Calculus Volume 3

# 5.3Double Integrals in Polar Coordinates

Calculus Volume 35.3 Double Integrals in Polar Coordinates

### Learning Objectives

• 5.3.1 Recognize the format of a double integral over a polar rectangular region.
• 5.3.2 Evaluate a double integral in polar coordinates by using an iterated integral.
• 5.3.3 Recognize the format of a double integral over a general polar region.
• 5.3.4 Use double integrals in polar coordinates to calculate areas and volumes.

Double integrals are sometimes much easier to evaluate if we change rectangular coordinates to polar coordinates. However, before we describe how to make this change, we need to establish the concept of a double integral in a polar rectangular region.

### Polar Rectangular Regions of Integration

When we defined the double integral for a continuous function in rectangular coordinates—say, $gg$ over a region $RR$ in the $xyxy$-plane—we divided $RR$ into subrectangles with sides parallel to the coordinate axes. These sides have either constant $xx$-values and/or constant $yy$-values. In polar coordinates, the shape we work with is a polar rectangle, whose sides have constant $rr$-values and/or constant $θθ$-values. This means we can describe a polar rectangle as in Figure 5.28(a), with $R={(r,θ)|a≤r≤b,α≤θ≤β}.R={(r,θ)|a≤r≤b,α≤θ≤β}.$

In this section, we are looking to integrate over polar rectangles. Consider a function $f(r,θ)f(r,θ)$ over a polar rectangle $R.R.$ We divide the interval $[a,b][a,b]$ into $mm$ subintervals $[ri−1,ri][ri−1,ri]$ of length $Δr=(b−a)/mΔr=(b−a)/m$ and divide the interval $[α,β][α,β]$ into $nn$ subintervals $[θj−1,θj][θj−1,θj]$ of width $Δθ=(β−α)/n.Δθ=(β−α)/n.$ This means that the circles $r=rir=ri$ and rays $θ=θjθ=θj$ for $1≤i≤m1≤i≤m$ and $1≤j≤n1≤j≤n$ divide the polar rectangle $RR$ into smaller polar subrectangles $RijRij$ (Figure 5.28(b)).

Figure 5.28 (a) A polar rectangle $RR$ (b) divided into subrectangles $Rij.Rij.$ (c) Close-up of a subrectangle.

As before, we need to find the area $ΔAΔA$ of the polar subrectangle $RijRij$ and the “polar” volume of the thin box above $Rij.Rij.$ Recall that, in a circle of radius $r,r,$ the length $ss$ of an arc subtended by a central angle of $θθ$ radians is $s=rθ.s=rθ.$ Notice that the polar rectangle $RijRij$ looks a lot like a trapezoid with parallel sides $ri−1Δθri−1Δθ$ and $riΔθriΔθ$ and with a width $Δr.Δr.$ Hence the area of the polar subrectangle $RijRij$ is

$ΔA=12Δr(ri−1Δθ+r1Δθ).ΔA=12Δr(ri−1Δθ+r1Δθ).$

Simplifying and letting $rij*=12(ri−1+ri),rij*=12(ri−1+ri),$ we have $ΔA=rij*ΔrΔθ.ΔA=rij*ΔrΔθ.$ Therefore, the polar volume of the thin box above $RijRij$ (Figure 5.29) is

$f(rij*,θij*)ΔA=f(rij*,θij*)rij*ΔrΔθ.f(rij*,θij*)ΔA=f(rij*,θij*)rij*ΔrΔθ.$
Figure 5.29 Finding the volume of the thin box above polar rectangle $Rij.Rij.$

Using the same idea for all the subrectangles and summing the volumes of the rectangular boxes, we obtain a double Riemann sum as

$∑i=1m∑j=1nf(rij*,θij*)rij*ΔrΔθ.∑i=1m∑j=1nf(rij*,θij*)rij*ΔrΔθ.$

As we have seen before, we obtain a better approximation to the polar volume of the solid above the region $RR$ when we let $mm$ and $nn$ become larger. Hence, we define the polar volume as the limit of the double Riemann sum,

$V=limm,n→∞∑i=1m∑j=1nf(rij*,θij*)rij*ΔrΔθ.V=limm,n→∞∑i=1m∑j=1nf(rij*,θij*)rij*ΔrΔθ.$

This becomes the expression for the double integral.

