x y -plane$ by the change of variables $x = g (u, v)$ and $y = h (u, v),$ or sometimes we write $x = x (u, v)$ and $y = y (u, v) .$ We shall typically assume that each of these functions has continuous first partial derivatives, which means $g_{u}, g_{v}, h_{u},$ and $h_{v}$ exist and are also continuous. The need for this requirement will become clear soon.

On the left-hand side of this figure, there is a region G with point (u, v) given in the Cartesian u v-plane. Then there is an arrow from this graph to the right-hand side of the figure marked with x = g(u, v) and y = h(u, v). On the right-hand side of this figure there is a region R with point (x, y) given in the Cartesian xy- plane. — Figure 5.71 The transformation of a region $G$ in the $u v -plane$ into a region $R$ in the $x y -plane .$

Definition

A transformation $T : G \to R,$ defined as $T (u, v) = (x, y),$ is said to be a one-to-one transformation if no two points map to the same image point.

To show that $T$ is a one-to-one transformation, we assume $T (u_{1}, v_{1}) = T (u_{2}, v_{2})$ and show that as a consequence we obtain $(u_{1}, v_{1}) = (u_{2}, v_{2}) .$ If the transformation $T$ is one-to-one in the domain $G,$ then the inverse $T^{−1}$ exists with the domain $R$ such that $T^{−1} \circ T$ and $T \circ T^{−1}$ are identity functions.

Figure 5.71 shows the mapping $T (u, v) = (x, y)$ where $x$ and $y$ are related to $u$ and $v$ by the equations $x = g (u, v)$ and $y = h (u, v) .$ The region $G$ is the domain of $T$ and the region $R$ is the range of $T,$ also known as the image of $G$ under the transformation $T .$

Example 5.65

Determining How the Transformation Works

Suppose a transformation $T$ is defined as $T (r, θ) = (x, y)$ where $x = r cos θ, y = r sin θ .$ Find the image of the polar rectangle $G = {(r, θ) | 0 < r \leq 1, 0 \leq θ \leq π / 2}$ in the $r θ -plane$ to a region $R$ in the $x y -plane .$ Show that $T$ is a one-to-one transformation in $G$ and find $T^{−1} (x, y) .$

Solution

Since $r$ varies from 0 to 1 in the $r θ -plane,$ we have a circular disc of radius 0 to 1 in the $x y -plane .$ Because $θ$ varies from 0 to $π /2$ in the $r θ -plane,$ we end up getting a quarter circle of radius $1$ in the first quadrant of the $x y -plane$ (Figure 5.72). Hence $R$ is a quarter circle bounded by $x^{2} + y^{2} = 1$ in the first quadrant.

On the left-hand side of this figure, there is a rectangle G with a marked subrectangle given in the first quadrant of the Cartesian r theta-plane. Then there is an arrow from this graph to the right-hand side of the figure marked with x = r cos theta and y = r sin theta. On the right-hand side of this figure there is a quarter circle R with a marked subannulus (analogous to the rectangle in the other graph) given in the Cartesian x y-plane. — Figure 5.72 A rectangle in the $r θ -plane$ is mapped into a quarter circle in the $x y -plane .$

In order to show that $T$ is a one-to-one transformation, assume $T (r_{1}, θ_{1}) = T (r_{2}, θ_{2})$ and show as a consequence that $(r_{1}, θ_{1}) = (r_{2}, θ_{2}) .$ In this case, we have

\begin{matrix} T (r_{1}, θ_{1}) & = & T (r_{2}, θ_{2}), \\ (x_{1}, y_{1}) & = & (x_{2}, y_{2}), \\ (r_{1} cos θ_{1}, r_{1} sin θ_{1}) & = & (r_{2} cos θ_{2}, r_{2} sin θ_{2}), \\ r_{1} cos θ_{1} & = & r_{2} cos θ_{2} \\ r_{1} sin θ_{1} & = & r_{2} sin θ_{2} \end{matrix}

Dividing, we obtain

\begin{matrix} \frac{r_{1} cos θ_{1}}{r_{1} sin θ_{1}} & = & \frac{r_{2} cos θ_{2}}{r_{2} sin θ_{2}} \\ \frac{cos θ_{1}}{sin θ_{1}} & = & \frac{cos θ_{2}}{sin θ_{2}} \\ cot θ_{1} & = & cot θ_{2} \\ θ_{1} & = & θ_{2} \end{matrix}

since the cotangent function is one-one function in the interval $0 \leq θ \leq π / 2 .$ Also, since $0 < r \leq 1,$ we have $r_{1} = r_{2}, θ_{1} = θ_{2} .$ Therefore, $(r_{1}, θ_{1}) = (r_{2}, θ_{2})$ and $T$ is a one-to-one transformation from $G$ into $R .$

To find $T^{−1} (x, y)$ solve for $r, θ$ in terms of $x, y .$ We already know that $r^{2} = x^{2} + y^{2}$ and $tan θ = \frac{y}{x} .$ Thus $T^{−1} (x, y) = (r, θ)$ is defined as $r = \sqrt{x^{2} + y^{2}}$ and $θ = {tan}^{−1} (\frac{y}{x}) .$

Example 5.66

Finding the Image under $T$

Let the transformation $T$ be defined by $T (u, v) = (x, y)$ where $x = u^{2} - v^{2}$ and $y = u v .$ Find the image of the triangle in the $u v -plane$ with vertices $(0, 0), (0, 1),$ and $(1, 1) .$

Solution

The triangle and its image are shown in Figure 5.73. To understand how the sides of the triangle transform, call the side that joins $(0, 0)$ and $(0, 1)$ side $A,$ the side that joins $(0, 0)$ and $(1, 1)$ side $B,$ and the side that joins $(1, 1)$ and $(0, 1)$ side $C .$

On the left-hand side of this figure, there is a triangular region given in the Cartesian uv-plane with boundaries A, B, and C represented by the v axis, the line u = v, and the line v = 1, respectively. Then there is an arrow from this graph to the right-hand side of the figure marked with x = u squared minus v squared and y = u v. On the right-hand side of this figure there is a complex region given in the Cartesian x y-plane with boundaries A’, B’, and C’ given by the x axis, y axis, and a line curving from (negative 1, 0) through (0, 1), namely x = y squared minus 1, respectively. — Figure 5.73 A triangular region in the $u v -plane$ is transformed into an image in the $x y -plane .$

For the side $A : u = 0, 0 \leq v \leq 1$ transforms to $x = − v^{2}, y = 0$ so this is the side $A'$ that joins $(−1, 0)$ and $(0, 0) .$

