# Double Integrals in Polar Coordinates (Exercises) - Mathematics

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## Terms and Concepts

1. When evaluating (displaystyle intint_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(rcos heta, rsin heta)) and (dA) is replaced with (r,dr,d heta).

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 = ig{(r, heta),|,0 leq r leq 2, space 0 leq heta leq frac{pi}{2}ig})

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 = ig{(r, heta),|,4 leq r leq 5, space frac{pi}{2} leq heta leq piig})

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 = ig{(r, heta),|,0 leq r leq sqrt{2}, space 0 leq heta leq piig})

8) (R = ig{(x,y),|,x^2 + y^2 leq 4xig})

9) (R = ig{(x,y),|,x^2 + y^2 leq 4yig})

(R = ig{(r, heta),|,0 leq r leq 4 space sin heta, space 0 leq heta leq piig})

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

10)

11)

(D = ig{(r, heta),|, 3 leq r leq 5, space frac{pi}{4} leq heta leq frac{pi}{2}ig})

12)

13)

(D = ig{(r, heta),|,3 leq r leq 5, space frac{3pi}{4} leq heta leq frac{5pi}{4}ig})

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 = ig{(r, heta),|,0 leq r leq an heta space sec heta, space 0 leq heta leq frac{pi}{4}ig})

## 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 = ig{(r, heta),|,3 leq r leq 5, space 0 leq heta leq 2piig})

17) (f(x,y) = x + y), (R = ig{(r, heta),|,3 leq r leq 5, space 0 leq heta leq 2piig})

(0)

18) (f(x,y) = x^2 + xy, space R = ig{(r, heta ),|,1 leq r leq 2, space pi leq heta leq 2piig})

19) (f(x,y) = x^4 + y^4, space R = ig{(r, heta),|,1 leq r leq 2, space frac{3pi}{2} leq heta leq 2piig})

(frac{63pi}{16})

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

21) (f(x,y) = x^4 + 2x^2y^2 + y^4), where (R = ig{(r, heta),|,3 leq r leq 4, space frac{pi}{3} leq heta leq frac{2pi}{3}ig}).

(frac{3367pi}{18})

22) (f(x,y) = sin (arctan frac{y}{x})), where (R = ig{(r, heta),|,1 leq r leq 2, space frac{pi}{6} leq heta leq frac{pi}{3}ig})

23) (f(x,y) = arctan left(frac{y}{x} ight)), where (R = ig{(r, heta),|,2 leq r leq 3, space frac{pi}{4} leq heta leq frac{pi}{3}ig})

(frac{35pi^2}{576})

24) (displaystyle iint_R e^{x^2+y^2} left[1 + 2 space arctan left(frac{y}{x} ight) ight] ,dA, space R = ig{(r, heta),|,1 leq r leq 2, space frac{pi}{6} leq heta frac{pi}{3}ig})

25) (displaystyle iint_R left(e^{x^2+y^2} + x^4 + 2x^2y^2 + y^4 ight) arctan left(frac{y}{x} ight) ,dA, space R = ig{(r, heta ),|,1 leq r leq 2, space frac{pi}{4} leq heta leq frac{pi}{3}ig})

(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 heta}^{2 space sec heta}r^3 ,dr space d heta)

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

(frac{5}{4} ln (3 + 2sqrt{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^{ an heta space sec heta} space dr space d heta)

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^{ an heta space sec heta} ,r space sin heta space dr space d heta)

(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 heta quad = quad frac{32pi}{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 heta 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{3pi}{2}} int_0^{4} ig( 2rsin heta - rcos hetaig) ,r,dr space d heta 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}^1int_{sqrt{1-x^2}}^{sqrt{4-x^2}}(x+5),dy,dx+int_1^2int_0^{sqrt{4-x^2}}(x+5),dy,dx)

(displaystyle int_{0}^{pi} int_1^{2} ig( rcos heta + 5ig) ,r,dr space d heta quad = quad frac{15pi}{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 heta).

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

(frac{3pi}{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 heta) 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 heta) (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}(4pi - 3sqrt{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 heta), where (-frac{pi}{4} leq heta 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 = 3sin 2 heta), where (0 leq heta leq frac{pi}{2}) (see the following graph).

