
# 14.3E: Double Integrals in Polar Coordinates (Exercises)

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#### Terms and Concepts

1. When evaluating $$\displaystyle \int\int_R f(x,y)\,dA$$ using polar coordinates, $$f(x,y)$$ is replaced with _______ and $$dA$$ is replaced with _______.

$$f(x,y)$$ is replaced with $$f(r\cos \theta, r\sin\theta)$$ and $$dA$$ is replaced with $$r\,dr\,d\theta$$.

2. Why would one be interested in evaluating a double integral with polar coordinates?

#### Defining Polar Regions

In exercises 3 - 6, express the region $$R$$ in polar coordinates.

3) $$R$$ is the region of the disk of radius 2 centered at the origin that lies in the first quadrant.

$$R = \big\{(r, \theta)\,|\,0 \leq r \leq 2, \space 0 \leq \theta \leq \frac{\pi}{2}\big\}$$

4) $$R$$ is the region of the disk of radius 3 centered at the origin.

5) $$R$$ is the region between the circles of radius 4 and radius 5 centered at the origin that lies in the second quadrant.

$$R = \big\{(r, \theta)\,|\,4 \leq r \leq 5, \space \frac{\pi}{2} \leq \theta \leq \pi\big\}$$

6) $$R$$ is the region bounded by the $$y$$-axis and $$x = \sqrt{1 - y^2}$$.

7) $$R$$ is the region bounded by the $$x$$-axis and $$y = \sqrt{2 - x^2}$$.

$$R = \big\{(r, \theta)\,|\,0 \leq r \leq \sqrt{2}, \space 0 \leq \theta \leq \pi\big\}$$

8) $$R = \big\{(x,y)\,|\,x^2 + y^2 \leq 4x\big\}$$

9) $$R = \big\{(x,y)\,|\,x^2 + y^2 \leq 4y\big\}$$

$$R = \big\{(r, \theta)\,|\,0 \leq r \leq 4 \space \sin \theta, \space 0 \leq \theta \leq \pi\big\}$$

In exercises 10 - 15, the graph of the polar rectangular region $$D$$ is given. Express $$D$$ in polar coordinates.

10)
11)
$$D = \big\{(r, \theta)\,|\, 3 \leq r \leq 5, \space \frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}\big\}$$

12)

13)

$$D = \big\{(r, \theta)\,|\,3 \leq r \leq 5, \space \frac{3\pi}{4} \leq \theta \leq \frac{5\pi}{4}\big\}$$
14)  In the following graph, the region $$D$$ is situated below $$y = x$$ and is bounded by $$x = 1, \space x = 5$$, and $$y = 0$$.

15)  In the following graph, the region $$D$$ is bounded by $$y = x$$ and $$y = x^2$$.

$$D = \big\{(r, \theta)\,|\,0 \leq r \leq \tan \theta \space \sec \theta, \space 0 \leq \theta \leq \frac{\pi}{4}\big\}$$

#### Evaluating Polar Double Integrals

In exercises 16 - 25, evaluate the double integral $$\displaystyle \iint_R f(x,y) \,dA$$ over the polar rectangular region $$R$$.

16)  $$f(x,y) = x^2 + y^2$$, $$R = \big\{(r, \theta)\,|\,3 \leq r \leq 5, \space 0 \leq \theta \leq 2\pi\big\}$$

17)  $$f(x,y) = x + y$$, $$R = \big\{(r, \theta)\,|\,3 \leq r \leq 5, \space 0 \leq \theta \leq 2\pi\big\}$$

$$0$$

18)  $$f(x,y) = x^2 + xy, \space R = \big\{(r, \theta )\,|\,1 \leq r \leq 2, \space \pi \leq \theta \leq 2\pi\big\}$$

19)  $$f(x,y) = x^4 + y^4, \space R = \big\{(r, \theta)\,|\,1 \leq r \leq 2, \space \frac{3\pi}{2} \leq \theta \leq 2\pi\big\}$$

$$\frac{63\pi}{16}$$

20)  $$f(x,y) = \sqrt[3]{x^2 + y^2}$$, where $$R = \big\{(r, \theta)\,|\,0 \leq r \leq 1, \space \frac{\pi}{2} \leq \theta \leq \pi\big\}$$.

