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Double Integrals Part 2 (Exercises)

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1) The region D bounded by y=x3, y=x3+1, x=0, and x=1 as given in the following figure.

A region is bounded by y = 1 + x cubed, y = x cubed, x = 0, and x = 1.

a. Classify this region as vertically simple (Type I) or horizontally simple (Type II).

Type:
Type I but not Type II

b. Find the area of the region D.

c. Find the average value of the function f(x,y)=3xy on the region graphed in the previous exercise.

Answer
2720

2) The region D bounded by y=sinx, y=1+sinx, x=0, and x=π2 as given in the following figure.

A region is bounded by y = 1 + sin x, y = sin x, x = 0, and x = pi/2.

a. Classify this region as vertically simple (Type I) or horizontally simple (Type II).

Type:
Type I but not Type II

b. Find the area of the region D.

Answer
π2units2

c. Find the average value of the function f(x,y)=cosx on the region D.

3) The region D bounded by x=y21 and x=1y2 as given in the following figure.

A region is bounded by x = negative 1 + y squared and x = the square root of the quantity (1 minus y squared).

a. Classify this region as vertically simple (Type I) or horizontally simple (Type II).

Type:
Type II but not Type I

b. Find the volume of the solid under the graph of the function f(x,y)=xy+1 and above the region D.

Answer
16(8+3π)units3

4) The region D bounded by y=0, x=10+y, and x=10y as given in the following figure.

A region is bounded by x = negative 10 + y, x = 10 minus y, and y = 0.

a. Classify this region as vertically simple (Type I) or horizontally simple (Type II).

Type:
Type II but not Type I

b. Find the volume of the solid under the graph of the function f(x,y)=x+y and above the region in the figure from the previous exercise.

Answer
10003units3

5) The region D bounded by y=0, x=y1, x=π2 as given in the following figure.

A region is bounded by x = pi/2, y = 0, and x = negative 1 + y.

Classify this region as vertically simple (Type I) or horizontally simple (Type II).

Type:
Type I and Type II

6) The region D bounded by y=0 and y=x21 as given in the following figure.

A region is bounded by y = 0 and y = negative 1 + x squared.

Classify this region as vertically simple (Type I) or horizontally simple (Type II).

Type:
Type I and Type II

7) Let D be the region bounded by the curves of equations y=cosx and y=4x2 and the x-axis. Explain why D is neither of Type I nor II.

Answer
The region D is not of Type I: it does not lie between two vertical lines and the graphs of two continuous functions g1(x) and g2(x). The region is not of Type II: it does not lie between two horizontal lines and the graphs of two continuous functions h1(y) and h2(y).

8) Let D be the region bounded by the curves of equations y=x, y=x and y=2x2. Explain why D is neither of Type I nor II.

In exercises 9 - 14, evaluate the double integral Df(x,y)dA over the region D.

9) f(x,y)=1 and

D={(x,y)|0xπ2, sinxy1+sinx}

Answer
π2

10) f(x,y)=2 and

D={(x,y)|0y1, y1xarccosy}

11) f(x,y)=xy and

D={(x,y)|1y1, y21x1y2}

Answer
0

12) f(x,y)=siny and D is the triangular region with vertices (0,0), (0,3), and (3,0)

13) f(x,y)=x+1 and D is the triangular region with vertices (0,0), (0,2), and (2,2)

Answer
23

14) f(x,y)=2x+4y and

D={(x,y)|0x1, x3yx3+1}

In exercises 15 - 20, evaluate the iterated integrals.

15) 102x+12x(xy+1)dy dx

Answer
4120

16) 303x2x(x+y2)dy dx

17) 21uu21(8uv)dv du

Answer
63

18) e2e2lnu(v+lnu)dv du

19) 1/2014y214y24dx dy

Answer
π

20) 101y21y2(2x+4y3)dx dy

21) Let D be the region bounded by y=1x2, y=4x2, and the x- and y-axes.

a. Show that DxdA=104x21x2x dy dx+214x20x dy dx by dividing the region D into two regions of Type I.

b. Evaluate the integral DxdA.

22) Let D be the region bounded by y=1, y=x, y=lnx, and the x-axis.

a. Show that Dy2dA=012x2xy2dy dx+102x2xy2dy dx by dividing the region D into two regions of Type I, where D={(x,y)|yx,yx, y2x2}.

b. Evaluate the integral Dy2dA.