(frac{16}{3pi})

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{2pi}{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-hsqrt{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 intint_R f(x,y),dA) in polar coordinates.

51) Consider (displaystyle intint_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 o 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.

(frac{256pi}{3} space ext{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{3pi}{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).

(4pi)

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 = ( heta_2 - heta_1) [G(R_2) - G(R_1)], space where space G'(r) = rg(r)] and ((x,y) in D = {(r, heta)|R_1 leq r leq R_2, space 0 leq heta leq 2pi}), with (0 leq R_1 < R_2) and (0 leq heta_1 < heta_2 leq 2pi).

59) Use the information from the preceding exercise to calculate the integral (displaystyle 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, heta)|R_1 leq r leq R_2, space 0 leq heta 2pi}), 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 = 2pi [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 ( heta) only; that is, (f(x,y) = h( heta)), where ((x,y) in D = {(r, heta)|R_1 leq r leq R_2, space heta_1 leq heta leq heta_2}), with (0 leq R_1 < R_2) and (0 leq heta_1 < heta_2 leq 2pi).

Show that (displaystyle iint_D f(x,y) ,dA = frac{1}{2} (R_2^2 - R_1^2) [H( heta_2) - H( heta_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 = ig{(r, heta),|, 1 leq r leq 2, space frac{pi}{6} leq heta leq frac{pi}{3}ig}.)

(sqrt{3} - frac{pi}{4})

64) Let (f) be a continuous function that can be expressed in polar coordinates as a function of ( heta) only; that is (f(x,y) = g(r)h( heta)), where ((x,y) in ig{(r, heta ),|,R_1 leq r leq R_2, space heta_1 leq heta leq heta_2ig}) with (0 leq R_1 < R_2) and (0 leq heta_1 < heta_2 leq 2pi). Show that [iint_D f(x,y),dA = [G(R_2) - G(R_1)] space [H( heta_2) - H( heta_1)],] where (G) and (H) are antiderivatives of (g) and (h), respectively.

65) Evaluate (displaystyle iint_D arctan left(frac{y}{x} ight) sqrt{x^2 + y^2},dA,) where (D = ig{(r, heta),|, 2 leq r leq 3, space frac{pi}{4} leq heta leq frac{pi}{3}ig}).

(frac{133pi^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}{2pisigma^2} e^{-frac{x^2+y^2}{2sigma^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^2/2sigma^2}.]

68) The double improper integral [int_{-infty}^{infty} int_{-infty}^{infty} e^{-x^2+y^2/2},dy , dx] may be defined as the limit value of the double integrals (displaystyle 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 ightarrowinfty} iint_{D_a} e^{-x^2+y^2/2},dA.]

Use polar coordinates to show that (displaystyle int_{-infty}^{infty} int_{-infty}^{infty} e^{-x^2+y^2/2},dy , dx = 2pi.)

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

[ int_{-infty}^{infty} int_{-infty}^{infty} e^{-x^2+y^2/2},dy ,dx = left(int_{-infty}^{infty} e^{-x^2/2}dx ight) left( int_{-infty}^{infty} e^{-y^2/2}dy ight).]

## Contributors

• Problems 1, 2, 34 - 37 and 50 - 51 are from Apex Calculus, Chapter 13.3
• Edited by Paul Seeburger (Monroe Community College)

## Session 50: Double Integrals in Polar Coordinates

The following images show the chalkboard contents from these video excerpts. Click each image to enlarge.

## Exercise 2 DOUBLE INTEGRAL IN POLAR COORDINAȚES. V1-y² dædy V2/2 V2/2 intergal " dydx + Consider the (i) Use the polar coordinates to combine the integrals into a single double integral. (ii) Evaluate the integral. (Assume -T <0<1)

Use polar coordinates to combine the integrals into one double integral.

help_outline

#### Image Transcriptionclose

Exercise 2 DOUBLE INTEGRAL IN POLAR COORDINAȚES. V1-y² dædy V2/2 V2/2 intergal " dydx + Consider the (i) Use the polar coordinates to combine the integrals into a single double integral. (ii) Evaluate the integral. (Assume -T <0<1)

## Multivariable Calculus

###### Objectives

After completing this section you should.