21)  $$f(x,y) = x^4 + 2x^2y^2 + y^4$$, where $$R = \big\{(r,\theta)\,|\,3 \leq r \leq 4, \space \frac{\pi}{3} \leq \theta \leq \frac{2\pi}{3}\big\}$$.

$$\frac{3367\pi}{18}$$

22)  $$f(x,y) = \sin (\arctan \frac{y}{x})$$, where $$R = \big\{(r, \theta)\,|\,1 \leq r \leq 2, \space \frac{\pi}{6} \leq \theta \leq \frac{\pi}{3}\big\}$$

23)  $$f(x,y) = \arctan \left(\frac{y}{x}\right)$$, where $$R = \big\{(r, \theta)\,|\,2 \leq r \leq 3, \space \frac{\pi}{4} \leq \theta \leq \frac{\pi}{3}\big\}$$

$$\frac{35\pi^2}{576}$$

24)  $$\displaystyle \iint_R e^{x^2+y^2} \left[1 + 2 \space \arctan \left(\frac{y}{x}\right)\right] \,dA, \space R = \big\{(r,\theta)\,|\,1 \leq r \leq 2, \space \frac{\pi}{6} \leq \theta \frac{\pi}{3}\big\}$$

25)  $$\displaystyle \iint_R \left(e^{x^2+y^2} + x^4 + 2x^2y^2 + y^4 \right) \arctan \left(\frac{y}{x}\right) \,dA, \space R = \big\{(r, \theta )\,|\,1 \leq r \leq 2, \space \frac{\pi}{4} \leq \theta \leq \frac{\pi}{3}\big\}$$

$$\frac{7}{576}\pi^2 (21 - e + e^4)$$

#### Converting Double Integrals to Polar Form

In exercises 26 - 29, 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.

26)  $$\displaystyle \int_1^2 \int_0^x (x^2 + y^2)\,dy \space dx = \int_0^{\frac{\pi}{4}} \int_{\sec \theta}^{2 \space \sec \theta}r^3 \,dr \space d\theta$$

27)  $$\displaystyle \int_2^3 \int_0^x \frac{x}{\sqrt{x^2 + y^2}}\,dy \space dx = \int_0^{\pi/4} \int_0^{\tan \theta \space \sec \theta} \,r \space \cos \theta \space dr \space d\theta$$

$$\frac{5}{4} \ln (3 + 2\sqrt{2})$$

28)  $$\displaystyle \int_0^1 \int_{x^2}^x \frac{1}{\sqrt{x^2 + y^2}}\,dy \space dx = \int_0^{\pi/4} \displaystyle \int_0^{\tan \theta \space \sec \theta} \space dr \space d\theta$$

29)  $$\displaystyle \int_0^1 \int_{x^2}^x \frac{y}{\sqrt{x^2 + y^2}}\,dy \space dx = \int_0^{\pi/4} \displaystyle \int_0^{\tan \theta \space \sec \theta} \,r \space \sin \theta \space dr \space d\theta$$

$$\frac{1}{6}(2 - \sqrt{2})$$

In exercises 30 - 37, draw the region of integration, $$R$$, labeling all limits of integration, convert the integrals to polar coordinates and evaluate them.

30)  $$\displaystyle \int_0^3 \int_0^{\sqrt{9-y^2}}\,dx \space dy$$

31)  $$\displaystyle \int_0^2 \int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}}\,dx \space dy$$

$$\displaystyle \int_0^{\pi} \int_0^2 r^5 \,dr \space d\theta \quad = \quad \frac{32\pi}{3}$$

32)  $$\displaystyle \int_0^1 \int_0^{\sqrt{1-x^2}} (x + y) \space dy \space dx$$

33)  $$\displaystyle \int_0^4 \int_{-\sqrt{16-x^2}}^{\sqrt{16-x^2}} sin (x^2 + y^2) \space dy \space dx$$

$$\displaystyle \int_{-\pi/2}^{\pi/2} \int_0^4 \,r \space \sin (r^2) \space dr \space d\theta \quad = \quad \pi \space \sin^2 8$$

34)  $$\displaystyle \int_0^5 \int_{-\sqrt{25-x^2}}^{\sqrt{25-x^2}}\sqrt{x^2+y^2}\,dy\,dx$$

