# 4.E: Multiple Integration (Exercises)

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## 15.1: Double Integrals over Rectangular Regions

In the following exercises, use the midpoint rule with $$m = 4$$ and $$n = 2$$ to estimate the volume of the solid bounded by the surface $$z = f(x,y)$$, the vertical planes $$x = 1$$, $$x = 2$$, $$y = 1$$, and $$y = 2$$, and the horizontal plane $$x = 0$$.

$$f(x,y) = 4x + 2y + 8xy$$

[Hide Solution]

27.

$$f(x,y) = 16x^2 + \frac{y}{2}$$

In the following exercises, estimate the volume of the solid under the surface $$z = f(x,y)$$ and above the rectangular region R by using a Riemann sum with $$m = n = 2$$ and the sample points to be the lower left corners of the subrectangles of the partition.

$$f(x,y) = sin \space x - cos \space y$$, $$R = [0, \pi] \times [0, \pi]$$

[Hide Solution]

0.

$$f(x,y) = cos \space x + cos \space y$$, $$R = [0, \pi] \times [0, \frac{\pi}{2}]$$

Use the midpoint rule with $$m = n = 2$$ to estimate $$\iint_R f(x,y) dA$$, where the values of the function f on $$R = [8,10] \times [9,11]$$ are given in the following table.

y
x 9 9.5 10 10.5 11
8 9.8 5 6.7 5 5.6
8.5 9.4 4.5 8 5.4 3.4
9 8.7 4.6 6 5.5 3.4
9.5 6.7 6 4.5 5.4 6.7
10 6.8 6.4 5.5 5.7 6.8

[Hide Solution]

21.3.

The values of the function f on the rectangle $$R = [0,2] \times [7,9]$$ are given in the following table. Estimate the double integral $$\iint_R f(x,y)dA$$ by using a Riemann sum with $$m = n = 2$$. Select the sample points to be the upper right corners of the subsquares of R.

$$y_0 = 7$$ $$y_1 = 8$$ $$y_2 = 9$$
$$x_0 = 0$$ 10.22 10.21 9.85
$$x_1 = 1$$ 6.73 9.75 9.63
$$x_2 = 2$$ 5.62 7.83 8.21

The depth of a children’s 4-ft by 4-ft swimming pool, measured at 1-ft intervals, is given in the following table.

1. Estimate the volume of water in the swimming pool by using a Riemann sum with $$m = n = 2$$. Select the sample points using the midpoint rule on $$R = [0,4] \times [0,4]$$.
2. Find the average depth of the swimming pool.
y
x 0 1 2 3 4
0 1 1.5 2 2.5 3
1 1 1.5 2 2.5 3
2 1 1.5 1.5 2.5 3
3 1 1 1.5 2 2.5
4 1 1 1 1.5 2

[Hide Solution]

a. 28 $$ft^3$$ b. 1.75 ft.

The depth of a 3-ft by 3-ft hole in the ground, measured at 1-ft intervals, is given in the following table.

1. Estimate the volume of the hole by using a Riemann sum with $$m = n = 3$$ and the sample points to be the upper left corners of the subsquares of $$R$$.
2. Find the average depth of the hole.
y
x 0 1 2 3
0 6 6.5 6.4 6
1 6.5 7 7.5 6.5
2 6.5 6.7 6.5 6
3 6 6.5 5 5.6

The level curves $$f(X,Y) = K$$ of the function f are given in the following graph, where k is a constant.

1. Apply the midpoint rule with $$M = N = 2$$m=n=2m=n=2 to estimate the double integral $$\iint_R f(x,y)dA$$, where $$R = [0.2,1] \times [0,0.8]$$.
2. Estimate the average value of the function f on $$R$$.

a. 0.112 b. $$f_{ave} ≃ 0.175$$; here $$f(0.4,0.2) ≃ 0.1$$, $$f(0.2,0.6) ≃− 0.2$$, $$f(0.8,0.2) ≃ 0.6$$, and $$f(0.8,0.6) ≃ 0.2$$.