Be able to change coordinates of a double integral between Cartesian and polar coordinates

We now want to explore how to perform (u)-substitution in high dimensions. Let's start with a review from first semester calculus.

###### Review 11.3.1

Consider the integral (dsint_<-1>^4 e^ <-3x>dx ext<.>)

Let (u=-3x ext<.>) Solve for (x) and then compute (dx ext<.>)

Explain why (dsint_<-1>^4 e^ <-3x>dx=int_<3>^<-12>e^u left(-frac<1><3> ight)du ext<.>)

Explain why (dsint_<-1>^4 e^ <-3x>dx=int_<-12>^<3>e^u left|-frac<1><3> ight| du ext<.>)

If the (u)-values are between (-3) and (2 ext<,>) what would the (x)-values be between? How does the length of the (u) interval ([-3,2]) relate to the length of the corresponding (x) interval?

In the exercise above, we used a change of coordinates (u=-3x ext<,>) or (x=-1/3 u ext<.>) By taking derivatives, we found that (dx=-frac<1><3>du ext<.>) The negative means that the orientation of the interval was reversed. The fraction (frac13) tells us that lengths (dx) using (x) coordinates will be (1/3)rd as long as lengths (du) using (u) coordinates. When we write (dx = fracdu ext<,>) the number (frac) is called the Jacobian of (x) with respect to (u ext<.>) The Jacobian tells us how lengths are altered when we change coordinate systems. We now generalize this to polar coordinates. Before we're done with this section, we'll generalize the Jacobian to any change of coordinates.

###### Exercise 11.3.2

Consider the polar change of coordinates (x=rcos heta) and (y=rsin heta ext<,>) which we could just write as

If you need a reminder of how to compute determinants, refer to Section 2.3.1

Compute the derivative (Dvec T(r, heta) ext<.>) You should have a 2 by 2 matrix.

We need a single number from this matrix that tells us something about area. Determinants are connected to area.

Compute the determinant of (Dvec T(r, heta)) and simplify.

The determinant you found above is called the Jacobian of the polar coordinate transformation. Let's summarize these results in a theorem.

###### Theorem 11.3.1

If we use the polar coordinate transformation (x=rcos heta, y=rsin heta ext<,>) then we can switch from ((x,y)) coordinates to ((r, heta)) coordinates if we use

Ask me in class to give you an informal picture approach that explains why (dxdy=rdrd heta ext<.>)

The number (|r|) is called the Jacobian of (x) and (y) with respect to (r) and ( heta ext<.>) If we require all bounds for (r) to be nonnegative, we can ignore the absolute value. If (R_) is a region in the (xy) plane that corresponds to the region (R_) in the (r heta) plane (where (rgeq 0)), then we can write

egin iint_<>> f(x,y) dxdy = iint_<>> f(rcos heta,rsin heta) r drd heta. end

### Subsection 11.3.1 Practice Changing Coordinates

We need some practice using this idea. We'll start by describing regions using inequalities on (r) and ( heta ext<.>)

###### Exercise 11.3.3

For each region (R) below, draw the region in the (xy)-plane. Then give a set of inequalities of the form (aleq rleq b, alpha(r)leq heta leq eta(r)) or (alphalt hetalt eta, a( heta)leq rleq b( heta) ext<.>) For example, if the region is the inside of the circle (x^2+y^2=9 ext<,>) then we could write (0leq hetaleq 2pi ext<,>) (0leq rleq 3 ext<.>)

The region (R) is the quarter circle in the first quadrant inside the circle (x^2+y^2=25 ext<.>)

The region (R) is below (y=sqrt<9-x^2> ext<,>) above (y=x ext<,>) and to the right of (x=0 ext<.>)

The region (R) is the triangular region below (y=sqrt 3 x ext<,>) above the (x)-axis, and to the left of (x=1 ext<.>)

###### Exercise 11.3.4

Consider the opening exercise for this unit. We want to find the volume under (f(x,y)=9-x^2-y^2) where (xgeq0) and (zgeq 0 ext<.>) We obtained the integral formula

Write bounds for the region (R) by giving bounds for (r) and ( heta ext<.>)

Rewrite the double integral as an iterated integral using the bounds for (r) and ( heta ext<.>) Don't forget the Jacobian (as (dxdy=rdrd heta)).