35)  $$\displaystyle \int_{-4}^4 \int_{-\sqrt{16-y^2}}^{0}(2y-x)\,dx\,dy$$

$$\displaystyle \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \int_0^{4} \big( 2r\sin \theta - r\cos\theta\big) \,r\,dr \space d\theta \quad = \quad \frac{128}{3}$$

36)  $$\displaystyle \int_0^2 \int_{y}^{\sqrt{8-y^2}}(x+y)\,dx\,dy$$

37)  $$\displaystyle \int_{-2}^{-1} \int_{0}^{\sqrt{4-x^2}}(x+5)\,dy\,dx+\int_{-1}^1\int_{\sqrt{1-x^2}}^{\sqrt{4-x^2}}(x+5)\,dy\,dx+\int_1^2\int_0^{\sqrt{4-x^2}}(x+5)\,dy\,dx$$

$$\displaystyle \int_{0}^{\pi} \int_1^{2} \big( r\cos \theta + 5\big) \,r\,dr \space d\theta \quad = \quad \frac{15\pi}{2}$$

38)  Evaluate the integral $$\displaystyle \iint_D r \,dA$$ where $$D$$ is the region bounded by the polar axis and the upper half of the cardioid $$r = 1 + \cos \theta$$.

39)  Find the area of the region $$D$$ bounded by the polar axis and the upper half of the cardioid $$r = 1 + \cos \theta$$.

$$\frac{3\pi}{4}$$

40)  Evaluate the integral $$\displaystyle \iint_D r \,dA,$$ where $$D$$ is the region bounded by the part of the four-leaved rose $$r = \sin 2\theta$$ situated in the first quadrant (see the following figure).

41)  Find the total area of the region enclosed by the four-leaved rose $$r = \sin 2\theta$$ (see the figure in the previous exercise).

$$\frac{\pi}{2}$$

42)  Find the area of the region $$D$$ which is the region bounded by $$y = \sqrt{4 - x^2}$$, $$x = \sqrt{3}$$, $$x = 2$$, and $$y = 0$$.

43)  Find the area of the region $$D$$, which is the region inside the disk $$x^2 + y^2 \leq 4$$ and to the right of the line $$x = 1$$.

$$\frac{1}{3}(4\pi - 3\sqrt{3})$$

44)  Determine the average value of the function $$f(x,y) = x^2 + y^2$$ over the region $$D$$ bounded by the polar curve $$r = \cos 2\theta$$, where $$-\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4}$$ (see the following graph).

45)  Determine the average value of the function $$f(x,y) = \sqrt{x^2 + y^2}$$ over the region $$D$$ bounded by the polar curve $$r = 3\sin 2\theta$$, where $$0 \leq \theta \leq \frac{\pi}{2}$$ (see the following graph).

$$\frac{16}{3\pi}$$

46)  Find the volume of the solid situated in the first octant and bounded by the paraboloid $$z = 1 - 4x^2 - 4y^2$$ and the planes $$x = 0, \space y = 0$$, and $$z = 0$$.

47)  Find the volume of the solid bounded by the paraboloid $$z = 2 - 9x^2 - 9y^2$$ and the plane $$z = 1$$.

$$\frac{\pi}{18}$$

48)

1. Find the volume of the solid $$S_1$$ bounded by the cylinder $$x^2 + y^2 = 1$$ and the planes $$z = 0$$ and $$z = 1$$.
2. Find the volume of the solid $$S_2$$ outside the double cone $$z^2 = x^2 + y^2$$ inside the cylinder $$x^2 + y^2 = 1$$, and above the plane $$z = 0$$.
3. Find the volume of the solid inside the cone $$z^2 = x^2 + y^2$$ and below the plane $$z = 1$$ by subtracting the volumes of the solids $$S_1$$ and $$S_2$$.

49)

1. Find the volume of the solid $$S_1$$ inside the unit sphere $$x^2 + y^2 + z^2 = 1$$ and above the plane $$z = 0$$.
2. Find the volume of the solid $$S_2$$ inside the double cone $$(z - 1)^2 = x^2 + y^2$$ and above the plane $$z = 0$$.
3. Find the volume of the solid outside the double cone $$(z - 1)^2 = x^2 + y^2$$ and inside the sphere $$x^2 + y^2 + z^2 = 1$$.
a. $$\frac{2\pi}{3}$$; b. $$\frac{\pi}{2}$$; c. $$\frac{\pi}{6}$$

In Exercises 50-51, special double integrals are presented that are especially well suited for evaluation in polar coordinates.

