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Calculus Volume 3

5.3 Double Integrals in Polar Coordinates

Calculus Volume 35.3 Double Integrals in Polar Coordinates
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  1. Preface
  2. 1 Parametric Equations and Polar Coordinates
    1. Introduction
    2. 1.1 Parametric Equations
    3. 1.2 Calculus of Parametric Curves
    4. 1.3 Polar Coordinates
    5. 1.4 Area and Arc Length in Polar Coordinates
    6. 1.5 Conic Sections
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  3. 2 Vectors in Space
    1. Introduction
    2. 2.1 Vectors in the Plane
    3. 2.2 Vectors in Three Dimensions
    4. 2.3 The Dot Product
    5. 2.4 The Cross Product
    6. 2.5 Equations of Lines and Planes in Space
    7. 2.6 Quadric Surfaces
    8. 2.7 Cylindrical and Spherical Coordinates
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  4. 3 Vector-Valued Functions
    1. Introduction
    2. 3.1 Vector-Valued Functions and Space Curves
    3. 3.2 Calculus of Vector-Valued Functions
    4. 3.3 Arc Length and Curvature
    5. 3.4 Motion in Space
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  5. 4 Differentiation of Functions of Several Variables
    1. Introduction
    2. 4.1 Functions of Several Variables
    3. 4.2 Limits and Continuity
    4. 4.3 Partial Derivatives
    5. 4.4 Tangent Planes and Linear Approximations
    6. 4.5 The Chain Rule
    7. 4.6 Directional Derivatives and the Gradient
    8. 4.7 Maxima/Minima Problems
    9. 4.8 Lagrange Multipliers
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Chapter Review Exercises
  6. 5 Multiple Integration
    1. Introduction
    2. 5.1 Double Integrals over Rectangular Regions
    3. 5.2 Double Integrals over General Regions
    4. 5.3 Double Integrals in Polar Coordinates
    5. 5.4 Triple Integrals
    6. 5.5 Triple Integrals in Cylindrical and Spherical Coordinates
    7. 5.6 Calculating Centers of Mass and Moments of Inertia
    8. 5.7 Change of Variables in Multiple Integrals
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  7. 6 Vector Calculus
    1. Introduction
    2. 6.1 Vector Fields
    3. 6.2 Line Integrals
    4. 6.3 Conservative Vector Fields
    5. 6.4 Green’s Theorem
    6. 6.5 Divergence and Curl
    7. 6.6 Surface Integrals
    8. 6.7 Stokes’ Theorem
    9. 6.8 The Divergence Theorem
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Chapter Review Exercises
  8. 7 Second-Order Differential Equations
    1. Introduction
    2. 7.1 Second-Order Linear Equations
    3. 7.2 Nonhomogeneous Linear Equations
    4. 7.3 Applications
    5. 7.4 Series Solutions of Differential Equations
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  9. A | Table of Integrals
  10. B | Table of Derivatives
  11. C | Review of Pre-Calculus
  12. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
  13. Index

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)|arb,αθβ}.R={(r,θ)|arb,αθβ}.

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 [ri1,ri][ri1,ri] of length Δr=(ba)/mΔr=(ba)/m and divide the interval [α,β][α,β] into nn subintervals [θi1,θi][θi1,θi] of width Δθ=(βα)/n.Δθ=(βα)/n. This means that the circles r=rir=ri and rays θ=θiθ=θi for 1im1im and 1jn1jn divide the polar rectangle RR into smaller polar subrectangles RijRij (Figure 5.28(b)).

This figure consists of three figures labeled a, b, and c. In figure a, a sector of an annulus is shown in the polar coordinate plane with radii a and b and angles alpha and beta from the theta = 0 axis. In figure b, this sector of an annulus is cut up into subsectors in a manner similar to the way in which previous spaces were cut up into subrectangles. In figure c, one of these subsectors is shown with angle Delta theta, distance between inner and outer radii Delta r, and area Delta A = r* sub theta Delta r Delta theta, where the center point is given as (r* sub i j, theta* sub i j).
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 ri1Δθri1Δθ and riΔθriΔθ and with a width Δr.Δr. Hence the area of the polar subrectangle RijRij is

ΔA=12Δr(ri1Δθ+r1Δθ).ΔA=12Δr(ri1Δθ+r1Δθ).