For the side $B : u = v, 0 \leq u \leq 1$ transforms to $x = 0, y = u^{2}$ so this is the side $B^{'}$ that joins $(0, 0)$ and $(0, 1) .$

For the side $C : 0 \leq u \leq 1, v = 1$ transforms to $x = u^{2} - 1, y = u$ (hence $x = y^{2} - 1)$ so this is the side $C^{'}$ that makes the upper half of the parabolic arc joining $(−1, 0)$ and $(0, 1) .$

All the points in the entire region of the triangle in the $u v -plane$ are mapped inside the parabolic region in the $x y -plane .$

Checkpoint 5.43

Let a transformation $T$ be defined as $T (u, v) = (x, y)$ where $x = u + v, y = 3 v .$ Find the image of the rectangle $G = {(u, v) : 0 \leq u \leq 1, 0 \leq v \leq 2}$ from the $u v -plane$ after the transformation into a region $R$ in the $x y -plane .$ Show that $T$ is a one-to-one transformation and find $T^{−1} (x, y) .$

Jacobians

Recall that we mentioned near the beginning of this section that each of the component functions must have continuous first partial derivatives, which means that $g_{u}, g_{v}, h_{u},$ and $h_{v}$ exist and are also continuous. A transformation that has this property is called a $C^{1}$ transformation (here $C$ denotes continuous). Let $T (u, v) = (g (u, v), h (u, v)),$ where $x = g (u, v)$ and $y = h (u, v),$ be a one-to-one $C^{1}$ transformation. We want to see how it transforms a small rectangular region $S,$ $Δ u$ units by $Δ v$ units, in the $u v -plane$ (see the following figure).

On the left-hand side of this figure, there is a region S with lower right corner point (u sub 0, v sub 0), height Delta v, and length Delta u given in the Cartesian u v-plane. Then there is an arrow from this graph to the right-hand side of the figure marked with T. On the right-hand side of this figure there is a region R with point (x sub 0, y sub 0) given in the Cartesian x y-plane with sides r(u, v sub 0) along the bottom and r(u sub 0, v) along the left. — Figure 5.74 A small rectangle $S$ in the $u v -plane$ is transformed into a region $R$ in the $x y -plane .$

Since $x = g (u, v)$ and $y = h (u, v),$ we have the position vector $r (u, v) = g (u, v) i + h (u, v) j$ of the image of the point $(u, v) .$ Suppose that $(u_{0}, v_{0})$ is the coordinate of the point at the lower left corner that mapped to $(x_{0}, y_{0}) = T (u_{0}, v_{0}) .$ The line $v = v_{0}$ maps to the image curve with vector function $r (u, v_{0}),$ and the tangent vector at $(x_{0}, y_{0})$ to the image curve is

r_{u} = g_{u} (u_{0}, v_{0}) i + h_{u} (u_{0}, v_{0}) j = \frac{\partial x}{\partial u} i + \frac{\partial y}{\partial u} j .

Similarly, the line $u = u_{0}$ maps to the image curve with vector function $r (u_{0}, v),$ and the tangent vector at $(x_{0}, y_{0})$ to the image curve is

r_{v} = g_{v} (u_{0}, v_{0}) i + h_{v} (u_{0}, v_{0}) j = \frac{\partial x}{\partial v} i + \frac{\partial y}{\partial v} j .

Now, note that

r_{u} = lim_{Δ u \to 0} \frac{r (u_{0} + Δ u, v_{0}) - r (u_{0}, v_{0})}{Δ u} so r (u_{0} + Δ u, v_{0}) - r (u_{0}, v_{0}) \approx Δ u r_{u} .

Similarly,

r_{v} = lim_{Δ v \to 0} \frac{r (u_{0}, v_{0} + Δ v) - r (u_{0}, v_{0})}{Δ v} so r (u_{0}, v_{0} + Δ v) - r (u_{0}, v_{0}) \approx Δ v r_{v} .

This allows us to estimate the area $Δ A$ of the image $R$ by finding the area of the parallelogram formed by the sides $Δ v r_{v}$ and $Δ u r_{u} .$ By using the cross product of these two vectors by adding the k component as $0,$ the area $Δ A$ of the image $R$ (refer to The Cross Product) is approximately $|| Δ {ur}_{u} \times Δ {vr}_{v} || = || r_{u} \times r_{v} || Δ u Δ v .$ In determinant form, the cross product is

r_{u} \times r_{v} = | \begin{matrix} i & j & k \\ \frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} & 0 \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v} & 0 \end{matrix} | = | \begin{array}{l} \frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v} \end{array} | k = (\frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} \frac{\partial y}{\partial u}) k .

Since $|| k || = 1,$ we have $Δ A \approx || r_{u} \times r_{v} || Δ u Δ v = (\frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} \frac{\partial y}{\partial u}) Δ u Δ v .$

Definition

The Jacobian of the $C^{1}$ transformation $T (u, v) = (g (u, v), h (u, v))$ is denoted by $J (u, v)$ and is defined by the $2 \times 2$ determinant

J (u, v) = | \frac{\partial (x, y)}{\partial (u, v)} | = | \begin{array}{l} \frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v} \end{array} | = (\frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} \frac{\partial y}{\partial u}) .

Using the definition, we have

Δ A \approx J (u, v) Δ u Δ v = | \frac{\partial (x, y)}{\partial (u, v)} | Δ u Δ v .

Note that the Jacobian is frequently denoted simply by

J (u, v) = \frac{\partial (x, y)}{\partial (u, v)} .

Note also that

| \begin{array}{l} \frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v} \end{array} | = (\frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} \frac{\partial y}{\partial u}) = | \begin{array}{l} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{array} | .

Hence the notation $J (u, v) = \frac{\partial (x, y)}{\partial (u, v)}$ suggests that we can write the Jacobian determinant with partials of $x$ in the first row and partials of $y$ in the second row.

Example 5.67

Finding the Jacobian

Find the Jacobian of the transformation given in Example 5.65.

Solution

The transformation in the example is $T (r, θ) = (r cos θ, r sin θ)$ where $x = r cos θ$ and $y = r sin θ .$ Thus the Jacobian is

\begin{matrix} J (r, θ) & = \frac{\partial (x, y)}{\partial (r, θ)} = | \begin{array}{l} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial θ} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial θ} \end{array} | = | \begin{array}{l} cos θ & - r sin θ \\ sin θ & r cos θ \end{array} | \\ = r {cos}^{2} θ + r {sin}^{2} θ = r ({cos}^{2} θ + {sin}^{2} θ) = r . \end{matrix}

Example 5.68

Finding the Jacobian

Find the Jacobian of the transformation given in Example 5.66.