### Definition

The double integral of the function $f(r,θ)f(r,θ)$ over the polar rectangular region $RR$ in the $rθrθ$-plane is defined as

$∬Rf(r,θ)dA=limm,n→∞∑i=1m∑j=1nf(rij*,θij*)ΔA=limm,n→∞∑i=1m∑j=1nf(rij*,θij*)rij*ΔrΔθ.∬Rf(r,θ)dA=limm,n→∞∑i=1m∑j=1nf(rij*,θij*)ΔA=limm,n→∞∑i=1m∑j=1nf(rij*,θij*)rij*ΔrΔθ.$
(5.8)

Again, just as in Double Integrals over Rectangular Regions, the double integral over a polar rectangular region can be expressed as an iterated integral in polar coordinates. Hence,

$∬Rf(r,θ)dA=∬Rf(r,θ)rdrdθ=∫θ=αθ=β∫r=ar=bf(r,θ)rdrdθ.∬Rf(r,θ)dA=∬Rf(r,θ)rdrdθ=∫θ=αθ=β∫r=ar=bf(r,θ)rdrdθ.$

Notice that the expression for $dAdA$ is replaced by $rdrdθrdrdθ$ when working in polar coordinates. Another way to look at the polar double integral is to change the double integral in rectangular coordinates by substitution. When the function $ff$ is given in terms of $xx$ and $y,y,$ using $x=rcosθ,y=rsinθ,anddA=rdrdθx=rcosθ,y=rsinθ,anddA=rdrdθ$ changes it to

$∬Rf(x,y)dA=∬Rf(rcosθ,rsinθ)rdrdθ.∬Rf(x,y)dA=∬Rf(rcosθ,rsinθ)rdrdθ.$

Note that all the properties listed in Double Integrals over Rectangular Regions for the double integral in rectangular coordinates hold true for the double integral in polar coordinates as well, so we can use them without hesitation.

### Example 5.24

#### Sketching a Polar Rectangular Region

Sketch the polar rectangular region $R={(r,θ)|1≤r≤3,0≤θ≤π}.R={(r,θ)|1≤r≤3,0≤θ≤π}.$

Now that we have sketched a polar rectangular region, let us demonstrate how to evaluate a double integral over this region by using polar coordinates.

### Example 5.25

#### Evaluating a Double Integral over a Polar Rectangular Region

Evaluate the integral $∬R3xdA∬R3xdA$ over the region $R={(r,θ)|1≤r≤2,0≤θ≤π}.R={(r,θ)|1≤r≤2,0≤θ≤π}.$

### Checkpoint5.17

Sketch the region $R={(r,θ)|1≤r≤2,−π2≤θ≤π2},R={(r,θ)|1≤r≤2,−π2≤θ≤π2},$ and evaluate $∬RxdA.∬RxdA.$

### Example 5.26

#### Evaluating a Double Integral by Converting from Rectangular Coordinates

Evaluate the integral $∬R(1−x2−y2)dA∬R(1−x2−y2)dA$ where $RR$ is the unit circle on the $xyxy$-plane.

### Example 5.27

#### Evaluating a Double Integral by Converting from Rectangular Coordinates

Evaluate the integral $∬R(x+y)dA∬R(x+y)dA$ where $R={(x,y)|1≤x2+y2≤4,x≤0}.R={(x,y)|1≤x2+y2≤4,x≤0}.$

### Checkpoint5.18

Evaluate the integral $∬R(4−x2−y2)dA∬R(4−x2−y2)dA$ where $RR$ is the circle of radius $22$ on the $xyxy$-plane.

### General Polar Regions of Integration

To evaluate the double integral of a continuous function by iterated integrals over general polar regions, we consider two types of regions, analogous to Type I and Type II as discussed for rectangular coordinates in Double Integrals over General Regions. It is more common to write polar equations as $r=f(θ)r=f(θ)$ than $θ=f(r),θ=f(r),$ so we describe a general polar region as $R={(r,θ)|α≤θ≤β,h1(θ)≤r≤h2(θ)}R={(r,θ)|α≤θ≤β,h1(θ)≤r≤h2(θ)}$ (see the following figure).