Compute the integral in the previous part by hand. [Suggestion: you'll want to simplify (9-x^2-y^2) to (9-r^2) before integrating.]

###### Exercise 11.3.5

Find the centroid of a semicircular disc of radius (a) ((ygeq 0)). Actually compute any integrals.

###### Exercise 11.3.6

try switching coordinate systems to polar coordinates. This will require you to first draw the region of integration, and then then obtain bounds for the region in polar coordinates.

We're now ready to define the Jacobian of any transformation.

### Subsection 11.3.2 Computational Practice

Q: If we need to find the equation of a plane when given a point in 3-dimensions and two planes that in.

A: Click to see the answer

Q: Find the directional derivative of f(x, y) = 3x2 − 2y2 at (−3/ 4, 0) in the direction from P(−3/4, 0.

A: The directional derivative is the scalar product of the derivative of the function to the vector alo.

Q: For Exercises 17 and 18, find the acute angle between the given lines by using vectors parallel to t.

A: Since you have asked multiple questions, we will solve the first question for you. If you want any s.

Q: Show that among all rectangles with an 8-m perimeter, the one with largest area is a square.

A: Click to see the answer

Q: Determine the slope of 3(x2+y2)2=100xy At the point (3,1)

A: Click to see the answer

Q: Inflection points Does ƒ(x) = 2x5 - 10x4 + 20x3 + x + 1have any inflection points? If so, identify t.

A: Click to see the answer

Q: A. Solve and check each linear equation. 6x - 7 3x - 5 5x - 78 %3D a 7 28

A: Click to see the answer

Q: Determine the limit of the sequence or show that the sequence diverges by using the appropriate Limi.

A: Click to see the answer

Q: Explain why the antiderivative y1 = ex+C1 is equivalent to the antiderivative y2 = Cex

## 7.4 Area and Arc Length in Polar Coordinates

We have studied the formulas for area under a curve defined in rectangular coordinates and parametrically defined curves. Now we turn our attention to deriving a formula for the area of a region bounded by a polar curve. Recall that the proof of the Fundamental Theorem of Calculus used the concept of a Riemann sum to approximate the area under a curve by using rectangles. For polar curves we use the Riemann sum again, but the rectangles are replaced by sectors of a circle.

The line segments are connected by arcs of constant radius. This defines sectors whose areas can be calculated by using a geometric formula. The area of each sector is then used to approximate the area between successive line segments. We then sum the areas of the sectors to approximate the total area. This approach gives a Riemann sum approximation for the total area. The formula for the area of a sector of a circle is illustrated in the following figure.

Since the radius of a typical sector in Figure 7.39 is given by r i = f ( θ i ) , r i = f ( θ i ) , the area of the ith sector is given by

## Multivariable Calculus

###### Objectives

After finishing this section, you should.

Be able to change between standard coordinate systems for triple integrals:

Just as we did with polar coordinates in two dimensions, we can compute a Jacobian for any change of coordinates in three dimensions. We will focus on cylindrical and spherical coordinate systems.

Remember that the Jacobian of a transformation is found by first taking the derivative of the transformation, then finding the determinant, and finally computing the absolute value.

###### Exercise 13.2.1

The cylindrical change of coordinates is:

egin xamp =rcos heta, y=rsin heta, z=z ext < or in vector form >amp vec C(r, heta,z) amp = (rcos heta,rsin heta, z) end

The spherical change of coordinates is:

egin xamp = hosinphicos heta, y= hosinphisin heta, z= hocosphi ext < or in vector form >amp vec S( ho,phi, heta) amp = ( hosinphicos heta, hosinphisin heta, hocosphi). end

Verify that the Jacobian of the cylindrical transformation is (dsfrac = |r| ext<.>)

If you want to make sure you don't have to use absolute values, what must you require?