The level curves $$f(x,y) = k$$ of the function f are given in the following graph, where k is a constant.

1. Apply the midpoint rule with $$m = n = 2$$ to estimate the double integral $$\iint_R f(x,y)dA$$, where $$R = [0.1,0.5] \times [0.1,0.5]$$.
2. Estimate the average value of the function f on $$R$$.

The solid lying under the surface $$z = \sqrt{4 - y^2}$$ and above the rectangular region$$R = [0,2] \times [0,2]$$ is illustrated in the following graph. Evaluate the double integral $$\iint_Rf(x,y)$$, where $$f(x,y) = \sqrt{4 - y^2}$$ by finding the volume of the corresponding solid.

[Hide Solution]

$$2\pi$$

The solid lying under the plane $$z = y + 4$$ and above the rectangular region $$R = [0,2] \times [0,4]$$ is illustrated in the following graph. Evaluate the double integral $$\iint_R f(x,y)dA$$, where $$f(x,y) = y + 4$$, by finding the volume of the corresponding solid.

In the following exercises, calculate the integrals by interchanging the order of integration.

$\int_{-1}^1\left(\int_{-2}^2 (2x + 3y + 5)dx \right) \space dy$

[Hide Solution]

40.

$\int_0^2\left(\int_0^1 (x + 2e^y + 3)dx \right) \space dy$

$\int_1^{27}\left(\int_1^2 (\sqrt[3]{x} + \sqrt[3]{y})dy \right) \space dx$

[Hide Solution]

$$\frac{81}{2} + 39\sqrt[3]{2}$$.

$\int_1^{16}\left(\int_1^8 (\sqrt[4]{x} + 2\sqrt[3]{y})dy \right) \space dx$

$\int_{ln \space 2}^{ln \space 3}\left(\int_0^1 e^{x+y}dy \right) \space dx$

[Hide Solution]

$$e - 1$$.

$\int_0^2\left(\int_0^1 3^{x+y}dy \right) \space dx$

$\int_1^6\left(\int_2^9 \frac{\sqrt{y}}{y^2}dy \right) \space dx$

[Hide Solution]

$$15 - \frac{10\sqrt{2}}{9}$$.

$\int_1^9 \left(\int_4^2 \frac{\sqrt{x}}{y^2}dy \right) dx$

In the following exercises, evaluate the iterated integrals by choosing the order of integration.

$\int_0^{\pi} \int_0^{\pi/2} sin(2x)cos(3y)dx \space dy$

[Hide Solution]

0.

$\int_{\pi/12}^{\pi/8}\int_{\pi/4}^{\pi/3} [cot \space x + tan(2y)]dx \space dy$

$\int_1^e \int_1^e \left[\frac{1}{x}sin(ln \space x) + \frac{1}{y}cos (ln \space y)\right] dx \space dy$

[Hide Solution]

$$(e − 1)(1 + sin1 − cos1)$$

$\int_1^e \int_1^e \frac{sin(ln \space x)cos (ln \space y)}{xy} dx \space dy$

$\int_1^2 \int_1^2 \left(\frac{ln \space y}{x} + \frac{x}{2y + 1}\right) dy \space dx$

[Hide solution]

$$\frac{3}{4}ln \left(\frac{5}{3}\right) + 2b \space ln^2 2 - ln \space 2$$

$\int_1^e \int_1^2 x^2 ln(x) dy \space dx$

$\int_1^{\sqrt{3}} \int_1^2 y \space arctan \left(\frac{1}{x}\right) dy \space dx$

[Hide Solution]

$$\frac{1}{8}[(2\sqrt{3} - 3) \pi + 6 \space ln \space 2]$$.

$\int_0^1 \int_0^{1/2} (arcsin \space x + arcsin \space y) dy \space dx$

$\int_0^1 \int_0^2 xe^{x+4y} dy \space dx$

[Hide Solution]

$$\frac{1}{4}e^4 (e^4 - 1)$$.