Solution

The transformation in the example is $T (u, v) = (u^{2} - v^{2}, u v)$ where $x = u^{2} - v^{2}$ and $y = u v .$ Thus the Jacobian is

J (u, v) = \frac{\partial (x, y)}{\partial (u, v)} = | \begin{array}{l} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{array} | = | \begin{array}{l} 2 u & v \\ - 2 v & u \end{array} | = 2 u^{2} + 2 v^{2} .

Checkpoint 5.44

Find the Jacobian of the transformation: $T (u, v) = (u + v, 2 v) .$

Change of Variables for Double Integrals

We have already seen that, under the change of variables $T (u, v) = (x, y)$ where $x = g (u, v)$ and $y = h (u, v),$ a small region $Δ A$ in the $x y -plane$ is related to the area formed by the product $Δ u Δ v$ in the $u v -plane$ by the approximation

Δ A \approx J (u, v) Δ u, Δ v .

Now let’s go back to the definition of double integral for a minute:

\iint_{R} f (x, y) d A = lim_{m, n \to \infty} \sum_{i = 1}^{m} \sum_{j = 1}^{n} f (x_{i j}, y_{i j}) Δ A .

Referring to Figure 5.75, observe that we divided the region $S$ in the $u v -plane$ into small subrectangles $S_{i j}$ and we let the subrectangles $R_{i j}$ in the $x y -plane$ be the images of $S_{i j}$ under the transformation $T (u, v) = (x, y) .$

On the left-hand side of this figure, there is a rectangle S with an inscribed red oval and a subrectangle with lower right corner point (u sub ij, v sub ij), height Delta v, and length Delta u given in the Cartesian u v-plane. Then there is an arrow from this graph to the right-hand side of the figure marked with T. On the right-hand side of this figure there is a region R with inscribed (deformed) red oval and a subrectangle R sub ij with corner point (x sub ij, y sub ij) given in the Cartesian x y-plane. The subrectangle is blown up and shown with vectors pointing along the edge from the corner point. — Figure 5.75 The subrectangles $S_{i j}$ in the $u v -plane$ transform into subrectangles $R_{i j}$ in the $x y -plane .$

Then the double integral becomes

\iint_{R} f (x, y) d A = lim_{m, n \to \infty} \sum_{i = 1}^{m} \sum_{j = 1}^{n} f (x_{i j}, y_{i j}) Δ A = lim_{m, n \to \infty} \sum_{i = 1}^{m} \sum_{j = 1}^{n} f (g (u_{i j}, v_{i j}), h (u_{i j}, v_{i j})) | J (u_{i j}, v_{i j}) | Δ u Δ v .

Notice this is exactly the double Riemann sum for the integral

\iint_{S} f (g (u, v), h (u, v)) | \frac{\partial (x, y)}{\partial (u, v)} | d u d v .

Theorem 5.14

Change of Variables for Double Integrals

Let $T (u, v) = (x, y)$ where $x = g (u, v)$ and $y = h (u, v)$ be a one-to-one $C^{1}$ transformation, with a nonzero Jacobian on the interior of the region $S$ in the $u v -plane;$ it maps $S$ into the region $R$ in the $x y -plane .$ If $f$ is continuous on $R,$ then

\iint_{R} f (x, y) d A = \iint_{S} f (g (u, v), h (u, v)) | \frac{\partial (x, y)}{\partial (u, v)} | d u d v .

With this theorem for double integrals, we can change the variables from $(x, y)$ to $(u, v)$ in a double integral simply by replacing

d A = d x d y = | \frac{\partial (x, y)}{\partial (u, v)} | d u d v

when we use the substitutions $x = g (u, v)$ and $y = h (u, v)$ and then change the limits of integration accordingly. This change of variables often makes any computations much simpler.

Example 5.69

Changing Variables from Rectangular to Polar Coordinates

Consider the integral

\int_{0}^{2} \int_{0}^{\sqrt{2 x - x^{2}}} \sqrt{x^{2} + y^{2}} d y d x .

Use the change of variables $x = r cos θ$ and $y = r sin θ,$ and find the resulting integral.

Solution

First we need to find the region of integration. This region is bounded below by $y = 0$ and above by $y = \sqrt{2 x - x^{2}}$ (see the following figure).

A semicircle in the first quadrant of the xy plane with radius 1 and center (1, 0). The equation for this curve is given as y = the square root of (2x minus x squared) — Figure 5.76 Changing a region from rectangular to polar coordinates.

Squaring and collecting terms, we find that the region is the upper half of the circle $x^{2} + y^{2} - 2 x = 0,$ that is, $y^{2} + {(x - 1)}^{2} = 1 .$ In polar coordinates, the circle is $r = 2 cos θ$ so the region of integration in polar coordinates is bounded by $0 \leq r \leq 2 cos θ$ and $0 \leq θ \leq \frac{π}{2} .$

The Jacobian is $J (r, θ) = r,$ as shown in Example 5.67. Since $r \geq 0,$ we have $| J (r, θ) | = r .$

The integrand $\sqrt{x^{2} + y^{2}}$ changes to $r$ in polar coordinates, so the double iterated integral is

\int_{0}^{2} \int_{0}^{\sqrt{2 x - x^{2}}} \sqrt{x^{2} + y^{2}} d y d x = \int_{0}^{π / 2} \int_{0}^{2 cos θ} r | J (r, θ) | d r d θ = \int_{0}^{π / 2} \int_{0}^{2 cos θ} r^{2} d r d θ .

Checkpoint 5.45

Considering the integral $\int_{0}^{1} \int_{0}^{\sqrt{1 - x^{2}}} (x^{2} + y^{2}) d y d x,$ use the change of variables $x = r cos θ$ and $y = r sin θ,$ and find the resulting integral.

Notice in the next example that the region over which we are to integrate may suggest a suitable transformation for the integration. This is a common and important situation.

Example 5.70

Changing Variables

Consider the integral $\iint_{R} (x - y) d y d x,$ where $R$ is the parallelogram joining the points $(1, 2),$ $(3, 4), (4, 3),$ and $(6, 5)$ (Figure 5.77). Make appropriate changes of variables, and write the resulting integral.

A parallelogram R with corners (1, 2), (3, 4), (6, 5), and (4, 3). — Figure 5.77 The region of integration for the given integral.