Figure 5.32 A general polar region between $α<θ<βα<θ<β$ and $h1(θ)

### Theorem5.8

#### Double Integrals over General Polar Regions

If $f(r,θ)f(r,θ)$ is continuous on a general polar region $DD$ as described above, then

$∬Df(r,θ)rdrdθ=∫θ=αθ=β∫r=h1(θ)r=h2(θ)f(r,θ)rdrdθ∬Df(r,θ)rdrdθ=∫θ=αθ=β∫r=h1(θ)r=h2(θ)f(r,θ)rdrdθ$
(5.9)

### Example 5.28

#### Evaluating a Double Integral over a General Polar Region

Evaluate the integral $∬Dr2sinθrdrdθ∬Dr2sinθrdrdθ$ where $DD$ is the region bounded by the polar axis and the upper half of the cardioid $r=1+cosθ.r=1+cosθ.$

### Checkpoint5.19

Evaluate the integral

### Polar Areas and Volumes

As in rectangular coordinates, if a solid $SS$ is bounded by the surface $z=f(r,θ),z=f(r,θ),$ as well as by the surfaces $r=a,r=b,θ=α,r=a,r=b,θ=α,$ and $θ=β,θ=β,$ we can find the volume $VV$ of $SS$ by double integration, as

$V=∬Rf(r,θ)rdrdθ=∫θ=αθ=β∫r=ar=bf(r,θ)rdrdθ.V=∬Rf(r,θ)rdrdθ=∫θ=αθ=β∫r=ar=bf(r,θ)rdrdθ.$

If the base of the solid can be described as $D={(r,θ)|α≤θ≤β,h1(θ)≤r≤h2(θ)},D={(r,θ)|α≤θ≤β,h1(θ)≤r≤h2(θ)},$ then the double integral for the volume becomes

$V=∬Df(r,θ)rdrdθ=∫θ=αθ=β∫r=h1(θ)r=h2(θ)f(r,θ)rdrdθ.V=∬Df(r,θ)rdrdθ=∫θ=αθ=β∫r=h1(θ)r=h2(θ)f(r,θ)rdrdθ.$

We illustrate this idea with some examples.

### Example 5.29

#### Finding a Volume Using a Double Integral

Find the volume of the solid that lies under the paraboloid $z=1−x2−y2z=1−x2−y2$ and above the unit circle on the $xyxy$-plane (see the following figure).

Figure 5.34 The paraboloid $z=1−x2−y2z=1−x2−y2$.

### Example 5.30

#### Finding a Volume Using Double Integration

Find the volume of the solid that lies under the paraboloid $z=4−x2−y2z=4−x2−y2$ and above the disk $(x−1)2+y2=1(x−1)2+y2=1$ on the $xyxy$-plane. See the paraboloid in Figure 5.35 intersecting the cylinder $(x−1)2+y2=1(x−1)2+y2=1$ above the $xyxy$-plane.

Figure 5.35 Finding the volume of a solid with a paraboloid cap and a circular base.

Notice in the next example that integration is not always easy with polar coordinates. Complexity of integration depends on the function and also on the region over which we need to perform the integration. If the region has a more natural expression in polar coordinates or if $ff$ has a simpler antiderivative in polar coordinates, then the change in polar coordinates is appropriate; otherwise, use rectangular coordinates.

### Example 5.31

#### Finding a Volume Using a Double Integral

Find the volume of the region that lies under the paraboloid $z=x2+y2z=x2+y2$ and above the triangle enclosed by the lines $y=x,x=0,y=x,x=0,$ and $x+y=2x+y=2$ in the $xyxy$-plane (Figure 5.36).

To answer the question of how the formulas for the volumes of different standard solids such as a sphere, a cone, or a cylinder are found, we want to demonstrate an example and find the volume of an arbitrary cone.