50)   The surface of a right circular cone with height $$h$$ and base radius $$a$$ can be described by the equation $$f(x,y)=h-h\sqrt{\frac{x^2}{a^2}+\frac{y^2}{a^2}}$$, where the tip of the cone lies at $$(0,0,h)$$ and the circular base lies in the $$xy$$-plane, centered at the origin.
Confirm that the volume of a right circular cone with height h and base radius a is $$V=\frac{1}{3}\pi a^2h$$ by evaluating $$\displaystyle \int\int_R f(x,y)\,dA$$ in polar coordinates.

51)  Consider $$\displaystyle \int\int_R e^{-(x^2+y^2)}\,dA.$$
(a) Why is this integral difficult to evaluate in rectangular coordinates, regardless of the region $$R$$?
(b) Let $$R$$ be the region bounded by the circle of radius a centered at the origin. Evaluate the double integral using polar coordinates.
(c) Take the limit of your answer from (b), as $$a\to \infty$$. What does this imply about the volume under the surface of $$e^{-(x^2+y^2)}$$ over the entire $$xy)-plane? 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). 52) If the sphere has radius 4 and the cylinder has radius 2 find the volume of the spherical ring. 53) A cylindrical hole of diameter 6 cm is bored through a sphere of radius 5 cm such that the axis of the cylinder passes through the center of the sphere. Find the volume of the resulting spherical ring. Answer: \(\frac{256\pi}{3} \space \text{cm}^3$$

54)  Find the volume of the solid that lies under the double cone $$z^2 = 4x^2 + 4y^2$$, inside the cylinder $$x^2 + y^2 = x$$, and above the plane $$z = 0$$.

55)  Find the volume of the solid that lies under the paraboloid $$z = x^2 + y^2$$, inside the cylinder $$x^2 + y^2 = 1$$ and above the plane $$z = 0$$.

$$\frac{3\pi}{32}$$

56)  Find the volume of the solid that lies under the plane $$x + y + z = 10$$ and above the disk $$x^2 + y^2 = 4x$$.

57)  Find the volume of the solid that lies under the plane $$2x + y + 2z = 8$$ and above the unit disk $$x^2 + y^2 = 1$$.

$$4\pi$$

58)  A radial function $$f$$ 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, $$f (x,y) = g(r)$$, where $$r = \sqrt{x^2 + y^2}$$. Show that if $$f$$ is a continuous radial function, then

$\iint_D f(x,y)dA = (\theta_2 - \theta_1) [G(R_2) - G(R_1)], \space where \space G'(r) = rg(r)$ and $$(x,y) \in D = {(r, \theta)|R_1 \leq r \leq R_2, \space 0 \leq \theta \leq 2\pi}$$, with $$0 \leq R_1 < R_2$$ and $$0 \leq \theta_1 < \theta_2 \leq 2\pi$$.

59)  Use the information from the preceding exercise to calculate the integral $\iint_D (x^2 + y^2)^3 dA,$ where $$D$$ is the unit disk.

$$\frac{\pi}{4}$$

60)  Let $$f(x,y) = \frac{F'(r)}{r}$$ be a continuous radial function defined on the annular region $$D = {(r,\theta)|R_1 \leq r \leq R_2, \space 0 \leq \theta 2\pi}$$, where $$r = \sqrt{x^2 + y^2}$$, $$0 < R_1 < R_2$$, and $$F$$ is a differentiable function.

Show that $$\displaystyle \iint_D f(x,y)\,dA = 2\pi [F(R_2) - F(R_1)].$$

61)  Apply the preceding exercise to calculate the integral $$\displaystyle \iint_D \frac{e^{\sqrt{x^2+y^2}}}{\sqrt{x^2 + y^2}} \,dx \space dy$$ where D is the annular region between the circles of radii 1 and 2 situated in the third quadrant.

$$\frac{1}{2} \pi e(e - 1)$$

62)  Let $$f$$ be a continuous function that can be expressed in polar coordinates as a function of $$\theta$$ only; that is, $$f(x,y) = h(\theta)$$, where $$(x,y) \in D = {(r, \theta)|R_1 \leq r \leq R_2, \space \theta_1 \leq \theta \leq \theta_2}$$, with $$0 \leq R_1 < R_2$$ and $$0 \leq \theta_1 < \theta_2 \leq 2\pi$$.

