 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2001/paper-2/8/b/solution

d V = r^{2} sin θ d r d θ d ϕ

. Since

1 + cos θ \geq 0

throughout

0 \leq θ \leq π

, the solid has radial range

0 \leq r \leq 1 + cos θ

and full azimuthal range

0 \leq ϕ < 2 π

. Its volume is

V = \int_{0}^{2 π} \int_{0}^{π} \int_{0}^{1 + c o s θ} r^{2} sin θ d r d θ d ϕ = \frac{2 π}{3} \int_{0}^{π} (1 + cos θ)^{3} sin θ d θ .

Put

u = 1 + cos θ

, so

d u = - sin θ d θ

and the endpoints become

2

and

0

. Consequently

V = \frac{2 π}{3} \int_{0}^{2} u^{3} d u = \frac{8 π}{3} .

The polar equation describes the entire three-dimensional surface, so the spherical polar coordinates volume element, rather than a two-dimensional polar area element, is required.

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