23) Let D be the region bounded by y=x2, y=x+2, and y=x.

a. Show that DxdA=10yyx dx dy+41yy2x dx dy by dividing the region D into two regions of Type II, where D={(x,y)|yx2, yx, yx+2}.

b. Evaluate the integral DxdA.

Answer
a. Answers may vary;
b. 73

24) The region D bounded by x=0,y=x5+1, and y=3x2 is shown in the following figure. Find the area A(D) of the region D.

A region is bounded by y = 1 + x to the fifth power, y = 3 minus x squared, and x = 0.

25) The region D bounded by y=cosx, y=4+cosx, and x=±π3 is shown in the following figure. Find the area A(D) of the region D.

A region is bounded by y = cos x, y = 4 + cos x, x = negative 1, and x = 1.

Answer
8π3

26) Find the area A(D) of the region D={(x,y)|y1x2,y4x2, y0, x0}.

27) Let D be the region bounded by y=1, y=x, y=lnx, and the x-axis. Find the area A(D) of the region D.

Answer
(e32)units2

28) Find the average value of the function f(x,y)=siny on the triangular region with vertices (0,0), (0,3), and (3,0).

29) Find the average value of the function f(x,y)=x+1 on the triangular region with vertices (0,0), (0,2), and (2,2).

Answer
The average value of f on this triangular region is 13.

In exercises 30 - 33, change the order of integration and evaluate the integral.

30) π/21x+10sinxdydx

31) 101xx1xdydx

Answer
101xx1x dy dx=01y+10x dx dy+101y0x dx dy=13

32) 01y+1y+1y2dx dy

33) 1/21/2y2+1y2+1y dx dy

Answer
1/21/2y2+1y2+1y dx dy=21x21x21y dy dx=0

34) The region D is shown in the following figure. Evaluate the double integral D(x2+y)dA by using the easier order of integration.

A region is bounded by y = negative 4 + x squared and y = 4 minus x squared.

35) The region D is shown in the following figure. Evaluate the double integral D(x2y2)dA by using the easier order of integration.

A region is bounded by y to the fourth power = 1 minus x and y to the fourth power = 1 + x.

Answer
D(x2y2)dA=111y4y41(x2y2)dx dy=4644095

36) Find the volume of the solid under the surface z=2x+y2 and above the region bounded by y=x5 and y=x.

37) Find the volume of the solid under the plane z=3x+y and above the region determined by y=x7 and y=x.

Answer
45units3

38) Find the volume of the solid under the plane z=3x+y and above the region bounded by x=tany, x=tany, and x=1.

39) Find the volume of the solid under the surface z=x3 and above the plane region bounded by x=siny, x=siny, and x=1.

Answer
5π32units3

40) Let g be a positive, increasing, and differentiable function on the interval [a,b]. Show that the volume of the solid under the surface z=g(x) and above the region bounded by y=0, y=g(x), x=a, and x=b is given by 12(g2(b)g2(a)).

41) Let g be a positive, increasing, and differentiable function on the interval [a,b] and let k be a positive real number. Show that the volume of the solid under the surface z=g(x) and above the region bounded by y=g(x), y=g(x)+k, x=a, and x=b is given by k(g(b)g(a)).

42) Find the volume of the solid situated in the first octant and determined by the planes z=2, z=0, x+y=1, x=0, and y=0.

43) Find the volume of the solid situated in the first octant and bounded by the planes x+2y=1, x=0, z=4, and z=0.

Answer
1units3

44) Find the volume of the solid bounded by the planes x+y=1, xy=1, x=0, z=0, and z=10.

45) Find the volume of the solid bounded by the planes x+y=1, xy=1, x+y=1 xy=1, z=1, and z=0

Answer
2units3

46) Let S1 and S2 be the solids situated in the first octant under the planes x+y+z=1 and x+y+2z=1 respectively, and let S be the solid situated between S1, S2, x=0, and y=0.

  1. Find the volume of the solid S1.
  2. Find the volume of the solid S2.
  3. Find the volume of the solid S by subtracting the volumes of the solids S1 and S2.

47) Let S1 and S2 be the solids situated in the first octant under the planes 2x+2y+z=2 and x+y+z=1 respectively, and let S be the solid situated between S1, S2, x=0, and y=0.