Solution

First, we need to understand the region over which we are to integrate. The sides of the parallelogram are $x - y + 1 = 0, x - y - 1 = 0,$ $x - 3 y + 5 = 0, and x - 3 y + 9 = 0$ (Figure 5.78). Another way to look at them is $x - y = −1, x - y = 1,$ $x - 3 y = −5,$ and $x - 3 y = - 9 .$

Clearly the parallelogram is bounded by the lines $y = x + 1, y = x - 1, y = \frac{1}{3} (x + 5),$ and $y = \frac{1}{3} (x + 9) .$

Notice that if we were to make $u = x - y$ and $v = x - 3 y,$ then the limits on the integral would be $−1 \leq u \leq 1$ and $−9 \leq v \leq - 5 .$

To solve for $x$ and $y,$ we multiply the first equation by $3$ and subtract the second equation, $3 u - v = (3 x - 3 y) - (x - 3 y) = 2 x .$ Then we have $x = \frac{3 u - v}{2} .$ Moreover, if we simply subtract the second equation from the first, we get $u - v = (x - y) - (x - 3 y) = 2 y$ and $y = \frac{u - v}{2} .$

A parallelogram R with corners (1, 2), (3, 4), (6, 5), and (4, 3) formed by the lines y = x + 1, y = x minus 1, y = (x + 9)/3, and y = (x + 5)/3. — Figure 5.78 A parallelogram in the $x y -plane$ that we want to transform by a change in variables.

Thus, we can choose the transformation

T (u, v) = (\frac{3 u - v}{2}, \frac{u - v}{2})

and compute the Jacobian $J (u, v) .$ We have

J (u, v) = \frac{\partial (x, y)}{\partial (u, v)} = | \begin{array}{l} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{array} | = | \begin{array}{l} 3 / 2 & - 1 / 2 \\ 1 / 2 & - 1 / 2 \end{array} | = - \frac{3}{4} + \frac{1}{4} = - \frac{1}{2} .

Therefore, $| J (u, v) | = \frac{1}{2} .$ Also, the original integrand becomes

x - y = \frac{1}{2} [3 u - v - u + v] = \frac{1}{2} [3 u - u] = \frac{1}{2} [2 u] = u .

Therefore, by the use of the transformation $T,$ the integral changes to

\iint_{R} (x - y) d y d x = \int_{−9}^{−5} \int_{−1}^{1} J (u, v) u d u d v = \int_{−9}^{−5} \int_{−1}^{1} (\frac{1}{2}) u d u d v,

which is much simpler to compute. In fact, it is easily found to be zero. And this is just one example of why we transform integrals like this.

Checkpoint 5.46

Make appropriate changes of variables in the integral $\iint_{R} \frac{4}{{(x - y)}^{2}} d y d x,$ where $R$ is the trapezoid bounded by the lines $x - y = 2, x - y = 4, x = 0, and y = 0 .$ Write the resulting integral.

We are ready to give a problem-solving strategy for change of variables.

Problem-Solving Strategy

Change of Variables

Sketch the region given by the problem in the $x y -plane$ and then write the equations of the curves that form the boundary.
Depending on the region or the integrand, choose the transformations $x = g (u, v)$ and $y = h (u, v) .$
Determine the new limits of integration in the $u v -plane .$
Find the Jacobian $J (u, v) .$
In the integrand, replace the variables to obtain the new integrand.
Replace $d y d x$ or $d x d y,$ whichever occurs, by $J (u, v) d u d v .$

In the next example, we find a substitution that makes the integrand much simpler to compute.

Example 5.71

Evaluating an Integral

Using the change of variables $u = x - y$ and $v = x + y,$ evaluate the integral

\iint_{R} (x - y) e^{x^{2} - y^{2}} d A,

where $R$ is the region bounded by the lines $x + y = 1$ and $x + y = 3$ and the curves $x^{2} - y^{2} = −1$ and $x^{2} - y^{2} = 1$ (see the first region in Figure 5.79).

Solution

As before, first find the region $R$ and picture the transformation so it becomes easier to obtain the limits of integration after the transformations are made (Figure 5.79).

On the left-hand side of this figure, there is a complex region R in the Cartesian x y-plane bounded by x squared minus y squared = negative 1, x squared minus y squared = 1, x + y = 3, and x + y = 1. Then there is an arrow from this graph to the right-hand side of the figure marked with x = (u + v)/2 and y = (v minus u)/2. On the right-hand side of this figure there is a simpler region S in the Cartesian u v-plane bounded by u v = negative 1, u v = 1, v = 1, and v = 3. — Figure 5.79 Transforming the region $R$ into the region $S$ to simplify the computation of an integral.

Given $u = x - y$ and $v = x + y,$ we have $x = \frac{u + v}{2}$ and $y = \frac{v - u}{2}$ and hence the transformation to use is $T (u, v) = (\frac{u + v}{2}, \frac{v - u}{2}) .$ The lines $x + y = 1$ and $x + y = 3$ become $v = 1$ and $v = 3,$ respectively. The curves $x^{2} - y^{2} = 1$ and $x^{2} - y^{2} = −1$ become $u v = 1$ and $u v = −1,$ respectively.

Thus we can describe the region $S$ (see the second region Figure 5.79) as

S = {(u, v) | 1 \leq v \leq 3, \frac{−1}{v} \leq u \leq \frac{1}{v}} .

The Jacobian for this transformation is

J (u, v) = \frac{\partial (x, y)}{\partial (u, v)} = | \begin{array}{l} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{array} | = | \begin{array}{l} 1 / 2 & - 1 / 2 \\ 1 / 2 & 1 / 2 \end{array} | = \frac{1}{2} .

Therefore, by using the transformation $T,$ the integral changes to

\iint_{R} (x - y) e^{x^{2} - y^{2}} d A = \frac{1}{2} \int_{1}^{3} \int_{−1 / v}^{1 / v} u e^{u v} d u d v .

Doing the evaluation, we have

\frac{1}{2} \int_{1}^{3} \int_{−1 / v}^{1 / v} u e^{u v} d u d v = \frac{2}{3 e} \approx 0.245 .

Checkpoint 5.47

Using the substitutions $x = v$ and $y = \sqrt{u + v},$ evaluate the integral $\iint_{R} y sin (y^{2} - x) d A$ where $R$ is the region bounded by the lines $y = \sqrt{x}, x = 2, and y = 0 .$

Change of Variables for Triple Integrals

Changing variables in triple integrals works in exactly the same way. Cylindrical and spherical coordinate substitutions are special cases of this method, which we demonstrate here.