Simplifying and letting rij*=12(ri1+ri),rij*=12(ri1+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.
In x y z space, there is a surface f (r, theta). On the x y plane, a series of subsectors of annuli are drawn as in the previous figure with radius between annuli Delta r and angle between subsectors Delta theta. A subsector from the surface f(r, theta) is projected down onto one of these subsectors. This subsector has center point marked (r* sub i j, theta* sub i j).
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=1mj=1nf(rij*,θij*)rij*ΔrΔθ.i=1mj=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,ni=1mj=1nf(rij*,θij*)rij*ΔrΔθ.V=limm,ni=1mj=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,ni=1mj=1nf(rij*,θij*)ΔA=limm,ni=1mj=1nf(rij*,θij*)rij*ΔrΔθ.Rf(r,θ)dA=limm,ni=1mj=1nf(rij*,θij*)ΔA=limm,ni=1mj=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=βr=ar=bf(r,θ)rdrdθ.Rf(r,θ)dA=Rf(r,θ)rdrdθ=θ=αθ=βr=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)|1r3,0θπ}.R={(r,θ)|1r3,0θπ}.

Solution

As we can see from Figure 5.30, r=1r=1 and r=3r=3 are circles of radius 1and31and3 and 0θπ0θπ covers the entire top half of the plane. Hence the region RR looks like a semicircular band.

Half an annulus R is drawn with inner radius 1 and outer radius 3. That is, the inner semicircle is given by x squared + y squared = 1, whereas the outer semicircle is given by x squared + y squared = 9.
Figure 5.30 The polar region RR lies between two semicircles.

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 R3xdAR3xdA over the region R={(r,θ)|1r2,0θπ}.R={(r,θ)|1r2,0θπ}.

Solution

First we sketch a figure similar to Figure 5.30 but with outer radius 2.2. From the figure we can see that we have

R3xdA=θ=0θ=πr=1r=23rcosθrdrdθUse an iterated integral with correct limitsof integration.=θ=0θ=πcosθ[r3|r=1r=2]dθIntegrate first with respect tor.=θ=0θ=π7cosθdθ=7sinθ|θ=0θ=π=0.R3xdA=θ=0θ=πr=1r=23rcosθrdrdθUse an iterated integral with correct limitsof integration.=θ=0θ=πcosθ[r3|r=1r=2]dθIntegrate first with respect tor.=θ=0θ=π7cosθdθ=7sinθ|θ=0θ=π=0.
Checkpoint 5.17

Sketch the region R={(r,θ)|1r2,π2θπ2},R={(r,θ)|1r2,π2θπ2}, and evaluate RxdA.RxdA.

Example 5.26

Evaluating a Double Integral by Converting from Rectangular Coordinates

Evaluate the integral R(1x2y2)dAR(1x2y2)dA where RR is the unit circle on the xyxy-plane.

Solution

The region RR is a unit circle, so we can describe it as R={(r,θ)|0r1,0θ2π}.R={(r,θ)|0r1,0θ2π}.

Using the conversion x=rcosθ,y=rsinθ,x=rcosθ,y=rsinθ, and dA=rdrdθ,dA=rdrdθ, we have

R(1x2y2)dA=02π01(1r2)rdrdθ=02π01(rr3)drdθ=02π[r22r44]01dθ=02π14dθ=π2.R(1x2y2)dA=02π01(1r2)rdrdθ=02π01(rr3)drdθ=02π[r22r44]01dθ=02π14dθ=π2.

Example 5.27

Evaluating a Double Integral by Converting from Rectangular Coordinates

Evaluate the integral R(x+y)dAR(x+y)dA where R={(x,y)|1x2+y24,x0}.R={(x,y)|1x2+y24,x0}.

Solution

We can see that RR is an annular region that can be converted to polar coordinates and described as R={(r,θ)|1r2,π2θ3π2}R={(r,θ)|1r2,π2θ3π2} (see the following graph).

Two semicircles are drawn in the second and third quadrants, with equations x squared + y squared = 1 and x squared + y squared = 2.
Figure 5.31 The annular region of integration R.R.