### Example 5.32

#### Finding a Volume Using a Double Integral

Use polar coordinates to find the volume inside the cone $z=2−x2+y2z=2−x2+y2$ and above the $xy-plane.xy-plane.$

#### Analysis

Note that if we were to find the volume of an arbitrary cone with radius $aa$ units and height $hh$ units, then the equation of the cone would be $z=h−hax2+y2.z=h−hax2+y2.$

We can still use Figure 5.37 and set up the integral as $∫θ=0θ=2π∫r=0r=a(h−har)rdrdθ.∫θ=0θ=2π∫r=0r=a(h−har)rdrdθ.$

Evaluating the integral, we get $13πa2h.13πa2h.$

### Checkpoint5.20

Use polar coordinates to find an iterated integral for finding the volume of the solid enclosed by the paraboloids $z=x2+y2z=x2+y2$ and $z=16−x2−y2.z=16−x2−y2.$

As with rectangular coordinates, we can also use polar coordinates to find areas of certain regions using a double integral. As before, we need to understand the region whose area we want to compute. Sketching a graph and identifying the region can be helpful to realize the limits of integration. Generally, the area formula in double integration will look like

$AreaA=∫αβ∫h1(θ)h2(θ)1rdrdθ.AreaA=∫αβ∫h1(θ)h2(θ)1rdrdθ.$

### Example 5.33

#### Finding an Area Using a Double Integral in Polar Coordinates

Evaluate the area bounded by the curve $r=cos4θ.r=cos4θ.$

### Example 5.34

#### Finding Area Between Two Polar Curves

Find the area enclosed by the circle $r=3cosθr=3cosθ$ and the cardioid $r=1+cosθ.r=1+cosθ.$

### Checkpoint5.21

Find the area enclosed inside the cardioid $r=3−3sinθr=3−3sinθ$ and outside the cardioid $r=1+sinθ.r=1+sinθ.$

### Example 5.35

#### Evaluating an Improper Double Integral in Polar Coordinates

Evaluate the integral $∬R2e−10(x2+y2)dxdy.∬R2e−10(x2+y2)dxdy.$

### Checkpoint5.22

Evaluate the integral $∬R2e−4(x2+y2)dxdy.∬R2e−4(x2+y2)dxdy.$

### Section 5.3 Exercises

In the following exercises, express the region $DD$ in polar coordinates.

122.

$DD$ is the region of the disk of radius $22$ centered at the origin that lies in the first quadrant.

123.

$DD$ is the region between the circles of radius $44$ and radius $55$ centered at the origin that lies in the second quadrant.

124.

$DD$ is the region bounded by the $yy$-axis and $x=1−y2.x=1−y2.$

125.

$DD$ is the region bounded by the $xx$-axis and $y=2−x2.y=2−x2.$

126.

$D = { ( x , y ) | x 2 + y 2 ≤ 4 x } D = { ( x , y ) | x 2 + y 2 ≤ 4 x }$

127.

$D = { ( x , y ) | x 2 + y 2 ≤ 4 y } D = { ( x , y ) | x 2 + y 2 ≤ 4 y }$

In the following exercises, the graph of the polar rectangular region $DD$ is given. Express $DD$ in polar coordinates.

128.
129.
130.
131.
132.

In the following graph, the region $DD$ is situated below $y=xy=x$ and is bounded by $x=1,x=5,x=1,x=5,$ and $y=0.y=0.$

133.

In the following graph, the region $DD$ is bounded by $y=xy=x$ and $y=x2.y=x2.$

In the following exercises, evaluate the double integral $∬Rf(x,y)dA∬Rf(x,y)dA$ over the polar rectangular region $D.D.$

134.

$f ( x , y ) = x 2 + y 2 , D = { ( r , θ ) | 3 ≤ r ≤ 5 , 0 ≤ θ ≤ 2 π } f ( x , y ) = x 2 + y 2 , D = { ( r , θ ) | 3 ≤ r ≤ 5 , 0 ≤ θ ≤ 2 π }$

135.