Verify that the Jacobian of the spherical transformation is (dsfrac = | ho^2sinphi| ext<.>)

If you want to make sure you don't have to use absolute values, what must you require?

The previous exercise shows us that, provided we require (rgeq0) and (0leq phileq pi ext<,>) we can write:

egin dV=dxdydz = rdrd heta dz = ho^2sinphi d ho dphi d heta, end

Cylindrical coordinates are extremely useful for problems which involve:

Spherical coordinates are extremely useful for problems which involve:

### Subsection 13.2.1 Using the 3-D Jacobian

###### Exercise 13.2.2

The double cone (z^2=x^2+y^2) has two halves. Each half is called a nappe. Set up an integral in the coordinate system of your choice that would give the volume of the region that is between the (xy) plane and the upper nappe of the double cone (z^2=x^2+y^2 ext<,>) and between the cylinders (x^2+y^2=4) and (x^2+y^2=16 ext<.>) Then evaluate the integral.

###### Exercise 13.2.3

Set up an integral in the coordinate system of your choice that would give the volume of the solid ball that is inside the sphere (a^2=x^2+y^2+z^2 ext<.>) Compute the integral to give a formula for the volume of a sphere of radius (a ext<.>)

###### Exercise 13.2.4

Find the volume of the solid domain (D) in space which is above the cone (z=sqrt) and below the paraboloid (z=6-x^2-y^2 ext<.>) Use cylindrical coordinates to set up and then evaluate your integral.

You'll need to find where the surface intersect, as their intersection will help you determine the appropriate bounds.

For the next several exercises be sure to check that you've correctly swapped bounds by having Sage or WolframAlpha actually compute all of the integrals.

###### Exercise 13.2.5

Consider the region (D) in space that is inside both the sphere (x^2+y^2+z^2=9) and the cylinder (x^2+y^2=4 ext<.>)

Set up an iterated integral in Cartesian (rectangular) coordinates that would give the volume of (D ext<.>)

Start by drawing the region.

Set up an iterated integral in cylindrical coordinates that would give the volume of (D ext<.>)

###### Exercise 13.2.6

Consider the region (D) in space that is both inside the sphere (x^2+y^2+z^2=9) and yet outside the cylinder (x^2+y^2=4 ext<.>)

Set up two iterated integrals in cylindrical coordinates that would give the volume of (D ext<.>)

For the first integral use the order (dzdrd heta ext<.>)

For the second, use the order (d heta dr dz ext<.>)

Set up an iterated integral in spherical coordinates that would give the volume of (D ext<.>)

###### Exercise 13.2.7

The integral (dsint_<0>^int_<0>^<1>int_r>^>rdzdrd heta) represents the volume of solid domain (D) in space. Set up integrals in both rectangular coordinates and spherical coordinates that would give the volume of the exact same region.

###### Exercise 13.2.8

The temperature at each point in space of a solid occupying the region <(D)>, which is the upper portion of the ball of radius 4 centered at the origin, is given by (T(x,y,z) = sin(xy+z) ext<.>) Set up an iterated integral formula that would give the average temperature.

## Double Integrals in Polar Coordinates (Exercises) - Mathematics

Continuation of MATH 140, including most topics from Chapters 6-10 in the same text as the MATH 140 text. Note: Credit will only be given for one of Math 141 and Math 121.

#### Prerequisites

MATH 140 with a C- or better, or Math130 with a B- or better.

#### Topics

COMPLEX NUMBERS AND SERIES:

SIMPSON'S AND TRAPEZOIDAL RULE PROGRAM, FOR:

P-SERIES AND SUMMATION PROGRAM, FOR:

Applications of the Integral

Volume
Length of a curve
Area of a surface
Work
Moments and centers of gravity
Parametrized curves, and lengths of curves given parametrically

Inverse Functions, l'Hôpital's Rule, and Differential Equations

Inverse functions
Exponential and logarithmic functions,
L'Hôpital's rule
Introduction to differential equations

Techniques of Integration

Techniques of integration, including integration by parts, trigonometric
substitutions, and partial fractions
Trapezoidal and Simpson's rules
Improper integrals

Sequences and Series

Sequences and convergence of sequences
Infinite series and convergence tests for series
Taylor polynomials and Taylor series
Complex numbers and series

## Multivariable Calculus Online

Multivariable Calculus Online is adapted from the textbook Calculus: A Modern Approach by Kevin Shirley and Jeff Knisley.