  1. Find the volume of the solid S1.
  2. Find the volume of the solid S2.
  3. Find the volume of the solid S by subtracting the volumes of the solids S1 and S2.
Answer
a. 13units3
b. 16units3
c. 16units3

48) Let S1 and S2 be the solids situated in the first octant under the plane x+y+z=2 and under the sphere x2+y2+z2=4, respectively. If the volume of the solid S2 is 4π3 determine the volume of the solid S situated between S1 and S2 by subtracting the volumes of these solids.

49) Let S1 and S2 be the solids situated in the first octant under the plane x+y+z=2 and bounded by the cylinder x2+y2=4, respectively.

  1. Find the volume of the solid S1.
  2. Find the volume of the solid S2.
  3. Find the volume of the solid S situated between S1 and S2 by subtracting the volumes of the solids S1 and S2.
Answer
a. 43units3
b. 2πunits3
c. 6π43units3

50) [T] The following figure shows the region D bounded by the curves y=sinx, x=0, and y=x4. Use a graphing calculator or CAS to find the x-coordinates of the intersection points of the curves and to determine the area of the region D. Round your answers to six decimal places.

A region is bounded by y = sin x, y = x to the fourth power, and x = 0.

51) [T] The region D bounded by the curves y=cosx, x=0, and y=x3 is shown in the following figure. Use a graphing calculator or CAS to find the x-coordinates of the intersection points of the curves and to determine the area of the region D. Round your answers to six decimal places.

A region is bounded by y = cos x, y = x cubed, and x = 0.

Answer
0 and 0.865474; A(D)=0.621135units3

52) Suppose that (X,Y) is the outcome of an experiment that must occur in a particular region S in the xy-plane. In this context, the region S is called the sample space of the experiment and X and Y are random variables. If D is a region included in S, then the probability of (X,Y) being in D is defined as P[(X,Y)D]=Dp(x,y)dx dy, where p(x,y) is the joint probability density of the experiment. Here, p(x,y) is a nonnegative function for which Sp(x,y)dx dy=1. Assume that a point (X,Y) is chosen arbitrarily in the square [0,3]×[0,3] with the probability density

p(x,y)=19(x,y)[0,3]×[0,3],

p(x,y)=0 otherwise

Find the probability that the point (X,Y) is inside the unit square and interpret the result.

53) Consider X and Y two random variables of probability densities p1(x) and p2(x), respectively. The random variables X and Y are said to be independent if their joint density function is given by p(x,y)=p1(x)p2(y). At a drive-thru restaurant, customers spend, on average, 3 minutes placing their orders and an additional 5 minutes paying for and picking up their meals. Assume that placing the order and paying for/picking up the meal are two independent events X and Y. If the waiting times are modeled by the exponential probability densities

p1(x)=13ex/3 x0,

p1(x)=0 otherwise

p2(y)=15ey/5 y0

p2(y)=0 otherwise

respectively, the probability that a customer will spend less than 6 minutes in the drive-thru line is given by P[X+Y6]=Dp(x,y)dx dy, where D={(x,y)|x0, y0, x+y6}. Find P[X+Y6] and interpret the result.

Answer
P[X+Y6]=1+32e25e6/50.45; there is a 45% chance that a customer will spend 6 minutes in the drive-thru line.

54) [T] The Reuleaux triangle consists of an equilateral triangle and three regions, each of them bounded by a side of the triangle and an arc of a circle of radius s centered at the opposite vertex of the triangle. Show that the area of the Reuleaux triangle in the following figure of side length s is s22(π3).

An equilateral triangle with additional regions consisting of three arcs of a circle with radius equal to the length of the side of the triangle. These arcs connect two adjacent vertices, and the radius is taken from the opposite vertex.

55) [T] Show that the area of the lunes of Alhazen, the two blue lunes in the following figure, is the same as the area of the right triangle ABC. The outer boundaries of the lunes are semicircles of diameters AB and AC respectively, and the inner boundaries are formed by the circumcircle of the triangle ABC.

A right triangle with points A, B, and C. Point B has the right angle. There are two lunes drawn from A to B and from B to C with outer diameters AB and AC, respectively, and with the inner boundaries formed by the circumcircle of the triangle ABC.

 


Double Integrals Part 2 (Exercises) is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by LibreTexts.

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