Solutions to Homework 8

15.1

#2 [5] All rectangles are size 1 by 2, so that the lower bound is

2f(0, 0) + 2f(1, 0) + 2f(0, 2) + 2f(1, 2) = 20,

and the upper bound is

2f(1, 2) + 2f(2, 2) + 2f(1, 4) + 2f(2, 4) = 52.

#4 Here all rectangles are size 2 by 2, so that (starting at the lower left corner and continuing along the bottom before going to the next row) the lower bound is

(2 · 2)(75 + 74 + 73 + 72 + 71 + 73 + 72 + 71 + 71 + 70 + 71 + 69

+69 + 69 + 69 + 68 + 66 + 66 + 66 + 67 + 66 + 62 + 61 + 62 + 65) = 6872,

and the upper bound is

(2 · 2)(82 + 78 + 75 + 74 + 73 + 77 + 75 + 74 + 73 + 72 + 74 + 73

+72 + 71 + 71 + 72 + 71 + 69 + 69 + 69 + 70 + 68 + 66 + 66 + 68) = 7208.

To get bounds for the average temperature divide the total area (100), so that our average temperature is somewhere between 68.7 and 72.1 degrees. Averaging those two values we get, that the average temperature in the room should be around 70.4^{o}F.

#9 Here V = ∫_{R} w(x, y) dA and the lower bound is roughly

(10 · 2)(1 + 4 + 8 + 10 + 10 + 8 + 0 + 3 + 4 + 6 + 6 + 4 + 0 + 1
+2 + 3 + 3 + 2 + 0 + 0 + 1 + 1 + 1 + 1) = 1580,

and the upper bound is

(10 · 2)(8 + 13 + 16 + 17 + 17 + 16 + 4 + 8 + 10 + 11 + 11 + 10
+3 + 4 + 6 + 7 + 7 + 6 + 1 + 2 + 3 + 4 + 4 + 3) = 3820.

Averaging these two we get 2700 cubic feet of volume. a) The graph of f looks like the graph of g in the yz-plane slid parallel to itself

forward and backward in x-direction, since the value of x doesn’t affect the value of g. So the graph is a cylinder in the sense of page 21.

b) ∫_{R} f dA = Volume under the surface = (length in x-direction) · (Cross-sectional

area in yz-plane)= (b − a) ∫_{c}^{d} g(y) dy.

15.2

#2 ∫_{−1}^{1} ∫_{0}^{π/3}(5x^{2} + 1) sin 3y dy dx = ∫_{−1}^{1}(5x^{2} + 1)(2/3) dx = 32/9.

#6

∫_{1}^{4} ∫_{(x−1)/3}^{2} f dy dx.

#11 [5]

∫_{1}^{5} ∫_{x}^{2x} sin x dy dx = ∫_{1}^{5} y sin x|_{x}^{2x} dx = ∫_{1}^{5} x sin x dx = (sin x − x cos x)|_{1}^{5}

= sin 5 − 5 cos 5 − sin 1 + cos 1 ≈ −2.68.

The second integration step uses integration by parts (differentiate x, integrate sin x).

#13

∫_{−2}^{0} ∫_{−\sqrt{9−x^{2}}}^{0} 2xy dy dx = ∫_{−2}^{0} −x(9 − x^{2}) dx = ∫_{−2}^{0} x^{3} − 9x dx = 14.

#16 [5] This region lies between x-axis and y = \sqrt{x} and is bounded on the right by x = 9. So reversing limits we get

∫_{0}^{9} ∫_{0}^{√\overline{x}} y sin x^{2} dy dx = ∫_{0}^{9} \frac{1}{2}x sin x^{2} dx = −\frac{cos 81}{4} + \frac{1}{4}.

The second integration step uses substitution (u = x^{2}).

The plane has x-intercept 1/a, y-intercept 1/b and z-intercept 1/c. Hence the Volume is

∫_{0}^{1/a} ∫_{0}^{(1−ax)/b} z dy dx = … = \frac{1}{6abc}.

#27 [5] Say the lengths of the legs of the triangle are a and b. Put the triangle so that the legs are in positive x-direction (a) and in positive y-direction (b). Then the distance of any point to the leg on the y-axis is x and the hypothenuse of the triangle is given by y = −\frac{b}{a}x + b. So the average distance is given by

\frac{1}{A} ∫_{0}^{a} ∫_{0}^{− \frac{b}{a} x+b} x dy dx = \frac{1}{ab/2} \frac{a^{2}b}{6} = \frac{a}{3},

which is exactly 1/3 the length of the leg on the x-axis. Similarly when the roles of x and y are interchanged.

#29 This region is between x-axis and y = ln x from x = 1 to x = e, so that we get (after reversing the order of integration):

∫_{1}^{e} ∫_{0}^{ln x} \frac{x}{ln x} dy dx = ∫_{1}^{e} x dx = \frac{1}{2}(e^{2} − 1).