Hence, using the conversion x=rcosθ,y=rsinθ,x=rcosθ,y=rsinθ, and dA=rdrdθ,dA=rdrdθ, we have

R(x+y)dA=θ=π/2θ=3π/2r=1r=2(rcosθ+rsinθ)rdrdθ=(r=1r=2r2dr)(π/23π/2(cosθ+sinθ)dθ)=[r33]12[sinθcosθ]|π/23π/2=143.R(x+y)dA=θ=π/2θ=3π/2r=1r=2(rcosθ+rsinθ)rdrdθ=(r=1r=2r2dr)(π/23π/2(cosθ+sinθ)dθ)=[r33]12[sinθcosθ]|π/23π/2=143.
Checkpoint 5.18

Evaluate the integral R(4x2y2)dAR(4x2y2)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)|αθβ,h1(θ)rh2(θ)}R={(r,θ)|αθβ,h1(θ)rh2(θ)} (see the following figure).

A region D is shown in polar coordinates with edges given by theta = alpha, theta = beta, r = h2(theta), and r = h1(theta).
Figure 5.32 A general polar region between α<θ<βα<θ<β and h1(θ)<r<h2(θ).h1(θ)<r<h2(θ).
Theorem 5.8

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=βr=h1(θ)r=h2(θ)f(r,θ)rdrdθDf(r,θ)rdrdθ=θ=αθ=βr=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 Dr2sinθrdrdθDr2sinθrdrdθ where DD is the region bounded by the polar axis and the upper half of the cardioid r=1+cosθ.r=1+cosθ.

Solution

We can describe the region DD as {(r,θ)|0θπ,0r1+cosθ}{(r,θ)|0θπ,0r1+cosθ} as shown in the following figure.

A region D is given as the top half of a cardioid with equation r = 1 + cos theta.
Figure 5.33 The region DD is the top half of a cardioid.

Hence, we have

Dr2sinθrdrdθ=θ=0θ=πr=0r=1+cosθ(r2sinθ)rdrdθ=14θ=0θ=π[r4]r=0r=1+cosθsinθdθ=14θ=0θ=π(1+cosθ)4sinθdθ=14[(1+cosθ)55]0π=85.Dr2sinθrdrdθ=θ=0θ=πr=0r=1+cosθ(r2sinθ)rdrdθ=14θ=0θ=π[r4]r=0r=1+cosθsinθdθ=14θ=0θ=π(1+cosθ)4sinθdθ=14[(1+cosθ)55]0π=85.
Checkpoint 5.19

Evaluate the integral

Dr2sin22θrdrdθwhereD={(r,θ)|0θπ,0r2cos2θ}.Dr2sin22θrdrdθwhereD={(r,θ)|0θπ,0r2cos2θ}.

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=βr=ar=bf(r,θ)rdrdθ.V=Rf(r,θ)rdrdθ=θ=αθ=βr=ar=bf(r,θ)rdrdθ.

If the base of the solid can be described as D={(r,θ)|αθβ,h1(θ)rh2(θ)},D={(r,θ)|αθβ,h1(θ)rh2(θ)}, then the double integral for the volume becomes

V=Df(r,θ)rdrdθ=θ=αθ=βr=h1(θ)r=h2(θ)f(r,θ)rdrdθ.V=Df(r,θ)rdrdθ=θ=αθ=βr=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=1x2y2z=1x2y2 and above the unit circle on the xyxy-plane (see the following figure).

The paraboloid z = 1 minus x squared minus y squared is shown, which in this graph looks like a sheet with the middle gently puffed up and the corners anchored.
Figure 5.34 The paraboloid z=1x2y2z=1x2y2.

Solution

By the method of double integration, we can see that the volume is the iterated integral of the form R(1x2y2)dAR(1x2y2)dA where R={(r,θ)|0r1,0θ2π}.R={(r,θ)|0r1,0θ2π}.

This integration was shown before in Example 5.26, so the volume is π2π2 cubic units.

Example 5.30

Finding a Volume Using Double Integration

Find the volume of the solid that lies under the paraboloid z=4x2y2z=4x2y2 and above the disk (x1)2+y2=1(x1)2+y2=1 on the xyxy-plane. See the paraboloid in Figure 5.35 intersecting the cylinder (x1)2+y2=1(x1)2+y2=1 above the xyxy-plane.