$\int_1^2 \int_0^1 xe^{x-y} dy \space dx$

$\int_1^e \int_1^e \left(\frac{ln \space y}{\sqrt{y}} + \frac{ln \space x}{\sqrt{x}}\right) dy \space dx$

[Hide Solution]

$$4(e - 1)(2 - \sqrt{e})$$.

$\int_1^e \int_1^e \left(\frac{x \space ln \space y}{\sqrt{y}} + \frac{y \space ln \space x}{\sqrt{x}}\right) dy \space dx$

$\int_0^1 \int_1^2 \left(\frac{x}{x^2 + y^2} \right) dy \space dx$

[Hide Solution]

$$-\frac{\pi}{4} + ln \left(\frac{5}{4}\right) - \frac{1}{2} ln \space 2 + arctan \space 2$$.

$\int_0^1 \int_1^2 \frac{y}{x + y^2} dy \space dx$

In the following exercises, find the average value of the function over the given rectangles.

$$f(x,y) = −x +2y$$, $$R = [0,1] \times [0,1]$$

[Hide Solution]

$$\frac{1}{2}$$.

$$f(x,y) = x^4 + 2y^3$$, $$R = [1,2] \times [2,3]$$

$$f(x,y) = sinh \space x + sinh \space y$$, $$R = [0,1] \times [0,2]$$

[Hide solution]

$$\frac{1}{2}(2 \space cosh \space 1 + cosh \space 2 - 3)$$.

$$f(x,y) = arctan(xy)$$, $$R = [0,1] \times [0,1]$$

Let f and g be two continuous functions such that $$0 \leq m_1 \leq f(x) \leq M_1$$ for any $$x ∈ [a,b]$$ and $$0 \leq m_2 \leq g(y) \leq M_2$$ for any$$y ∈ [c,d]$$. Show that the following inequality is true:

$m_1m_2(b-a)(c-d) \leq \int_a^b \int_c^d f(x) g(y) dy dx \leq M_1M_2 (b-a)(c-d).$

In the following exercises, use property v. of double integrals and the answer from the preceding exercise to show that the following inequalities are true.

$$\frac{1}{e^2} \leq \iint_R e^{-x^2 - y^2} \space dA \leq 1$$, where $$R = [0,1] \times [0,1]$$

$$\frac{\pi^2}{144} \leq \iint_R sin \space x \space cos y \space dA \leq \frac{\pi^2}{48}$$, where $$R = \left[ \frac{\pi}{6}, \frac{\pi}{3}\right] \times \left[ \frac{\pi}{6}, \frac{\pi}{3}\right]$$

$$0 \leq \iint_R e^{-y}\space cos x \space dA \leq \frac{\pi}{2}$$, where $$R = \left[0, \frac{\pi}{2}\right] \times \left[0, \frac{\pi}{2}\right]$$

$$0 \leq \iint_R (ln \space x)(ln \space y) dA \leq (e - 1)^2$$, where $$R = [1, e] \times [1, e]$$

Let f and g be two continuous functions such that $$0 \leq m_1 \leq f(x) \leq M_1$$ for any $$x ∈ [a,b]$$ and $$0 \leq m_2 \leq g(y) \leq M_2$$ for any $$y ∈ [c,d]$$. Show that the following inequality is true:

$$(m_1 + m_2) (b - a)(c - d) \leq \int_a^b \int_c^d |f(x) + g(y)| \space dy \space dx \leq (M_1 + M_2)(b - a)(c - d)$$.