$f ( x , y ) = x + y , D = { ( r , θ ) | 3 ≤ r ≤ 5 , 0 ≤ θ ≤ 2 π } f ( x , y ) = x + y , D = { ( r , θ ) | 3 ≤ r ≤ 5 , 0 ≤ θ ≤ 2 π }$

136.

$f ( x , y ) = x 2 + x y , D = { ( r , θ ) | 1 ≤ r ≤ 2 , π ≤ θ ≤ 2 π } f ( x , y ) = x 2 + x y , D = { ( r , θ ) | 1 ≤ r ≤ 2 , π ≤ θ ≤ 2 π }$

137.

$f ( x , y ) = x 4 + y 4 , D = { ( r , θ ) | 1 ≤ r ≤ 2 , 3 π 2 ≤ θ ≤ 2 π } f ( x , y ) = x 4 + y 4 , D = { ( r , θ ) | 1 ≤ r ≤ 2 , 3 π 2 ≤ θ ≤ 2 π }$

138.

$f(x,y)=x2+y23,f(x,y)=x2+y23,$ where $D={(r,θ)|0≤r≤1,π2≤θ≤π}.D={(r,θ)|0≤r≤1,π2≤θ≤π}.$

139.

$f(x,y)=x4+2x2y2+y4,f(x,y)=x4+2x2y2+y4,$ where $D={(r,θ)|3≤r≤4,π3≤θ≤2π3}.D={(r,θ)|3≤r≤4,π3≤θ≤2π3}.$

140.

$f(x,y)=sin(arctanyx),f(x,y)=sin(arctanyx),$ where $D={(r,θ)|1≤r≤2,π6≤θ≤π3}D={(r,θ)|1≤r≤2,π6≤θ≤π3}$

141.

$f(x,y)=arctan(yx),f(x,y)=arctan(yx),$ where $D={(r,θ)|2≤r≤3,π4≤θ≤π3}D={(r,θ)|2≤r≤3,π4≤θ≤π3}$

142.

$∬ D e x 2 + y 2 [ 1 + 2 arctan ( y x ) ] d A , D = { ( r , θ ) | 1 ≤ r ≤ 2 , π 6 ≤ θ ≤ π 3 } ∬ D e x 2 + y 2 [ 1 + 2 arctan ( y x ) ] d A , D = { ( r , θ ) | 1 ≤ r ≤ 2 , π 6 ≤ θ ≤ π 3 }$

143.

$∬ D ( e x 2 + y 2 + x 4 + 2 x 2 y 2 + y 4 ) arctan ( y x ) d A , D = { ( r , θ ) | 1 ≤ r ≤ 2 , π 4 ≤ θ ≤ π 3 } ∬ D ( e x 2 + y 2 + x 4 + 2 x 2 y 2 + y 4 ) arctan ( y x ) d A , D = { ( r , θ ) | 1 ≤ r ≤ 2 , π 4 ≤ θ ≤ π 3 }$

In the following exercises, the integrals have been converted to polar coordinates. Verify that the identities are true and choose the easiest way to evaluate the integrals, in rectangular or polar coordinates.

144.

$∫ 1 2 ∫ 0 x ( x 2 + y 2 ) d y d x = ∫ 0 π 4 ∫ sec θ 2 sec θ r 3 d r d θ ∫ 1 2 ∫ 0 x ( x 2 + y 2 ) d y d x = ∫ 0 π 4 ∫ sec θ 2 sec θ r 3 d r d θ$

145.

$∫ 2 3 ∫ 0 x x x 2 + y 2 d y d x = ∫ 0 π / 4 ∫ 2 sec θ 3secθ r cos θ d r d θ ∫ 2 3 ∫ 0 x x x 2 + y 2 d y d x = ∫ 0 π / 4 ∫ 2 sec θ 3secθ r cos θ d r d θ$

146.

$∫ 0 1 ∫ x 2 x 1 x 2 + y 2 d y d x = ∫ 0 π / 4 ∫ 0 tan θ sec θ d r d θ ∫ 0 1 ∫ x 2 x 1 x 2 + y 2 d y d x = ∫ 0 π / 4 ∫ 0 tan θ sec θ d r d θ$

147.