A paraboloid with equation z = 4 minus x squared minus y squared is intersected by a cylinder with equation (x minus 1) squared + y squared = 1.
Figure 5.35 Finding the volume of a solid with a paraboloid cap and a circular base.

Solution

First change the disk (x1)2+y2=1(x1)2+y2=1 to polar coordinates. Expanding the square term, we have x22x+1+y2=1.x22x+1+y2=1. Then simplify to get x2+y2=2x,x2+y2=2x, which in polar coordinates becomes r2=2rcosθr2=2rcosθ and then either r=0r=0 or r=2cosθ.r=2cosθ. Similarly, the equation of the paraboloid changes to z=4r2.z=4r2. Therefore we can describe the disk (x1)2+y2=1(x1)2+y2=1 on the xyxy-plane as the region

D={(r,θ)|0θπ,0r2cosθ}.D={(r,θ)|0θπ,0r2cosθ}.

Hence the volume of the solid bounded above by the paraboloid z=4x2y2z=4x2y2 and below by r=2cosθr=2cosθ is

V=Df(r,θ)rdrdθ=θ=0θ=πr=0r=2cosθ(4r2)rdrdθ=θ=0θ=π[4r22r44|02cosθ]dθ=0π[8cos2θ4cos2θ]dθ=[52θ+52sinθcosθsinθcos3θ]0π=52π.V=Df(r,θ)rdrdθ=θ=0θ=πr=0r=2cosθ(4r2)rdrdθ=θ=0θ=π[4r22r44|02cosθ]dθ=0π[8cos2θ4cos2θ]dθ=[52θ+52sinθcosθsinθcos3θ]0π=52π.

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).

Solution

First examine the region over which we need to set up the double integral and the accompanying paraboloid.

This figure consists of three figures. The first is simply a paraboloid that opens up. The second shows the region D bounded by x = 0, y = x, and x + y = 2 with a vertical double-sided arrow within the region. The second shows the same region but in polar coordinates, so the lines bounding D are theta = pi/2, r = 2/(cos theta + sin theta), and theta = pi/4, with a double-sided arrow that has one side pointed at the origin.
Figure 5.36 Finding the volume of a solid under a paraboloid and above a given triangle.

The region DD is {(x,y)|0x1,xy2x}.{(x,y)|0x1,xy2x}. Converting the lines y=x,x=0,y=x,x=0, and x+y=2x+y=2 in the xyxy-plane to functions of rr and θ,θ, we have θ=π/4,θ=π/4, θ=π/2,θ=π/2, and r=2/(cosθ+sinθ),r=2/(cosθ+sinθ), respectively. Graphing the region on the xyxy-plane, we see that it looks like D={(r,θ)|π/4θπ/2,0r2/(cosθ+sinθ)}.D={(r,θ)|π/4θπ/2,0r2/(cosθ+sinθ)}. Now converting the equation of the surface gives z=x2+y2=r2.z=x2+y2=r2. Therefore, the volume of the solid is given by the double integral

V=Df(r,θ)rdrdθ=θ=π/4θ=π/2r=0r=2/(cosθ+sinθ)r2rdrdθ=π/4π/2[r44]02/(cosθ+sinθ)dθ=14π/4π/2(2cosθ+sinθ)4dθ=164π/4π/2(1cosθ+sinθ)4dθ=4π/4π/2(1cosθ+sinθ)4dθ.V=Df(r,θ)rdrdθ=θ=π/4θ=π/2r=0r=2/(cosθ+sinθ)r2rdrdθ=π/4π/2[r44]02/(cosθ+sinθ)dθ=14π/4π/2(2cosθ+sinθ)4dθ=164π/4π/2(1cosθ+sinθ)4dθ=4π/4π/2(1cosθ+sinθ)4dθ.

As you can see, this integral is very complicated. So, we can instead evaluate this double integral in rectangular coordinates as

V=01x2x(x2+y2)dydx.V=01x2x(x2+y2)dydx.

Evaluating gives

V=01x2x(x2+y2)dydx=01[x2y+y33]|x2xdx=01834x+4x28x33dx=[8x32x2+4x332x43]|01=43.V=01x2x(x2+y2)dydx=01[x2y+y33]|x2xdx=01834x+4x28x33dx=[8x32x2+4x332x43]|01=43.

