Double Integrals Part 2 (Exercises)

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1) The region $D$ bounded by $y = x^3, \space y = x^3 + 1, \space x = 0,$ 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:: $\frac{27}{20}$

2) The region $D$ bounded by $y = \sin x, \space y = 1 + \sin x, \space x = 0$, and $x = \frac{\pi}{2}$ 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:: $\frac{\pi}{2}\, \text{units}^2$

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

3) The region $D$ bounded by $x = y^2 - 1$ and $x = \sqrt{1 - y^2}$ 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:: $\frac{1}{6}(8 + 3\pi)\, \text{units}^3$

4) The region $D$ bounded by $y = 0, \space x = -10 + y,$ and $x = 10 - y$ 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:: $\frac{1000}{3}\, \text{units}^3$

5) The region $D$ bounded by $y = 0, \space x = y - 1, \space x = \frac{\pi}{2}$ 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 $y = x^2 - 1$ 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 $y = \cos x$ and $y = 4 - x^2$ 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 $g_1(x)$ and $g_2(x)$. The region is not of Type II: it does not lie between two horizontal lines and the graphs of two continuous functions $h_1(y)$ and $h_2(y)$.

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

In exercises 9 - 14, evaluate the double integral $\displaystyle \iint_D f(x,y) \,dA$ over the region $D$.

9) $f(x,y) = 1$ and

$D = \big\{(x,y)| \, 0 \leq x \leq \frac{\pi}{2}, \space \sin x \leq y \leq 1 + \sin x \big\}$

Answer:: $\frac{\pi}{2}$

10) $f(x,y) = 2$ and

$D = \big\{(x,y)| \, 0 \leq y \leq 1, \space y - 1 \leq x \leq \arccos y \big\}$

11) $f(x,y) = xy$ and

$D = \big\{(x,y)| \, -1 \leq y \leq 1, \space y^2 - 1 \leq x \leq \sqrt{1 - y^2} \big\}$

Answer:: $0$

12) $f(x,y) = \sin y$ and $D$ is the triangular region with vertices $(0,0), \space (0,3)$, and $(3,0)$

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

Answer:: $\frac{2}{3}$

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

$D = \big\{(x,y)|\, 0 \leq x \leq 1, \space x^3 \leq y \leq x^3 + 1 \big\}$

In exercises 15 - 20, evaluate the iterated integrals.

15) $\displaystyle \int_0^1 \int_{2\sqrt{x}}^{2\sqrt{x}+1} (xy + 1) \,dy \space dx$

Answer:: $\frac{41}{20}$

16) $\displaystyle \int_0^3 \int_{2x}^{3x} (x + y^2) \,dy \space dx$

17) $\displaystyle \int_1^2 \int_{-u^2-1}^{-u} (8 uv) \,dv \space du$

Answer:: $-63$

18) $\displaystyle \int_e^{e^2} \int_{\ln u}^2 (v + \ln u) \,dv \space du$

19) $\displaystyle \int_0^{1/2} \int_{-\sqrt{1-4y^2}}^{\sqrt{1-4y^2}} 4 \,dx \space dy$

Answer:: $\pi$

20) $\displaystyle \int_0^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} (2x + 4y^3) \,dx \space dy$

21) Let $D$ be the region bounded by $y = 1 - x^2, \space y = 4 - x^2$, and the $x$- and $y$-axes.

a. Show that $\displaystyle \iint_D x\,dA = \int_0^1 \int_{1-x^2}^{4-x^2} x \space dy \space dx + \int_1^2 \int_0^{4-x^2} x \space dy \space dx$ by dividing the region $D$ into two regions of Type I.

b. Evaluate the integral $\displaystyle \iint_D x \,dA.$

22) Let $D$ be the region bounded by $y = 1, \space y = x, \space y = \ln x$, and the $x$-axis.

a. Show that $\displaystyle \iint_D y^2 \,dA = \int_{-1}^0 \int_{-x}^{2-x^2} y^2 dy \space dx + \int_0^1 \int_x^{2-x^2} y^2 dy \space dx$ by dividing the region $D$ into two regions of Type I, where $D = \big\{(x,y)\,|\,y \geq x, y \geq -x, \space y \leq 2-x^2\big\}$.