$∫ 0 1 ∫ x 2 x y x 2 + y 2 d y d x = ∫ 0 π / 4 ∫ 0 tan θ sec θ r sin θ d r d θ ∫ 0 1 ∫ x 2 x y x 2 + y 2 d y d x = ∫ 0 π / 4 ∫ 0 tan θ sec θ r sin θ d r d θ$

In the following exercises, convert the integrals to polar coordinates and evaluate them.

148.

$∫ 0 3 ∫ 0 9 − y 2 ( x 2 + y 2 ) d x d y ∫ 0 3 ∫ 0 9 − y 2 ( x 2 + y 2 ) d x d y$

149.

$∫ 0 2 ∫ − 4 − y 2 4 − y 2 ( x 2 + y 2 ) 2 d x d y ∫ 0 2 ∫ − 4 − y 2 4 − y 2 ( x 2 + y 2 ) 2 d x d y$

150.

$∫ 0 1 ∫ 0 1 − x 2 ( x + y ) d y d x ∫ 0 1 ∫ 0 1 − x 2 ( x + y ) d y d x$

151.

$∫ 0 4 ∫ − 16 − x 2 16 − x 2 sin ( x 2 + y 2 ) d y d x ∫ 0 4 ∫ − 16 − x 2 16 − x 2 sin ( x 2 + y 2 ) d y d x$

152.

Evaluate the integral $∬DrdA∬DrdA$ where $DD$ is the region bounded by the polar axis and the upper half of the cardioid $r=1+cosθ.r=1+cosθ.$

153.

Find the area of the region $DD$ bounded by the polar axis and the upper half of the cardioid $r=1+cosθ.r=1+cosθ.$

154.

Evaluate the integral $∬DrdA,∬DrdA,$ where $DD$ is the region bounded by the part of the four-leaved rose $r=sin2θr=sin2θ$ situated in the first quadrant (see the following figure).

155.

Find the total area of the region enclosed by the four-leaved rose $r=sin2θr=sin2θ$ (see the figure in the previous exercise).

156.

Find the area of the region $D,D,$ which is the region bounded by $y=4−x2,y=4−x2,$ $x=3,x=3,$ $x=2,x=2,$ and $y=0.y=0.$

157.

Find the area of the region $D,D,$ which is the region inside the disk $x2+y2≤4x2+y2≤4$ and to the right of the line $x=1.x=1.$

158.

Determine the average value of the function $f(x,y)=x2+y2f(x,y)=x2+y2$ over the region $DD$ bounded by the polar curve $r=cos2θ,r=cos2θ,$ where $−π4≤θ≤π4−π4≤θ≤π4$ (see the following graph).

159.

Determine the average value of the function $f(x,y)=x2+y2f(x,y)=x2+y2$ over the region $DD$ bounded by the polar curve $r=3sin2θ,r=3sin2θ,$ where $0≤θ≤π20≤θ≤π2$ (see the following graph).

160.

Find the volume of the solid situated in the first octant and bounded by the paraboloid $z=1−4x2−4y2z=1−4x2−4y2$ and the planes $x=0,y=0,x=0,y=0,$ and $z=0.z=0.$

161.

Find the volume of the solid bounded by the paraboloid $z=2−9x2−9y2z=2−9x2−9y2$ and the plane $z=1.z=1.$

162.
1. Find the volume of the solid $S1S1$ bounded by the cylinder $x2+y2=1x2+y2=1$ and the planes $z=0z=0$ and $z=1.z=1.$
2. Find the volume of the solid $S2S2$ outside the double cone $z2=x2+y2,z2=x2+y2,$ inside the cylinder $x2+y2=1,x2+y2=1,$ and above the plane $z=0.z=0.$
3. Find the volume of the solid inside the cone $z2=x2+y2z2=x2+y2$ and below the plane $z=1z=1$ by subtracting the volumes of the solids $S1S1$ and $S2.S2.$
163.
1. Find the volume of the solid $S1S1$ inside the unit sphere $x2+y2+z2=1x2+y2+z2=1$ and above the plane $z=0.z=0.$
2. Find the volume of the solid $S2S2$ inside the double cone $(z−1)2=x2+y2(z−1)2=x2+y2$ and above the plane $z=0.z=0.$
3. Find the volume of the solid outside the double cone $(z−1)2=x2+y2(z−1)2=x2+y2$ and inside the sphere $x2+y2+z2=1.x2+y2+z2=1.$

For the following two exercises, consider a spherical ring, which is a sphere with a cylindrical hole cut so that the axis of the cylinder passes through the center of the sphere (see the following figure).