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=2x2+y2z=2x2+y2 and above the xy-plane.xy-plane.

Solution

The region DD for the integration is the base of the cone, which appears to be a circle on the xy-planexy-plane (see the following figure).

A cone given by z = 2 minus the square root of (x squared plus y squared) and a circle given by x squared plus y squared = 4. The cone is above the circle in xyz space.
Figure 5.37 Finding the volume of a solid inside the cone and above the xyxy-plane.

We find the equation of the circle by setting z=0:z=0:

0=2x2+y22=x2+y2x2+y2=4.0=2x2+y22=x2+y2x2+y2=4.

This means the radius of the circle is 2,2, so for the integration we have 0θ2π0θ2π and 0r2.0r2. Substituting x=rcosθx=rcosθ and y=rsinθy=rsinθ in the equation z=2x2+y2z=2x2+y2 we have z=2r.z=2r. Therefore, the volume of the cone is

θ=0θ=2πr=0r=2(2r)rdrdθ=2π43=8π3θ=0θ=2πr=0r=2(2r)rdrdθ=2π43=8π3 cubic units.

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=hhax2+y2.z=hhax2+y2.

We can still use Figure 5.37 and set up the integral as θ=0θ=2πr=0r=a(hhar)rdrdθ.θ=0θ=2πr=0r=a(hhar)rdrdθ.

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

Checkpoint 5.20

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=16x2y2.z=16x2y2.

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=αβh1(θ)h2(θ)1rdrdθ.AreaA=αβh1(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.

Solution

Sketching the graph of the function r=cos4θr=cos4θ reveals that it is a polar rose with eight petals (see the following figure).

A rose with eight petals given by r = cos (4 theta).
Figure 5.38 Finding the area of a polar rose with eight petals.

Using symmetry, we can see that we need to find the area of one petal and then multiply it by 8.8. Notice that the values of θθ for which the graph passes through the origin are the zeros of the function cos4θ,cos4θ, and these are odd multiples of π/8.π/8. Thus, one of the petals corresponds to the values of θθ in the interval [π/8,π/8].[π/8,π/8]. Therefore, the area bounded by the curve r=cos4θr=cos4θ is

A=8θ=π/8θ=π/8r=0r=cos4θ1rdrdθ=8π/8π/8[12r2|0cos4θ]dθ=8π/8π/812cos24θdθ=8[14θ+116sin4θcos4θ|π/8π/8]=8[π16]=π2.A=8θ=π/8θ=π/8r=0r=cos4θ1rdrdθ=8π/8π/8[12r2|0cos4θ]dθ=8π/8π/812cos24θdθ=8[14θ+116sin4θcos4θ|π/8π/8]=8[π16]=π2.

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.

Solution

First and foremost, sketch the graphs of the region (Figure 5.39).

A cardioid with equation 1 + cos theta is shown overlapping a circle given by r = 3 cos theta, which is a circle of radius 3 with center (1.5, 0). The area bounded by the x axis, the cardioid, and the dashed line connecting the origin to the intersection of the cardioid and circle on the r = 2 line is shaded.
Figure 5.39 Finding the area enclosed by both a circle and a cardioid.

We can from see the symmetry of the graph that we need to find the points of intersection. Setting the two equations equal to each other gives

3cosθ=1+cosθ.3cosθ=1+cosθ.

One of the points of intersection is θ=π/3.θ=π/3. The area above the polar axis consists of two parts, with one part defined by the cardioid from θ=0θ=0 to θ=π/3θ=π/3 and the other part defined by the circle from θ=π/3θ=π/3 to θ=π/2.θ=π/2. By symmetry, the total area is twice the area above the polar axis. Thus, we have

A=2[θ=0θ=π/3r=0r=1+cosθ1rdrdθ+θ=π/3θ=π/2r=0r=3cosθ1rdrdθ].A=2[θ=0θ=π/3r=0r=1+cosθ1rdrdθ+θ=π/3θ=π/2r=0r=3cosθ1rdrdθ].