Suppose that $G$ is a region in $u v w -space$ and is mapped to $D$ in $x y z -space$ (Figure 5.80) by a one-to-one $C^{1}$ transformation $T (u, v, w) = (x, y, z)$ where $x = g (u, v, w),$ $y = h (u, v, w),$ and $z = k (u, v, w) .$

On the left-hand side of this figure, there is a region G in u v w space. Then there is an arrow from this graph to the right-hand side of the figure marked with x = g(u, v, w), y = h(u, v, w), and z = k(u, v, w). On the right-hand side of this figure there is a region D in xyz space. — Figure 5.80 A region $G$ in $u v w -space$ mapped to a region $D$ in $x y z -space .$

Then any function $F (x, y, z)$ defined on $D$ can be thought of as another function $H (u, v, w)$ that is defined on $G :$

F (x, y, z) = F (g (u, v, w), h (u, v, w), k (u, v, w)) = H (u, v, w) .

Now we need to define the Jacobian for three variables.

Definition

The Jacobian determinant $J (u, v, w)$ in three variables is defined as follows:

J (u, v, w) = | \begin{matrix} \frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} & \frac{\partial z}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v} & \frac{\partial z}{\partial v} \\ \frac{\partial x}{\partial w} & \frac{\partial y}{\partial w} & \frac{\partial z}{\partial w} \end{matrix} | .

This is also the same as

J (u, v, w) = | \begin{matrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \\ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w} \end{matrix} | .

The Jacobian can also be simply denoted as $\frac{\partial (x, y, z)}{\partial (u, v, w)} .$

With the transformations and the Jacobian for three variables, we are ready to establish the theorem that describes change of variables for triple integrals.

Theorem 5.15

Change of Variables for Triple Integrals

Let $T (u, v, w) = (x, y, z)$ where $x = g (u, v, w), y = h (u, v, w),$ and $z = k (u, v, w),$ be a one-to-one $C^{1}$ transformation, with a nonzero Jacobian, that maps the region $G$ in the $u v w -space$ into the region $R$ in the $x y z -space .$ As in the two-dimensional case, if $F$ is continuous on $R,$ then

\begin{matrix} \underset{R}{∭} F (x, y, z) d V & = \underset{G}{∭} F (g (u, v, w), h (u, v, w), k (u, v, w)) | \frac{\partial (x, y, z)}{\partial (u, v, w)} | d u d v d w \\ = \underset{G}{∭} H (u, v, w) | J (u, v, w) | d u d v d w . \end{matrix}

Let us now see how changes in triple integrals for cylindrical and spherical coordinates are affected by this theorem. We expect to obtain the same formulas as in Triple Integrals in Cylindrical and Spherical Coordinates.

Example 5.72

Obtaining Formulas in Triple Integrals for Cylindrical and Spherical Coordinates

Derive the formula in triple integrals for

cylindrical and
spherical coordinates.

Solution

For cylindrical coordinates, the transformation is $T (r, θ, z) = (x, y, z)$ from the Cartesian $r θ z -plane$ to the Cartesian $x y z -plane$ (Figure 5.81). Here $x = r cos θ,$ $y = r sin θ,$ and $z = z .$ The Jacobian for the transformation is

$\begin{matrix} J (r, θ, z) & = \frac{\partial (x, y, z)}{\partial (r, θ, z)} = | \begin{array}{l} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial θ} & \frac{\partial x}{\partial z} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial θ} & \frac{\partial y}{\partial z} \\ \frac{\partial z}{\partial r} & \frac{\partial z}{\partial θ} & \frac{\partial z}{\partial z} \end{array} | \\ = | \begin{array}{l} cos θ & - r sin θ & 0 \\ sin θ & r cos θ & 0 \\ 0 & 0 & 1 \end{array} | = r {cos}^{2} θ + r {sin}^{2} θ = r ({cos}^{2} θ + {sin}^{2} θ) = r . \end{matrix}$

We know that $r \geq 0,$ so $| J (r, θ, z) | = r .$ Then the triple integral is

$\underset{D}{∭} f (x, y, z) d V = \underset{G}{∭} f (r cos θ, r sin θ, z) r d r d θ d z .$

Figure 5.81 The transformation from rectangular coordinates to cylindrical coordinates can be treated as a change of variables from region $G$ in $r θ z -space$ to region $D$ in $x y z -space .$
For spherical coordinates, the transformation is $T (ρ, θ, φ) = (x, y, z)$ from the Cartesian $p θ φ -plane$ to the Cartesian $x y z -plane$ (Figure 5.82). Here $x = ρ sin φ cos θ,$ $y = ρ sin φ sin θ,$ and $z = ρ cos φ .$ The Jacobian for the transformation is

$J (ρ, θ, φ) = \frac{\partial (x, y, z)}{\partial (ρ, θ, φ)} = | \begin{array}{l} \frac{\partial x}{\partial ρ} & \frac{\partial x}{\partial θ} & \frac{\partial x}{\partial φ} \\ \frac{\partial y}{\partial ρ} & \frac{\partial y}{\partial θ} & \frac{\partial y}{\partial φ} \\ \frac{\partial z}{\partial ρ} & \frac{\partial z}{\partial θ} & \frac{\partial z}{\partial φ} \end{array} | = | \begin{array}{l} sin φ cos θ & ρ sin φ sin θ & ρ cos φ cos θ \\ sin φ sin θ & ρ sin φ cos θ & ρ cos φ sin θ \\ cos ϕ & 0 & - ρ sin φ \end{array} | .$

Expanding the determinant with respect to the third row:

$\begin{matrix} = cos φ | \begin{array}{l} - ρ sin φ sin θ & ρ cos φ cos θ \\ ρ sin φ sin θ & ρ cos φ sin θ \end{array} | - ρ sin φ | \begin{array}{l} sin φ cos θ & - ρ sin φ sin θ \\ sin φ sin θ & ρ sin φ cos θ \end{array} | \\ = cos φ (− ρ^{2} sin φ cos φ {sin}^{2} θ - ρ^{2} sin φ cos φ {cos}^{2} θ) \\ - ρ sin φ (ρ {sin}^{2} φ {cos}^{2} θ + ρ {sin}^{2} φ {sin}^{2} θ) \\ = − ρ^{2} sin φ {cos}^{2} φ ({sin}^{2} θ + {cos}^{2} θ) - ρ^{2} sin φ {sin}^{2} φ ({sin}^{2} θ + {cos}^{2} θ) \\ = − ρ^{2} sin φ {cos}^{2} φ - ρ^{2} sin φ {sin}^{2} φ \\ = − ρ^{2} sin φ ({cos}^{2} φ + {sin}^{2} φ) = − ρ^{2} sin φ . \end{matrix}$

Since $0 \leq φ \leq π,$ we must have $sin φ \geq 0 .$ Thus $| J (ρ, θ, φ) | = | − ρ^{2} sin φ | = ρ^{2} sin φ .$

Figure 5.82 The transformation from rectangular coordinates to spherical coordinates can be treated as a change of variables from region $G$ in $ρ θ φ -space$ to region $D$ in $x y z -space .$

Then the triple integral becomes

$\underset{D}{∭} f (x, y, z) d V = \underset{G}{∭} f (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ) ρ^{2} sin φ d ρ d φ d θ .$

Let’s try another example with a different substitution.