164.

If the sphere has radius $44$ and the cylinder has radius $2,2,$ find the volume of the spherical ring.

165.

A cylindrical hole of diameter $66$ cm is bored through a sphere of radius $55$ cm such that the axis of the cylinder passes through the center of the sphere. Find the volume of the resulting spherical ring.

166.

Find the volume of the solid that lies under the double cone $z2=4x2+4y2,z2=4x2+4y2,$ inside the cylinder $x2+y2=x,x2+y2=x,$ and above the plane $z=0.z=0.$

167.

Find the volume of the solid that lies under the paraboloid $z=x2+y2,z=x2+y2,$ inside the cylinder $x2+y2=x,x2+y2=x,$ and above the plane $z=0.z=0.$

168.

Find the volume of the solid that lies under the plane $x+y+z=10x+y+z=10$ and above the disk $x2+y2=4x.x2+y2=4x.$

169.

Find the volume of the solid that lies under the plane $2x+y+2z=82x+y+2z=8$ and above the unit disk $x2+y2=1.x2+y2=1.$

170.

A radial function $ff$ is a function whose value at each point depends only on the distance between that point and the origin of the system of coordinates; that is, $f(x,y)=g(r),f(x,y)=g(r),$ where $r=x2+y2.r=x2+y2.$ Show that if $ff$ is a continuous radial function, then $∬Df(x,y)dA=(θ2−θ1)[G(R2)−G(R1)],∬Df(x,y)dA=(θ2−θ1)[G(R2)−G(R1)],$ where $G′(r)=rg(r)G′(r)=rg(r)$ and $(x,y)∈D={(r,θ)|R1≤r≤R2,0≤θ≤2π},(x,y)∈D={(r,θ)|R1≤r≤R2,0≤θ≤2π},$ with $0≤R1 and $0≤θ1<θ2≤2π.0≤θ1<θ2≤2π.$

171.

Use the information from the preceding exercise to calculate the integral $∬D(x2+y2)3dA,∬D(x2+y2)3dA,$ where $DD$ is the unit disk.

172.

Let $f(x,y)=F′(r)rf(x,y)=F′(r)r$ be a continuous radial function defined on the annular region $D={(r,θ)|R1≤r≤R2,0≤θ≤2π},D={(r,θ)|R1≤r≤R2,0≤θ≤2π},$ where $r=x2+y2,r=x2+y2,$ $0 and $FF$ is a differentiable function. Show that $∬Df(x,y)dA=2π[F(R2)−F(R1)].∬Df(x,y)dA=2π[F(R2)−F(R1)].$

173.

Apply the preceding exercise to calculate the integral $∬Dex2+y2x2+y2dxdy,∬Dex2+y2x2+y2dxdy,$ where $DD$ is the annular region between the circles of radii $11$ and $22$ situated in the third quadrant.

174.

Let $ff$ be a continuous function that can be expressed in polar coordinates as a function of $θθ$ only; that is, $f(x,y)=h(θ),f(x,y)=h(θ),$ where $(x,y)∈D={(r,θ)|R1≤r≤R2,θ1≤θ≤θ2},(x,y)∈D={(r,θ)|R1≤r≤R2,θ1≤θ≤θ2},$ with $0≤R1 and $0≤θ1<θ2≤2π.0≤θ1<θ2≤2π.$ Show that $∬Df(x,y)dA=12(R22−R12)[H(θ2)−H(θ1)],∬Df(x,y)dA=12(R22−R12)[H(θ2)−H(θ1)],$ where $HH$ is an antiderivative of $h.h.$

175.