In the following exercises, use property v. of double integrals and the answer from the preceding exercise to show that the following inequalities are true.

$$\frac{2}{e} \leq \iint_R (e^{-x^2} + e^{-y^2}) dA \leq 2$$, where $$R = [0,1] \times [0,1]$$

$$\frac{\pi^2}{36}\iint_R (sin \space x + cos \space y)dA \leq \frac{\pi^2 \sqrt{3}}{36}$$, where $$R = [\frac{\pi}{6}, \frac{\pi}{3}] \times [\frac{\pi}{6}, \frac{\pi}{3}]$$

$$\frac{\pi}{2}e^{-\pi/2} \leq \iint_R (cos \space x + e^{-y})dA \leq \pi$$, where $$R = [0, \frac{\pi}{2}] \times [0, \frac{\pi}{2}]$$

$$\frac{1}{e} \leq \iint_R (e^{-y} - ln \space x) dA \leq 2$$, where $$R = [0, 1] \times [0, 1]$$

In the following exercises, the function f is given in terms of double integrals.

1. Determine the explicit form of the function f.
2. Find the volume of the solid under the surface $$z = f(x,y)$$ and above the region R.
3. Find the average value of the function f on R.
4. Use a computer algebra system (CAS) to plot $$z = f(x,y)$$ and $$z = f_{ave}$$ in the same system of coordinates.

[T] $$f(x,y) = \int_0^y \int_0^x (xs + yt) ds \space dt$$, where $$(x,y) \in R = [0,1] \times [0,1]$$

[Hide Solution]

a. $$f(x,y) = \frac{1}{2} xy (x^2 + y^2)$$; b. $$V = \int_0^1 \int_0^1 f(x,y) dx \space dy = \frac{1}{8}$$; c. $$f_{ave} = \frac{1}{8}$$;

d.

[T] $$f(x,y) = \int_0^x \int_0^y [cos \space (s) + cos \space (t)] dt \space ds$$, where $$(x,y) \in R = [0,3] \times [0,3]$$

Show that if f and g are continuous on $$[a,b]$$ and $$[c,d]$$, respectively, then

$$\int_a^b \int_c^d |f(x) + g(y)| dy \space dx = (d - c) \int_a^b f(x)dx$$

$$+ \int_a^b \int_c^d g(y)dy \space dx = (b - a) \int_c^d g(y)dy + \int_c^d \int_a^b f(x) dx \space dy$$.

Show that $$\int_a^b \int_c^d yf(x) + xg(y) dy \space dx = \frac{1}{2} (d^2 - c^2) \left(\int_a^b f(x)dx\right) + \frac{1}{2} (b^2 - a^2) \left(\int_c^d g(y)dy\right)$$.

[T] Consider the function $$f(x,y) = e^{-x^2-y^2}$$, where $$(x,y) \in R = [−1,1] \times [−1,1]$$.

1. Use the midpoint rule with $$m = n = 2,4,..., 10$$ to estimate the double integral $$I = \iint_R e^{-x^2 - y^2} dA$$. Round your answers to the nearest hundredths.
2. For $$m = n = 2$$, find the average value of f over the region R. Round your answer to the nearest hundredths.
3. Use a CAS to graph in the same coordinate system the solid whose volume is given by $$\iint_R e^{-x^2-y^2} dA$$ and the plane $$z = f_{ave}$$.

[Hide Solution]

a. For $$m = n = 2$$, $$I = 4e^{-0.5} \approx 2.43$$ b. $$f_{ave} = e^{-0.5} \simeq 0.61$$;

c.

[T] Consider the function $$f(x,y) = sin \space (x^2) \space cos \space (y^2)$$, where $$(x,y \in R = [−1,1] \times [−1,1]$$.

1. Use the midpoint rule with $$m = n = 2,4,..., 10$$ to estimate the double integral $$I = \iint_R sin \space (x^2) \space cos \space (y^2) \space dA$$. Round your answers to the nearest hundredths.
2. For $$m = n = 2$$, find the average value of f over the region R. Round your answer to the nearest hundredths.
3. Use a CAS to graph in the same coordinate system the solid whose volume is given by $$\iint_R sin \space (x^2) \space cos \space (y^2) \space dA$$ and the plane $$z = f_{ave}$$.

