Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2001 / Paper 2 / 8

0 \leq y \leq 1

y \leq x \leq 2 - y

. It is the closed triangle with vertices

(0, 0)

(2, 0)

, and

(1, 1)

, bounded by

y = 0

x = y

, and

x = 2 - y

. Each horizontal slice runs from the rising left edge to the falling right edge.

Integrate with respect to

x

I = \int_{0}^{1} [\frac{2}{3} x^{3} + y x]_{x = y}^{x = 2 - y} d y = \int_{0}^{1} [\frac{16}{3} - 6 y + 2 y^{2} - \frac{4}{3} y^{3}] d y = \frac{16}{3} - 3 + \frac{2}{3} - \frac{1}{3} = \frac{8}{3} .

The triangle has area one, but the nonconstant integrand makes the double integral a weighted area.

b

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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