b. Evaluate the integral $\displaystyle \iint_D y^2 \,dA.$

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

a. Show that $\displaystyle \iint_D x \, dA = \int_0^1 \int_{-y}^{\sqrt{y}} x \space dx \space dy + \int_1^4 \int_{y-2}^{\sqrt{y}} x \space dx \space dy$ by dividing the region $D$ into two regions of Type II, where $D = \big\{(x,y)\,|\,y \geq x^2, \space y \geq -x, \space y \leq x + 2\big\}$.

b. Evaluate the integral $\displaystyle \iint_D x \,dA.$

Answer:: a. Answers may vary;
b. $\frac{8}{12}$

24) The region $D$ bounded by $x = 0, y = x^5 + 1$, and $y = 3 - x^2$ 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 $y = \cos x, \space y = 4 + \cos x$, and $x = \pm \frac{\pi}{3}$ 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:: $\frac{8\pi}{3}$

26) Find the area $A(D)$ of the region $D = \big\{(x,y)| \, y \geq 1 - x^2, y \leq 4 - x^2, \space y \geq 0, \space x \geq 0 \big\}$.

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

Answer:: $\left(e - \frac{3}{2}\right)\, \text{units}^2$

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

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

Answer:: The average value of $f$ on this triangular region is $\frac{1}{3}.$

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

30) $\displaystyle \int_{-1}^{\pi/2} \int_0^{x+1} \sin x \, dy \, dx$

31) $\displaystyle \int_0^1 \int_{x-1}^{1-x} x \, dy \, dx$

Answer:: $\displaystyle \int_0^1 \int_{x-1}^{1-x} x \space dy \space dx = \int_{-1}^0 \int_0^{y+1} x \space dx \space dy + \int_0^1 \int_0^{1-y} x \space dx \space dy = \frac{1}{3}$

32) $\displaystyle \int_{-1}^0 \int_{-\sqrt{y+1}}^{\sqrt{y+1}} y^2 dx \space dy$

33) $\displaystyle \int_{-1/2}^{1/2} \int_{-\sqrt{y^2+1}}^{\sqrt{y^2+1}} y \space dx \space dy$

Answer:: $\displaystyle \int_{-1/2}^{1/2} \int_{-\sqrt{y^2+1}}^{\sqrt{y^2+1}} y \space dx \space dy = \int_1^2 \int_{-\sqrt{x^2-1}}^{\sqrt{x^2-1}} y \space dy \space dx = 0$

34) The region $D$ is shown in the following figure. Evaluate the double integral $\displaystyle \iint_D (x^2 + y) \,dA$ 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 $\displaystyle \iint_D (x^2 - y^2) \,dA$ 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:: $\displaystyle \iint_D (x^2 - y^2) dA = \int_{-1}^1 \int_{y^4-1}^{1-y^4} (x^2 - y^2)dx \space dy = \frac{464}{4095}$

36) Find the volume of the solid under the surface $z = 2x + y^2$ and above the region bounded by $y = x^5$ and $y = x$.

37) Find the volume of the solid under the plane $z = 3x + y$ and above the region determined by $y = x^7$ and $y = x$.

Answer:: $\frac{4}{5}\, \text{units}^3$

38) Find the volume of the solid under the plane $z = 3x + y$ and above the region bounded by $x = \tan y, \space x = -\tan y$, and $x = 1$.

39) Find the volume of the solid under the surface $z = x^3$ and above the plane region bounded by $x = \sin y, \space x = -\sin y$, and $x = 1$.

Answer:: $\frac{5\pi}{32}\, \text{units}^3$

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 $y = 0, \space y = g(x), \space x = a$, and $x = b$ is given by $\frac{1}{2}(g^2 (b) - g^2 (a))$.

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 $y = g(x), \space y = g(x) + k, \space x = a$, 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 $z = 2$, $z = 0, \space x + y = 1, \space x = 0$, and $y = 0$.