Show that $$\displaystyle \iint_D f(x,y) \,dA = \frac{1}{2} (R_2^2 - R_1^2) [H(\theta_2) - H(\theta_1)]$$, where H is an antiderivative of $$h$$.

63)  Apply the preceding exercise to calculate the integral $$\displaystyle \iint_D \frac{y^2}{x^2}\,dA,$$ where $$D = \big\{(r, \theta)\,|\, 1 \leq r \leq 2, \space \frac{\pi}{6} \leq \theta \leq \frac{\pi}{3}\big\}.$$

$$\sqrt{3} - \frac{\pi}{4}$$

64)  Let $$f$$ be a continuous function that can be expressed in polar coordinates as a function of $$\theta$$ only; that is $$f(x,y) = g(r)h(\theta)$$, where $$(x,y) \in \big\{(r, \theta )\,|\,R_1 \leq r \leq R_2, \space \theta_1 \leq \theta \leq \theta_2\big\}$$ with $$0 \leq R_1 < R_2$$ and $$0 \leq \theta_1 < \theta_2 \leq 2\pi$$. Show that $\iint_D f(x,y)\,dA = [G(R_2) - G(R_1)] \space [H(\theta_2) - H(\theta_1)],$ where $$G$$ and $$H$$ are antiderivatives of $$g$$ and $$h$$, respectively.

65)  Evaluate $$\displaystyle \iint_D \arctan \left(\frac{y}{x}\right) \sqrt{x^2 + y^2}\,dA,$$ where $$D = \big\{(r,\theta)\,|\, 2 \leq r \leq 3, \space \frac{\pi}{4} \leq \theta \leq \frac{\pi}{3}\big\}$$.

$$\frac{133\pi^3}{864}$$

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

a. Show that the volume of the spherical cap in the figure below is $$\frac{1}{6} \pi h (3a^2 + h^2)$$.

b. A spherical segment is the solid defined by intersecting a sphere with two parallel planes. If the distance between the planes is $$h$$ show that the volume of the spherical segment in the figure below is $$\frac{1}{6}\pi h (3a^2 + 3b^2 + h^2)$$.

67)  In statistics, the joint density for two independent, normally distributed events with a mean $$\mu = 0$$ and a standard distribution $$\sigma$$ is defined by $$p(x,y) = \frac{1}{2\pi\sigma^2} e^{-\frac{x^2+y^2}{2\sigma^2}}$$. Consider $$(X,Y)$$, the Cartesian coordinates of a ball in the resting position after it was released from a position on the z-axis toward the $$xy$$-plane. Assume that the coordinates of the ball are independently normally distributed with a mean $$\mu = 0$$ and a standard deviation of $$\sigma$$ (in feet). The probability that the ball will stop no more than $$a$$ feet from the origin is given by $P[X^2 + Y^2 \leq a^2] = \iint_D p(x,y) dy \space dx,$ where $$D$$ is the disk of radius a centered at the origin. Show that $P[X^2 + Y^2 \leq a^2] = 1 - e^{-a^2/2\sigma^2}.$

68)  The double improper integral $\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2+y^2/2}dy \space dx$ may be defined as the limit value of the double integrals $\iint_D e^{-x^2+y^2/2}dA$ over disks $$D_a$$ of radii a centered at the origin, as a increases without bound; that is,

$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2+y^2/2}dy \space dx = \lim_{a\rightarrow\infty} \iint_{D_a} e^{-x^2+y^2/2}dA.$

Use polar coordinates to show that $\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2+y^2/2}dy \space dx = 2\pi.$

69)  Show that $\int_{-\infty}^{\infty} e^{-x^2/2}dx = \sqrt{2\pi}$ by using the relation

$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2+y^2/2}dy \space dx = \left(\int_{-\infty}^{\infty} e^{-x^2/2}dx \right) \left( \int_{-\infty}^{\infty} e^{-y^2/2}dy \right).$