Double Integrals Part 1 (Exercises)
- Last updated
- Jun 14, 2019
- Save as PDF
- Page ID
- 21037
( \newcommand{\kernel}{\mathrm{null}\,}\)
In exercises 1 and 2, 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 z=0.
1) f(x,y)=4x+2y+8xy
- Answer
- 27
2) f(x,y)=16x2+y2
In exercises 3 and 4, 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 sub-rectangles of the partition.
3) f(x,y)=sinx−cosy, R=[0,π]×[0,π]
- Answer
- 0
4) f(x,y)=cosx+cosy, R=[0,π]×[0,π2]
5) Use the midpoint rule with m=n=2 to estimate ∬Rf(x,y)dA, where the values of the function f on R=[8,10]×[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 |
- Answer
- 21.3
6) The values of the function f on the rectangle R=[0,2]×[7,9] are given in the following table. Estimate the double integral ∬Rf(x,y)dA by using a Riemann sum with m=n=2. Select the sample points to be the upper right corners of the sub-squares of R.
y0=7 | y1=8 | y2=9 | |
---|---|---|---|
x0=0 | 10.22 | 10.21 | 9.85 |
x1=1 | 6.73 | 9.75 | 9.63 |
x2=2 | 5.62 | 7.83 | 8.21 |
7) The depth of a children’s 4-ft by 4-ft swimming pool, measured at 1-ft intervals, is given in the following table.
- 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]×[0,4].
- 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
- Answer
- a. 28 ft3
b. 1.75 ft.
8) The depth of a 3-ft by 3-ft hole in the ground, measured at 1-ft intervals, is given in the following table.
- 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 sub-squares of R.
- 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
9) The level curves f(x,y)=k of the function f are given in the following graph, where k is a constant.
- Apply the midpoint rule with m=n=2 to estimate the double integral ∬Rf(x,y)dA, where R=[0.2,1]×[0,0.8].
- Estimate the average value of the function f on R.
- Answer
- a. Answers will vary somewhat, but should be close to these: f(0.4,0.2)≈0.125, f(0.4,0.6)≈−0.2, f(0.8,0.2)≈0.6, and f(0.8,0.6)≈0.3.
Since the area of each rectangle is ΔAi=(0.4)(0.4)=0.16 units2, ∬Rf(x,y)dA=∫10.2∫0.80f(x,y)dydx≈f(0.4,0.2)(0.16)+f(0.4,0.6)(0.16)+f(0.8,0.2)(0.16)+f(0.8,0.6)(0.16)=0.132
b. fave=∫10.2∫0.80f(x,y)dydxarea of R≈∫10.2∫0.80f(x,y)dydx0.64≈0.206
10) The level curves f(x,y)=k of the function f are given in the following graph, where k is a constant.
- Apply the midpoint rule with m=n=2 to estimate the double integral ∬Rf(x,y)dA, where R=[0.1,0.5]×[0.1,0.5].
- Estimate the average value of the function f on R.
11) The solid lying under the surface z=√4−y2 and above the rectangular regionR=[0,2]×[0,2] is illustrated in the following graph. Evaluate the double integral ∬Rf(x,y), where f(x,y)=√4−y2 by finding the volume of the corresponding solid.
- Answer
- 2π
12) The solid lying under the plane z=y+4 and above the rectangular region R=[0,2]×[0,4] is illustrated in the following graph. Evaluate the double integral ∬Rf(x,y)dA, where f(x,y)=y+4, by finding the volume of the corresponding solid.
In the exercises 13 - 20, calculate the integrals by reversing the order of integration.
13) ∫1−1(∫2−2(2x+3y+5)dx) dy
- Answer
- 40
14) ∫20(∫10(x+2ey+3)dx) dy
15) ∫271(∫21(3√x+3√y)dy) dx
- Answer
- 812+393√2
16) ∫161(∫81(4√x+23√y)dy) dx
17) ∫ln3ln2(∫10ex+ydy) dx
- Answer
- e−1
18) ∫20(∫103x+ydy) dx
19) ∫61(∫92√yx2dy) dx
- Answer
- 15−10√29
20) ∫91(∫24√xy2dy)dx
In exercises 21 - 34, evaluate the iterated integrals by choosing the order of integration.
21) ∫π0∫π/20sin(2x)cos(3y)dx dy
- Answer
- 0
22) ∫π/8π/12∫π/3π/4[cotx+tan(2y)]dx dy
23) ∫e1∫e1[1xsin(lnx)+1ycos(lny)]dx dy
- Answer
- (e−1)(1+sin1−cos1)
24) ∫e1∫e1sin(lnx)cos(lny)xydx dy
25) ∫21∫21(lnyx+x2y+1)dy dx
- Answer
- 34ln(53)+2(ln2)2−ln2
26) ∫e1∫21x2ln(x)dy dx
27) ∫√31∫21y arctan(1x)dy dx
- Answer
- 18[(2√3−3)π+6 ln2]
28) ∫10∫1/20(arcsinx+arcsiny)dy dx
29) ∫10∫21xex+4ydy dx
- Answer
- 14e4(e4−1)
30) ∫21∫10xex−ydy dx
31) ∫e1∫e1(lny√y+lnx√x)dy dx
- Answer
- 4(e−1)(2−√e)
32) ∫e1∫e1(x lny√y+y lnx√x)dy dx
33) ∫10∫21(xx2+y2)dy dx
- Answer
- −π4+ln(54)−12ln2+arctan2
34) ∫10∫21yx+y2dy dx
In exercises 35 - 38, find the average value of the function over the given rectangles.