In the following exercises, the functions fnfn are given, where $$n \geq 1$$ is a natural number.

1. Find the volume of the solids $$S_n$$ under the surfaces $$z = f_n(x,y)$$ and above the region R.
2. Determine the limit of the volumes of the solids $$S_n$$ as n increases without bound.

$$f(x,y) = x^n + y^n + xy, \space (x,y) \in R = [0,1] \times [0,1]$$

[Hide Solution]

a. $$\frac{2}{n + 1} + \frac{1}{4}$$ b. $$\frac{1}{4}$$

$$f(x,y) = \frac{1}{x^n} + \frac{1}{y^n}, \space (x,y) \in R = [1,2] \times [1,2]$$

Show that the average value of a function f on a rectangular region $$R = [a,b] \times [c,d]$$ is $$f_{ave} \approx \frac{1}{mn} \sum_{i=1}^m \sum_{j=1}^n f(x_{ij}^*,y_{ij}^*)$$,where $$(x_{ij}^*,y_{ij}^*)$$ are the sample points of the partition of R, where $$1 \leq i \leq m$$ and $$1 \leq j \leq n$$.

Use the midpoint rule with $$m = n$$ to show that the average value of a function f on a rectangular region $$R = [a,b] \times [c,d]$$ is approximated by

$f_{ave} \approx \frac{1}{n^2} \sum_{i,j =1}^n f \left(\frac{1}{2} (x_{i=1} + x_i), \space \frac{1}{2} (y_{j=1} + y_j)\right).$

An isotherm map is a chart connecting points having the same temperature at a given time for a given period of time. Use the preceding exercise and apply the midpoint rule with $$m = n = 2$$ to find the average temperature over the region given in the following figure.

[Hide Solution]

$$56.5^{\circ}$$ F; here $$f(x_1^*,y_1^*) = 71, \space f(x_2^*, y_1^*) = 72, \space f(x_2^*,y_1^*) = 40, \space f(x_2^*,y_2^*) = 43$$, where $$x_i^*$$ and $$y_j^*$$ are the midpoints of the subintervals of the partitions of [a,b] and [c,d], respectively

## 15.2: Double Integrals over General Regions

In the following exercises, specify whether the region is of Type I or Type II.

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.

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

[Hide Solution]

$$\frac{27}{20}$$

Find the area of the region $$D$$ given in the previous exercise.

The region $$D$$ bounded by $$y = sin \space x, \space y = 1 + sin \space x, \space x = 0$$, and $$x = \frac{\pi}{2}$$ as given in the following figure.

[Hide Solution]

Type I but not Type II

Find the average value of the function $$f(x,y) = cos \space x$$ on the region graphed in the previous exercise.

Find the area of the region $$D$$ given in the previous exercise.

[Hide Solution]

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

The region $$D$$ bounded by $$x = y^2 - 1$$ and $$x = \sqrt{1 - y^2}$$ as given in the following figure.

Find the volume of the solid under the graph of the function $$f(x,y) = xy + 1$$ and above the region in the figure in the previous exercise.

[Hide Solution]

$$\frac{1}{6}(8 + 3\pi)$$

The region $$D$$ bounded by $$y = 0, \space x = -10 + y,$$ and $$x = 10 - y$$ as given in the following figure.

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.

[Hide Solution]

$$\frac{1000}{3}$$

The region $$D$$ bounded by $$y = 0, \space x = y - 1, \space x = \frac{\pi}{2}$$ as given in the following figure.

The region $$D$$ bounded by $$y = 0$$ and $$y = x^2 - 1$$ as given in the following figure.

Type I and Type II

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.

Let $$D$$ be the region bounded by the curves of equations $$y = cos \space x$$ and $$y = 4 - x^2$$ and the $$x$$-axis. Explain why $$D$$ is neither of Type I nor II.

[Hide Solution]

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

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