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

Answer:: $1\, \text{units}^3$

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

45) Find the volume of the solid bounded by the planes $x + y = 1, \space x - y = 1, \space x + y = -1\space x - y = -1, \space z = 1$, and $z = 0$

Answer:: $2\, \text{units}^3$

46) Let $S_1$ and $S_2$ 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 $S_1, \space S_2, \space x = 0$, and $y = 0$.

Find the volume of the solid $S_1$.
Find the volume of the solid $S_2$.
Find the volume of the solid $S$ by subtracting the volumes of the solids $S_1$ and $S_2$.

47) Let $S_1$ and $S_2$ 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 $S_1, \space S_2, \space x = 0$, and $y = 0$.

Find the volume of the solid $S_1$.
Find the volume of the solid $S_2$.
Find the volume of the solid $S$ by subtracting the volumes of the solids $S_1$ and $S_2$.

Answer:: a. $\frac{1}{3}\, \text{units}^3$
b. $\frac{1}{6}\, \text{units}^3$
c. $\frac{1}{6}\, \text{units}^3$

48) Let $S_1$ and $S_2$ be the solids situated in the first octant under the plane $x + y + z = 2$ and under the sphere $x^2 + y^2 + z^2 = 4$, respectively. If the volume of the solid $S_2$ is $\frac{4\pi}{3}$ determine the volume of the solid $S$ situated between $S_1$ and $S_2$ by subtracting the volumes of these solids.

49) Let $S_1$ and $S_2$ be the solids situated in the first octant under the plane $x + y + z = 2$ and bounded by the cylinder $x^2 + y^2 = 4$, respectively.

Find the volume of the solid $S_1$.
Find the volume of the solid $S_2$.
Find the volume of the solid $S$ situated between $S_1$ and $S_2$ by subtracting the volumes of the solids $S_1$ and $S_2$.

Answer:: a. $\frac{4}{3}\, \text{units}^3$
b. $2\pi\, \text{units}^3$
c. $\frac{6\pi - 4}{3}\, \text{units}^3$

50) [T] The following figure shows the region $D$ bounded by the curves $y = \sin x, \space x = 0$, and $y = x^4$. 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 $y = \cos x, \space x = 0$, and $y = x^3$ 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; $A(D) = 0.621135\, \text{units}^3$

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 $P[(X,Y) \in D] = \iint_D p(x,y)dx \space dy$, where $p(x,y)$ is the joint probability density of the experiment. Here, $p(x,y)$ is a nonnegative function for which $\iint_S p(x,y) dx \space dy = 1$. Assume that a point $(X,Y)$ is chosen arbitrarily in the square $[0,3] \times [0,3]$ with the probability density

\[p(x,y) = \frac{1}{9} (x,y) \in [0,3] \times [0,3],\nonumber\]

\[p(x,y) = 0 \space \text{otherwise}\nonumber\]

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 $p_1(x)$ and $p_2(x)$, respectively. The random variables $X$ and $Y$ are said to be independent if their joint density function is given by $p_(x,y) = p_1(x)p_2(y)$. 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

\[p_1(x) = \frac{1}{3}e^{-x/3} \space x\geq 0,\nonumber\]

\[p_1(x) = 0 \space \text{otherwise}\nonumber\]

\[p_2(y) = \frac{1}{5} e^{-y/5} \space y \geq 0\nonumber\]

\[p_2(y) = 0 \space \text{otherwise}\nonumber\]

respectively, the probability that a customer will spend less than 6 minutes in the drive-thru line is given by $P[X + Y \leq 6] = \iint_D p(x,y) dx \space dy$, where $D = {(x,y)|x \geq 0, \space y \geq 0, \space x + y \leq 6}$. Find $P[X + Y \leq 6]$ and interpret the result.

Answer:: $P[X + Y \leq 6] = 1 + \frac{3}{2e^2} - \frac{5}{e^{6/5}} \approx 0.45$; 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 $\frac{s^2}{2}(\pi - \sqrt{3})$.

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

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 $ABC$.

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