Example 5.73

Evaluating a Triple Integral with a Change of Variables

Evaluate the triple integral

\int_{0}^{3} \int_{0}^{4} \int_{y / 2}^{(y / 2) + 1} (x + \frac{z}{3}) d x d y d z

in $x y z -space$ by using the transformation

u = (2 x - y) / 2, v = y / 2, and w = z / 3 .

Then integrate over an appropriate region in $u v w -space .$

Solution

As before, some kind of sketch of the region $G$ in $x y z -space$ over which we have to perform the integration can help identify the region $D$ in $u v w -space$ (Figure 5.83). Clearly $G$ in $x y z -space$ is bounded by the planes $x = y / 2, x = (y / 2) + 1, y = 0,$ $y = 4,$ $z = 0, and z = 4 .$ We also know that we have to use $u = (2 x - y) / 2, v = y / 2, and w = z / 3$ for the transformations. We need to solve for $x, y, and z .$ Here we find that $x = u + v,$ $y = 2 v,$ and $z = 3 w .$

Using elementary algebra, we can find the corresponding surfaces for the region $G$ and the limits of integration in $u v w -space .$ It is convenient to list these equations in a table.

Equations in $x y z$ for the region $D$	Corresponding equations in $u v w$ for the region $G$	Limits for the integration in $u v w$
$x = y / 2$	$u + v = 2 v / 2 = v$	$u = 0$
$x = y / 2$	$u + v = (2 v / 2) + 1 = v + 1$	$u = 1$
$y = 0$	$2 v = 0$	$v = 0$
$y = 4$	$2 v = 4$	$v = 2$
$z = 0$	$3 w = 0$	$w = 0$
$z = 3$	$3 w = 3$	$w = 1$

On the left-hand side of this figure, there is a box G with sides 1, 2, and 1 along the u, v, and w axes, respectively. Then there is an arrow from this graph to the right-hand side of the figure marked with x = u + v, y = 2v, and z = 3w. On the right-hand side of this figure there is a region D in xyz space that is a rotated box with sides 1, 4, and 3 along the x, y, and z axes. The rear plane is marked x = y/2 or y = 2x. The front plane is marked x = y/2 + 1 or y = 2x minus 2. — Figure 5.83 The region $G$ in $u v w -space$ is transformed to region $D$ in $x y z -space .$

Now we can calculate the Jacobian for the transformation:

J (u, v, w) = | \begin{matrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \\ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w} \end{matrix} | = | \begin{matrix} 1 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{matrix} | = 6 .

The function to be integrated becomes

f (x, y, z) = x + \frac{z}{3} = u + v + \frac{3 w}{3} = u + v + w .

We are now ready to put everything together and complete the problem.

\begin{matrix} \int_{0}^{3} \int_{0}^{4} \int_{y / 2}^{(y / 2) + 1} (x + \frac{z}{3}) d x d y d z \\ = \int_{0}^{1} \int_{0}^{2} \int_{0}^{1} (u + v + w) | J (u, v, w) | d u d v d w = \int_{0}^{1} \int_{0}^{2} \int_{0}^{1} (u + v + w) | 6 | d u d v d w \\ = 6 \int_{0}^{1} \int_{0}^{2} \int_{0}^{1} (u + v + w) d u d v d w = 6 {\int_{0}^{1} \int_{0}^{2} [\frac{u^{2}}{2} + v u + w u]}_{0}^{1} d v d w \\ = 6 \int_{0}^{1} \int_{0}^{2} (\frac{1}{2} + v + w) d v d w = 6 {\int_{0}^{1} [\frac{1}{2} v + \frac{v^{2}}{2} + w v]}_{0}^{2} d w \\ = 6 \int_{0}^{1} (3 + 2 w) d w = 6 {[3 w + w^{2}]}_{0}^{1} = 24. \end{matrix}

Checkpoint 5.48

Let $D$ be the region in $x y z -space$ defined by $1 \leq x \leq 2, 0 \leq x y \leq 2, and 0 \leq z \leq 1 .$

Evaluate $\underset{D}{∭} (x^{2} y + 3 x y z) d x d y d z$ by using the transformation $u = x, v = x y,$ and $w = 3 z .$

Section 5.7 Exercises

In the following exercises, the function $T : S \to R, T (u, v) = (x, y)$ on the region $S = {(u, v) | 0 \leq u \leq 1, 0 \leq v \leq 1}$ bounded by the unit square is given, where $R \subset ℝ^{2}$ is the image of $S$ under $T .$

Justify that the function $T$ is a $C^{1}$ transformation.
Find the images of the vertices of the unit square $S$ through the function $T .$
Determine the image $R$ of the unit square $S$ and graph it.

356.

$x = 2 u, y = 3 v$

357.

$x = \frac{u}{2}, y = \frac{v}{3}$

358.

$x = u - v, y = u + v$

359.

$x = 2 u - v, y = u + 2 v$

360.

$x = u^{2}, y = v^{2}$

361.

$x = u^{3}, y = v^{3}$

In the following exercises, determine whether the transformations $T : S \to R$ are one-to-one or not.

362.

$x = u^{2}, y = v^{2}, where S$ is the rectangle of vertices $(−1, 0), (1, 0), (1, 1), and (−1, 1) .$

363.

$x = u^{4}, y = u^{2} + v, where S$ is the triangle of vertices $(−2, 0), (2, 0), and (0, 2) .$

364.

$x = 2 u, y = 3 v, where S$ is the square of vertices $(−1, 1), (−1, −1), (1, −1), and (1, 1) .$

365.

$T (u, v) = (2 u - v, u),$ where $S$ is the triangle of vertices $(−1, 1), (−1, −1), and (1, −1) .$

366.

$x = u + v + w, y = u + v, z = w,$ where $S = R = ℝ^{3} .$

367.