Development of Multivariable Calculus Online was funded in part by National Science Foundation grant DUE-9950600.

Many of the applets and images were prepared with Javaview, an excellent web-based visualization tool which is free for download and can be used with Maple 6 through 9.

Additional applets were prepared with LiveGraphics3d, an excellent complement to Mathematica featuring parameterized graphics.

Sections written in LaTeX2e and converted to html with TTHGold. Final editing was performed with Microsoft Frontpage and Eversoft 1stPage2000.

A special thanks to my wife, Debra, for all her support, suggestions, guidance, and encouragement.

Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author and do not necessarily reflect the
views of the National Science Foundation.

## Examples

### Arctangent

Let's plot the trapezoids for $displaystyle f(x)=frac<1><1 + x^2>$ on $[0,5]$ with $N=10$.

Let's compute the sum of areas of the trapezoids:

and we can compare the trapezoid rule to the value

### Approximate ln(2)

Find a value $N$ which guarantees that the trapezoid rule approximation $T_N(f)$ of the integral

For $f(x) = frac<1>$, we compute $f''(x) = frac<2> leq 2$ for all $x in [1,2]$ therefore the error formula implies

$left| , int_1^2 frac<1> , dx - T_N(f) , ight| leq frac<2> <12N^2>$

Then $E_N^T leq 10^<-8>$ is guaranteed if $frac<1> <6N^2>leq 10^<-8>$ which implies

We need 4083 subintervals to guarantee $E_N^T(f) leq 10^<-8>$. Compute the approximation using our own implementation of the trapezoid rule:

We could also use scipy.integrate.trapz to get the exact same result:

Let's verify that this is within $10^<-6>$:

Success! However, a natural question arises: what is the actual smallest $N$ such that the trapezoid rule gives the estimate of $ln (2)$ to within $10^<-8>$?

### Fresnel Integral

Fresnel integrals are examples of nonelementary integrals: antiderivatives which cannot be written in terms of elementary functions. There are two types of Fresnel integrals:

$S(t) = int_0^t sin(x^2) dx ext C(t) = int_0^t cos(x^2) dx$

Use the trapezoid rule to approximate the Fresnel integral

such that the error is less than $10^<-5>$.

Compute the derivatives of the integrand

$f(x) = sin(x^2) , f'(x) = 2xcos(x^2)$

$f''(x) = 2cos(x^2) - 4x^2sin(x^2) , f'''(x) = -12xsin(x^2) - 8x^3cos(x^2)$

Since $f'''(x) leq 0$ for $x in [0,1]$, we see that $f''(x)$ is decreasing on $[0,1]$. Values of $f''(x)$ at the endpoints of the interval are

Therefore $left| , f''(x) , ight| leq 2.2852793274953065$ for $x in [0,1]$. Use the error bound formula to find a good choice for $N$

Let's compute the integral using the trapezoid rule with $N=138$ subintervals

Therefore the Fresnel integral $S(1)$ is approximately

$S(1) = int_0^1 sin(x^2) , dx approx 0.310273030322$

### Logarithmic Integral

The Eulerian logarithmic integral is another nonelementary integral

Let's compute $Li(10)$ such that the error is less than $10^<-4>$. Compute derivatives of the integrand

Plot $f''(x)$ on the interval $[2,10]$.

Clearly $f''(x)$ is decreasing on $[2,10]$ (and bounded below by 0) therefore the absolute maximum occurs at the left endpoint:

for $x in [2,10]$ and we compute

$frac<(b-a)^3> <12 N^2>K_2 leq 10^ <-4>Rightarrow frac<8^3> <12 N^2>2.021732598829855 leq 10^ <-4>Rightarrow sqrt< frac<8^3 10^4> <12>2.021732598829855> leq N$

Compute the trapzoid rule with $N=929$

Therefore the Eulerian logarithmic integral is

such that the error is less than $10^<-4>$.