35)f(x,y)=−x+2y, R=[0,1]×[0,1]
- Answer
- 12
36) f(x,y)=x4+2y3, R=[1,2]×[2,3]
37) f(x,y)=sinhx+sinhy, R=[0,1]×[0,2]
- Answer
- 12(2 cosh1+cosh2−3).
38) f(x,y)=arctan(xy), R=[0,1]×[0,1]
39) Let f and g be two continuous functions such that 0≤m1≤f(x)≤M1 for any x∈[a,b] and 0≤m2≤g(y)≤M2 for anyy∈[c,d]. Show that the following inequality is true:
m1m2(b−a)(c−d)≤∫ba∫dcf(x)g(y)dydx≤M1M2(b−a)(c−d).
In exercises 40 - 43, use property v. of double integrals and the answer from the preceding exercise to show that the following inequalities are true.
40) 1e2≤∬Re−x2−y2 dA≤1, where R=[0,1]×[0,1]
41) π2144≤∬Rsinxcosy dA≤π248, where R=[π6,π3]×[π6,π3]
42) 0≤∬Re−y cosx dA≤π2, where R=[0,π2]×[0,π2]
43) 0≤∬R(lnx)(lny)dA≤(e−1)2, where R=[1,e]×[1,e]
44) Let f and g be two continuous functions such that 0≤m1≤f(x)≤M1 for any x∈[a,b] and 0≤m2≤g(y)≤M2 for any y∈[c,d]. Show that the following inequality is true:
(m1+m2)(b−a)(c−d)≤∫ba∫dc|f(x)+g(y)| dy dx≤(M1+M2)(b−a)(c−d)
In exercises 45 - 48, use property v. of double integrals and the answer from the preceding exercise to show that the following inequalities are true.
45) 2e≤∬R(e−x2+e−y2)dA≤2, where R=[0,1]×[0,1]
46) π236∬R(sinx+cosy)dA≤π2√336, where R=[π6,π3]×[π6,π3]
47) π2e−π/2≤∬R(cosx+e−y)dA≤π, where R=[0,π2]×[0,π2]
48) 1e≤∬R(e−y−lnx)dA≤2, where R=[0,1]×[0,1]
In exercises 49 - 50, the function f is given in terms of double integrals.
- Determine the explicit form of the function f.
- Find the volume of the solid under the surface z=f(x,y) and above the region R.
- Find the average value of the function f on R.
- Use a computer algebra system (CAS) to plot z = f(x,y) and z = f_{ave} in the same system of coordinates.
49) [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]
- Answer
-
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.
50) [T] f(x,y) = \int_0^x \int_0^y [\cos(s) + \cos(t)] \, dt \space ds, where (x,y) \in R = [0,3] \times [0,3]
51) Show that if f and g are continuous on [a,b] and [c,d], respectively, then
\displaystyle \int_a^b \int_c^d |f(x) + g(y)| dy \space dx = (d - c) \int_a^b f(x)\,dx
\displaystyle + \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.
52) Show that \displaystyle \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).
53) [T] Consider the function f(x,y) = e^{-x^2-y^2}, where (x,y) \in R = [−1,1] \times [−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.
- For m = n = 2, find the average value of f over the region R. Round your answer to the nearest hundredths.
- 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}.
- Answer
-
a. For m = n = 2, I = 4e^{-0.5} \approx 2.43
b. f_{ave} = e^{-0.5} \simeq 0.61;c.
54) [T] Consider the function f(x,y) = \sin (x^2) \space \cos (y^2), where (x,y \in R = [−1,1] \times [−1,1].
- Use the midpoint rule with m = n = 2,4,..., 10 to estimate the double integral I = \iint_R \sin (x^2) \cos (y^2) \space dA. Round your answers to the nearest hundredths.
- For m = n = 2, find the average value of fover the region R. Round your answer to the nearest hundredths.
- Use a CAS to graph in the same coordinate system the solid whose volume is given by \iint_R \sin(x^2) \cos(y^2) \space dA and the plane z = f_{ave}.
In exercises 55 - 56, the functions f_n are given, where n \geq 1 is a natural number.
- Find the volume of the solids S_n under the surfaces z = f_n(x,y) and above the region R.
- Determine the limit of the volumes of the solids S_n as n increases without bound.
55) f(x,y) = x^n + y^n + xy, \space (x,y) \in R = [0,1] \times [0,1]
- Answer
- a. \frac{2}{n + 1} + \frac{1}{4}
b. \frac{1}{4}
56) f(x,y) = \frac{1}{x^n} + \frac{1}{y^n}, \space (x,y) \in R = [1,2] \times [1,2]
57) 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.
58) 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). \nonumber
59) 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.
- Answer
- 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.