$x = u^{2} + v + w, y = u^{2} + v, z = w,$ where $S = R = ℝ^{3} .$

In the following exercises, the transformations $T : S \to R$ are one-to-one. Find their related inverse transformations $T^{−1} : R \to S .$

368.

$x = 4 u, y = 5 v,$ where $S = R = ℝ^{2} .$

369.

$x = u + 2 v, y = − u + v,$ where $S = R = ℝ^{2} .$

370.

$x = e^{2 u + v}, y = e^{u - v},$ where $S = ℝ^{2}$ and $R = {(x, y) | x > 0, y > 0}$

371.

$x = ln u, y = ln (u v),$ where $S = {(u, v) | u > 0, v > 0}$ and $R = ℝ^{2} .$

372.

$x = u + v + w, y = 3 v, z = 2 w,$ where $S = R = ℝ^{3} .$

373.

$x = u + v, y = v + w, z = u + w,$ where $S = R = ℝ^{3} .$

In the following exercises, the transformation $T : S \to R, T (u, v) = (x, y)$ and the region $R \subset ℝ^{2}$ are given. Find the region $S \subset ℝ^{2} .$

374.

$x = a u, y = b v, R = {(x, y) | x^{2} + y^{2} \leq a^{2} b^{2}},$ where $a, b > 0$

375.

$x = a u, y = b v, R = {(x, y) | \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} \leq 1},$ where $a, b > 0$

376.

$x = \frac{u}{a}, y = \frac{v}{b}, z = \frac{w}{c},$ $R = {(x, y) | x^{2} + y^{2} + z^{2} \leq 1},$ where $a, b, c > 0$

377.

$x = a u, y = b v, z = c w, R = {(x, y) | \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} - \frac{z^{2}}{c^{2}} \leq 1, z > 0},$ where $a, b, c > 0$

In the following exercises, find the Jacobian $J$ of the transformation.

378.

$x = u + 2 v, y = − u + v$

379.

$x = \frac{u^{3}}{2}, y = \frac{v}{u^{2}}$

380.

$x = e^{2 u - v}, y = e^{u + v}$

381.

$x = u e^{v}, y = e^{− v}$

382.

$x = u cos (e^{v}), y = u sin (e^{v})$

383.

$x = v sin (u^{2}), y = v cos (u^{2})$

384.

$x = u cosh v, y = u sinh v, z = w$

385.

$x = v cosh (\frac{1}{u}), y = v sinh (\frac{1}{u}), z = u + w^{2}$

386.

$x = u + v, y = v + w, z = u$

387.

$x = u - v, y = u + v, z = u + v + w$

388.

The triangular region $R$ with the vertices $(0, 0), (1, 1), and (1, 2)$ is shown in the following figure.

Find a transformation $T : S \to R,$ $T (u, v) = (x, y) = (a u + b v, c u + d v),$ where $a, b, c,$ and $d$ are real numbers with $a d - b c \neq 0$ such that $T^{−1} (0, 0) = (0, 0), T^{−1} (1, 1) = (1, 0),$ and $T^{−1} (1, 2) = (0, 1) .$
Use the transformation $T$ to find the area $A (R)$ of the region $R .$

389.

The triangular region $R$ with the vertices $(0, 0), (2, 0), and (1, 3)$ is shown in the following figure.

Find a transformation $T : S \to R,$ $T (u, v) = (x, y) = (a u + b v, c u + d v),$ where $a, b, c$ and $d$ are real numbers with $a d - b c \neq 0$ such that $T^{−1} (0, 0) = (0, 0),$ $T^{−1} (2, 0) = (1, 0),$ and $T^{−1} (1, 3) = (0, 1) .$
Use the transformation $T$ to find the area $A (R)$ of the region $R .$

In the following exercises, use the transformation $u = y - x, v = y,$ to evaluate the integrals on the parallelogram $R$ of vertices $(0, 0), (1, 0), (2, 1), and (1, 1)$ shown in the following figure.

390.

$\iint_{R} (y - x) d A$

391.

$\iint_{R} (y^{2} - x y) d A$

In the following exercises, use the transformation $y - x = u, x + y = v$ to evaluate the integrals on the square $R$ determined by the lines $y = x, y = − x + 2, y = x + 2,$ and $y = − x$ shown in the following figure.

392.

$\iint_{R} e^{x + y} d A$

393.

$\iint_{R} sin (x - y) d A$

In the following exercises, use the transformation $x = u, 5 y = v$ to evaluate the integrals on the region $R$ bounded by the ellipse $x^{2} + 25 y^{2} = 1$ shown in the following figure.

394.

$\iint_{R} \sqrt{x^{2} + 25 y^{2}} d A$

395.

${\iint_{R} (x^{2} + 25 y^{2})}^{2} d A$

In the following exercises, use the transformation $u = x + y, v = x - y$ to evaluate the integrals on the trapezoidal region $R$ determined by the points $(1, 0), (2, 0), (0, 2), and (0, 1)$ shown in the following figure.

396.

$\iint_{R} (x^{2} - 2 x y + y^{2}) e^{x + y} d A$

397.

$\iint_{R} (x^{3} + 3 x^{2} y + 3 x y^{2} + y^{3}) d A$

398.

The circular annulus sector $R$ bounded by the circles $4 x^{2} + 4 y^{2} = 1$ and $9 x^{2} + 9 y^{2} = 64,$ the line $x = y \sqrt{3},$ and the $y -axis$ is shown in the following figure. Find a transformation $T$ from a rectangular region $S$ in the $r θ -plane$ to the region $R$ in the $x y -plane.$ Graph $S .$

399.

The solid $R$ bounded by the circular cylinder $x^{2} + y^{2} = 9$ and the planes $z = 0, z = 1,$ $x = 0, and y = 0$ is shown in the following figure. Find a transformation $T$ from a cylindrical box $S$ in $r θ z -space$ to the solid $R$ in $x y z -space .$

400.

Show that $\iint_{R} f (\sqrt{\frac{x^{2}}{3} + \frac{y^{2}}{3}}) d A = 2 π \sqrt{15} \int_{0}^{1} f (ρ) ρ d ρ,$ where $f$ is a continuous function on $[0, 1]$ and $R$ is the region bounded by the ellipse $5 x^{2} + 3 y^{2} = 15 .$

401.

Show that $\underset{R}{∭} f (\sqrt{16 x^{2} + 4 y^{2} + z^{2}}) d V = \frac{π}{2} \int_{0}^{1} f (ρ) ρ^{2} d ρ,$ where $f$ is a continuous function on $[0, 1]$ and $R$ is the region bounded by the ellipsoid $16 x^{2} + 4 y^{2} + z^{2} = 1 .$

402.