Evaluating each piece separately, we find that the area is

A=2(14π+9163+38π9163)=2(58π)=54πsquare units.A=2(14π+9163+38π9163)=2(58π)=54πsquare units.
Checkpoint 5.21

Find the area enclosed inside the cardioid r=33sinθr=33sinθ 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.

Solution

This is an improper integral because we are integrating over an unbounded region R2.R2. In polar coordinates, the entire plane R2R2 can be seen as 0θ2π,0θ2π, 0r.0r.

Using the changes of variables from rectangular coordinates to polar coordinates, we have

R2e−10(x2+y2)dxdy=θ=0θ=2πr=0r=e−10r2rdrdθ=θ=0θ=2π(limar=0r=ae−10r2rdr)dθ=(θ=0θ=2πdθ)(limar=0r=ae−10r2rdr)=2π(limar=0r=ae−10r2rdr)=2πlima(120)(e−10r2|0a)=2π(120)lima(e−10a21)=π10.R2e−10(x2+y2)dxdy=θ=0θ=2πr=0r=e−10r2rdrdθ=θ=0θ=2π(limar=0r=ae−10r2rdr)dθ=(θ=0θ=2πdθ)(limar=0r=ae−10r2rdr)=2π(limar=0r=ae−10r2rdr)=2πlima(120)(e−10r2|0a)=2π(120)lima(e−10a21)=π10.
Checkpoint 5.22

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=1y2.x=1y2.

125.

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

126.

D={(x,y)|x2+y24x}D={(x,y)|x2+y24x}

127.

D={(x,y)|x2+y24y}D={(x,y)|x2+y24y}

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

128.
Half an annulus D is drawn in the first and second quadrants with inner radius 3 and outer radius 5.
129.
A sector of an annulus D is drawn between theta = pi/4 and theta = pi/2 with inner radius 3 and outer radius 5.
130.
Half of an annulus D is drawn between theta = pi/4 and theta = 5 pi/4 with inner radius 3 and outer radius 5.
131.
 A sector of an annulus D is drawn between theta = 3 pi/4 and theta = 5 pi/4 with inner radius 3 and outer radius 5.
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.

A region D is given that is bounded by y = 0, x = 1, x = 5, and y = x, that is, a right triangle with a corner cut off.
133.

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

A region D is drawn between y = x and y = x squared, which looks like a deformed lens, with the bulbous part below the straight part.

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

134.

f(x,y)=x2+y2,D={(r,θ)|3r5,0θ2π}f(x,y)=x2+y2,D={(r,θ)|3r5,0θ2π}

135.

f(x,y)=x+y,D={(r,θ)|3r5,0θ2π}f(x,y)=x+y,D={(r,θ)|3r5,0θ2π}

136.

f(x,y)=x2+xy,D={(r,θ)|1r2,πθ2π}f(x,y)=x2+xy,D={(r,θ)|1r2,πθ2π}

137.

f(x,y)=x4+y4,D={(r,θ)|1r2,3π2θ2π}f(x,y)=x4+y4,D={(r,θ)|1r2,3π2θ2π}

138.

f(x,y)=x2+y23,f(x,y)=x2+y23, where D={(r,θ)|0r1,π2θπ}.D={(r,θ)|0r1,π2θπ}.

139.

f(x,y)=x4+2x2y2+y4,f(x,y)=x4+2x2y2+y4, where D={(r,θ)|3r4,π3θ2π3}.D={(r,θ)|3r4,π3θ2π3}.

140.

f(x,y)=sin(arctanyx),f(x,y)=sin(arctanyx), where D={(r,θ)|1r2,π6θπ3}D={(r,θ)|1r2,π6θπ3}

141.

f(x,y)=arctan(yx),f(x,y)=arctan(yx), where D={(r,θ)|2r3,π4θπ3}D={(r,θ)|2r3,π4θπ3}

142.

Dex2+y2[1+2arctan(yx)]dA,D={(r,θ)|1r2,π6θπ3}Dex2+y2[1+2arctan(yx)]dA,D={(r,θ)|1r2,π6θπ3}

143.

D(ex2+y2+x4+2x2y2+y4)arctan(yx)dA,D={(r,θ)|1r2,π4θπ3}D(ex2+y2+x4+2x2y2+y4)arctan(yx)dA,D={(r,θ)|1r2,π4θπ3}

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.

