Solutions to Homework 9

15.3

#3

∫_{W} f dV = ∫_{0}^{1} ∫_{0}^{1} ∫_{0}^{2} ax + by + cz dz dy dx = a + b + 2c

#6 This region is bounded by the xz-plane, xy-plane and the planes x = 1, z − x = 0 and y − x = 0. It looks like a square pyramid with base at x = 1 and the top (origin) shifted over on top of one of the corners.

#10 This region is a trapezoidal prism, with top and bottom given by z = 0 and z = 3. The sides are given by the vertical planes x = 0, y = 0, x = 1 and x + y = 2.

#14 Average value=\frac{1}{vol(V)} ∫_{V} x^{2} + y^{2} + z^{2} dV = \frac{1}{8} ∫_{0}^{2} ∫_{0}^{2} ∫_{0}^{2} x^{2} + y^{2} + z^{2} dx dy dz = 4.

15.5

#2

∫_{π/4}^{3π/4} ∫_{0}^{2} f r dr dθ

#4

∫_{0}^{π/2} ∫_{0}^{1/2} f r dr dθ

#10 The angles of π/4 and π/2 yield the lines y = x and x = 0 and r = 2/ sin θ yields that y = r sin θ = 2, so that the region is a triangle between those lines.

#16 After drawing the region we see that it is a wedge of width π/4 and radius 2, so in polar coordinates:

∫_{0}^{π/4} ∫_{0}^{2} r cos θr sin θ rdr dθ = ∫_{0}^{π/4} 4 cos θ sin θ dθ = ∫_{0}^{π/4} 2 sin 2θ dθ = 1.

#21 a) Total population = ∫_{π/2}^{3π/2} ∫_{1}^{4} δ(r, θ) rdr dθ.

b) δ increases as r increases, eliminating (iii), and also as the x-coordinate decreases; and cos θ is smallest at θ = π, so (i) best fits the description.

c)

∫_{π/2}^{3π/2} ∫_{1}^{4}(4 − r)(2 + cos θ) rdr dθ = 18(π − 1) ≈ 39.

15.6

#6 In cylindrical coordinates:

∫_{0}^{4} ∫_{0}^{π/2} ∫_{0}^{2} δ rdr dθ dz.

#8 In spherical coordinates:

∫_{0}^{π} ∫_{0}^{π} ∫_{2}^{3} δ ρ^{2} sin ϕ dρ dϕ dθ.

#13 In cylindrical coordinates:

V = ∫_{0}^{2} ∫_{π/6}^{π/3} ∫_{0}^{5} 1 rdr dθ dz = … = 25π/6.

#15 In spherical coordinates:

M = ∫_{0}^{π} ∫_{0}^{2π} ∫_{0}^{3}(3 − ρ) ρ^{2} sin ϕ dρ dϕ dθ = … = 27π.