[T] Find the area of the region bounded by the curves $x y = 1, x y = 3, y = 2 x,$ and $y = 3 x$ by using the transformation $u = x y$ and $v = \frac{y}{x} .$ Use a computer algebra system (CAS) to graph the boundary curves of the region $R .$

403.

[T] Find the area of the region bounded by the curves $x y = 2, x y = 3, y = x,$ and $y = 2 x$ by using the transformation $u = x y$ and $v = \frac{y}{x} .$

404.

Evaluate the triple integral $\int_{0}^{1} \int_{1}^{2} \int_{z}^{z + 1} (y + 1) d x d y d z$ by using the transformation $u = x - z,$ $v = 3 y, and w = \frac{z}{2} .$

405.

Evaluate the triple integral $\int_{0}^{2} \int_{4}^{6} \int_{3 z}^{3 z + 2} (5 - 4 y) d x d z d y$ by using the transformation $u = x - 3 z, v = 4 y, and w = z .$

406.

A transformation $T : ℝ^{2} \to ℝ^{2}, T (u, v) = (x, y)$ of the form $x = a u + b v, y = c u + d v,$ where $a, b, c, and d$ are real numbers, is called linear. Show that a linear transformation for which $a d - b c \neq 0$ maps parallelograms to parallelograms.

407.

The transformation $T_{θ} : ℝ^{2} \to ℝ^{2}, T_{θ} (u, v) = (x, y),$ where $x = u cos θ - v sin θ,$ $y = u sin θ + v cos θ,$ is called a rotation of angle $θ .$ Show that the inverse transformation of $T_{θ}$ satisfies $T_{θ}^{−1} = T_{− θ},$ where $T_{− θ}$ is the rotation of angle $− θ .$

408.

[T] Use the results of the previous exercise to find the region $S$ in the $u v -plane$ whose image through a rotation of angle $\frac{π}{4}$ is the region $R$ enclosed by the ellipse $x^{2} + 4 y^{2} = 1 .$ Use a CAS to answer the following questions.

Graph the region $S .$
Evaluate the integral $\iint_{S} e^{−2 u v} d u d v .$ Round your answer to two decimal places.

409.

[T] The transformations $T_{i} : ℝ^{2} \to ℝ^{2},$ $i = 1,…, 4,$ defined by $T_{1} (u, v) = (u, − v),$ $T_{2} (u, v) = (− u, v), T_{3} (u, v) = (− u, − v),$ and $T_{4} (u, v) = (v, u)$ are called reflections about the $x -axis, y -axis,$ origin, and the line $y = x,$ respectively.

Find the image of the region $S = {(u, v) | u^{2} + v^{2} - 2 u - 4 v + 1 \leq 0}$ in the $x y -plane$ through the transformation $T_{1} \circ T_{2} \circ T_{3} \circ T_{4} .$
Use a CAS to graph the image of $S .$
Evaluate the integral $\iint_{S} sin (u^{2}) d u d v$ by using a CAS. Round your answer to two decimal places.

410.

[T] The transformation $T_{k, 1, 1} : ℝ^{3} \to ℝ^{3}, T_{k, 1, 1} (u, v, w) = (x, y, z)$ of the form $x = k u,$ $y = v, z = w,$ where $k \neq 1$ is a positive real number, is called a stretch if $k > 1$ and a compression if $0 < k < 1$ in the $x -direction .$ Use a CAS to evaluate the integral $\underset{S}{∭} e^{− (4 x^{2} + 9 y^{2} + 25 z^{2})} d x d y d z$ on the solid $S = {(x, y, z) | 4 x^{2} + 9 y^{2} + 25 z^{2} \leq 1}$ by considering the compression $T_{2, 3, 5} (u, v, w) = (x, y, z)$ defined by $x = \frac{u}{2}, y = \frac{v}{3},$ and $z = \frac{w}{5} .$ Round your answer to four decimal places.

411.

[T] The transformation $T_{a, 0} : ℝ^{2} \to ℝ^{2}, T_{a, 0} (u, v) = (u + a v, v),$ where $a \neq 0$ is a real number, is called a shear in the $x -direction .$ The transformation, $T_{0, b} : ℝ^{2} \to ℝ^{2}, T_{0, b} (u, v) = (u, b u + v),$ where $b \neq 0$ is a real number, is called a shear in the $y -direction .$

Find transformations $T_{0, 2} \circ T_{3, 0} .$
Find the image $R$ of the trapezoidal region $S$ bounded by $u = 0, v = 0, v = 1,$ and $v = 2 - u$ through the transformation $T_{0, 2} \circ T_{3, 0} .$
Use a CAS to graph the image $R$ in the $x y -plane .$
Find the area of the region $R$ by using the area of region $S .$

412.

Use the transformation, $x = a u, y = a v, z = c w$ and spherical coordinates to show that the volume of a region bounded by the spheroid $\frac{x^{2} + y^{2}}{a^{2}} + \frac{z^{2}}{c^{2}} = 1$ is $\frac{4 π a^{2} c}{3} .$

413.

Find the volume of a football whose shape is a spheroid $\frac{x^{2} + y^{2}}{a^{2}} + \frac{z^{2}}{c^{2}} = 1$ whose length from tip to tip is $11$ inches and circumference at the center is $22$ inches. Round your answer to two decimal places.

414.

[T] Lamé ovals (or superellipses) are plane curves of equations ${(\frac{x}{a})}^{n} + {(\frac{y}{b})}^{n} = 1,$ where a, b, and n are positive real numbers.

Use a CAS to graph the regions $R$ bounded by Lamé ovals for $a = 1, b = 2, n = 4$ and $n = 6,$ respectively.
Find the transformations that map the region $R$ bounded by the Lamé oval $x^{4} + y^{4} = 1,$ also called a squircle and graphed in the following figure, into the unit disk.
Use a CAS to find an approximation of the area $A (R)$ of the region $R$ bounded by $x^{4} + y^{4} = 1 .$ Round your answer to two decimal places.

415.

[T] Lamé ovals have been consistently used by designers and architects. For instance, Gerald Robinson, a Canadian architect, has designed a parking garage in a shopping center in Peterborough, Ontario, in the shape of a superellipse of the equation ${(\frac{x}{a})}^{n} + {(\frac{y}{b})}^{n} = 1$ with $\frac{a}{b} = \frac{9}{7}$ and $n = e .$ Use a CAS to find an approximation of the area of the parking garage in the case $a = 900$ yards, $b = 700$ yards, and $n = 2.72$ .

5.7 Change of Variables in Multiple Integrals