120x(x2+y2)dydx=0π4secθ2secθr3drdθ120x(x2+y2)dydx=0π4secθ2secθr3drdθ

145.

230xxx2+y2dydx=0π/40tanθsecθrcosθdrdθ230xxx2+y2dydx=0π/40tanθsecθrcosθdrdθ

146.

01x2x1x2+y2dydx=0π/40tanθsecθdrdθ01x2x1x2+y2dydx=0π/40tanθsecθdrdθ

147.

01x2xyx2+y2dydx=0π/40tanθsecθrsinθdrdθ01x2xyx2+y2dydx=0π/40tanθsecθrsinθdrdθ

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

148.

0309y2(x2+y2)dxdy0309y2(x2+y2)dxdy

149.

024y24y2(x2+y2)2dxdy024y24y2(x2+y2)2dxdy

150.

0101x2(x+y)dydx0101x2(x+y)dydx

151.

0416x216x2sin(x2+y2)dydx0416x216x2sin(x2+y2)dydx

152.

Evaluate the integral DrdADrdA 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).

A region D is drawn in the first quadrant petal of the four petal rose given by r = sin (2 theta).
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=4x2,y=4x2, 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+y24x2+y24 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 π4θπ4π4θπ4 (see the following graph).

The first/fourth-quadrant petal of the four-petal rose given by r = cos (2 theta) is shown.
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 0θπ20θπ2 (see the following graph).

The first-quadrant petal of the four-petal rose given by r = 3sin (2 theta) is shown.
160.

Find the volume of the solid situated in the first octant and bounded by the paraboloid z=14x24y2z=14x24y2 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=29x29y2z=29x29y2 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 (z1)2=x2+y2(z1)2=x2+y2 and above the plane z=0.z=0.
  3. Find the volume of the solid outside the double cone (z1)2=x2+y2(z1)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).

A spherical ring is shown, that is, a sphere with a cylindrical hole going all the way through it.
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=(θ2θ1)[G(R2)G(R1)],Df(x,y)dA=(θ2θ1)[G(R2)G(R1)], where G(r)=rg(r)G(r)=rg(r) and (x,y)D={(r,θ)|R1rR2,0θ2π},(x,y)D={(r,θ)|R1rR2,0θ2π}, with 0R1<R20R1<R2 and 0θ1<θ22π.0θ1<θ22π.

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)|R1rR2,0θ2π},D={(r,θ)|R1rR2,0θ2π}, where r=x2+y2,r=x2+y2, 0<R1<R2,0<R1<R2, 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)|R1rR2,θ1θθ2},(x,y)D={(r,θ)|R1rR2,θ1θθ2}, with 0R1<R20R1<R2 and 0θ1<θ22π.0θ1<θ22π. Show that Df(x,y)dA=12(R22R12)[H(θ2)H(θ1)],Df(x,y)dA=12(R22R12)[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)|1r2,π6θπ3}.D={(r,θ)|1r2,π6θπ3}.

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)|R1rR2,θ1θθ2}(x,y)D={(r,θ)|R1rR2,θ1θθ2} with 0R1<R20R1<R2 and 0θ1<θ22π.0θ1<θ22π. 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)|2r3,π4θπ3}.D={(r,θ)|2r3,π4θπ3}.

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).
    A sphere of radius R has a circle inside of it h units from the top of the sphere. This circle has radius a, which is less than R.
  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).
    A sphere has two parallel circles inside of it h units apart. The upper circle has radius b, and the lower circle has radius a. Note that a > b.
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)=12πσ2ex2+y22σ2.p(x,y)=12πσ2ex2+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+Y2a2]=Dp(x,y)dydx,P[X2+Y2a2]=Dp(x,y)dydx, where DD is the disk of radius a centered at the origin. Show that P[X2+Y2a2]=1ea2/2σ2.P[X2+Y2a2]=1ea2/2σ2.

180.

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

  1. Use polar coordinates to show that e(x2+y2/2)dydx=2π.e(x2+y2/2)dydx=2π.
  2. Show that ex2/2dx=2π,ex2/2dx=2π, by using the relation e(x2+y2/2)dydx=(ex2/2dx)(ey2/2dy).e(x2+y2/2)dydx=(ex2/2dx)(ey2/2dy).
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