Apply the preceding exercise to calculate the integral $∬Dy2x2dA,∬Dy2x2dA,$ where $D={(r,θ)|1≤r≤2,π6≤θ≤π3}.D={(r,θ)|1≤r≤2,π6≤θ≤π3}.$

176.

Let $ff$ be a continuous function that can be expressed in polar coordinates as a function of $θθ$ only; that is, $f(x,y)=g(r)h(θ),f(x,y)=g(r)h(θ),$ where $(x,y)∈D={(r,θ)|R1≤r≤R2,θ1≤θ≤θ2}(x,y)∈D={(r,θ)|R1≤r≤R2,θ1≤θ≤θ2}$ with $0≤R1 and $0≤θ1<θ2≤2π.0≤θ1<θ2≤2π.$ Show that $∬Df(x,y)dA=[G(R2)−G(R1)][H(θ2)−H(θ1)],∬Df(x,y)dA=[G(R2)−G(R1)][H(θ2)−H(θ1)],$ where $GG$ and $HH$ are antiderivatives of $gg$ and $h,h,$ respectively.

177.

Evaluate $∬Darctan(yx)x2+y2dA,∬Darctan(yx)x2+y2dA,$ where $D={(r,θ)|2≤r≤3,π4≤θ≤π3}.D={(r,θ)|2≤r≤3,π4≤θ≤π3}.$

178.

A spherical cap is the region of a sphere that lies above or below a given plane.

1. Show that the volume of the spherical cap in the figure below is $16πh(3a2+h2).16πh(3a2+h2).$
2. A spherical segment is the solid defined by intersecting a sphere with two parallel planes. If the distance between the planes is $h,h,$ show that the volume of the spherical segment in the figure below is $16πh(3a2+3b2+h2).16πh(3a2+3b2+h2).$
179.

In statistics, the joint density for two independent, normally distributed events with a mean $μ=0μ=0$ and a standard distribution $σσ$ is defined by $p(x,y)=12πσ2e−x2+y22σ2.p(x,y)=12πσ2e−x2+y22σ2.$ Consider $(X,Y),(X,Y),$ the Cartesian coordinates of a ball in the resting position after it was released from a position on the z-axis toward the $xyxy$-plane. Assume that the coordinates of the ball are independently normally distributed with a mean $μ=0μ=0$ and a standard deviation of $σσ$ (in feet). The probability that the ball will stop no more than $aa$ feet from the origin is given by $P[X2+Y2≤a2]=∬Dp(x,y)dydx,P[X2+Y2≤a2]=∬Dp(x,y)dydx,$ where $DD$ is the disk of radius a centered at the origin. Show that $P[X2+Y2≤a2]=1−e−a2/2σ2.P[X2+Y2≤a2]=1−e−a2/2σ2.$

180.

The double improper integral $∫−∞∞∫−∞∞e−(x2+y2)/2dydx∫−∞∞∫−∞∞e−(x2+y2)/2dydx$ may be defined as the limit value of the double integrals $∬Dae−(x2+y2)/2dA∬Dae−(x2+y2)/2dA$ over disks $DaDa$ of radii a centered at the origin, as a increases without bound; that is, $∫−∞∞∫−∞∞e−(x2+y2)/2dydx=lima→∞∬Dae−(x2+y2)/2dA.∫−∞∞∫−∞∞e−(x2+y2)/2dydx=lima→∞∬Dae−(x2+y2)/2dA.$

1. Use polar coordinates to show that $∫−∞∞∫−∞∞e−(x2+y2)/2dydx=2π.∫−∞∞∫−∞∞e−(x2+y2)/2dydx=2π.$
2. Show that $∫−∞∞e−x2/2dx=2π,∫−∞∞e−x2/2dx=2π,$ by using the relation $∫−∞∞∫−∞∞e−(x2+y2)/2dydx=(∫−∞∞e−x2/2dx)(∫−∞∞e−y2/2dy).∫−∞∞∫−∞∞e−(x2+y2)/2dydx=(∫−∞∞e−x2/2dx)(∫−∞∞e−y2/2dy).$
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