Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2008 / Paper 1 / 14

0 \leq x \leq 2

0 \leq y \leq x^{3}

. Reversing the order in the double integral and then putting

u = x^{4} + 1

gives

\int_{0}^{2} \int_{0}^{x^{3}} x^{4} + 1 d y d x = \int_{0}^{2} x^{3} x^{4} + 1 d x = [\frac{( x ^{4} + 1 ) ^{3/2}}{6}]_{0}^{2} .

Therefore

I = \frac{17 17 - 1}{6} .

b

The base is the disk

D : (x - a)^{2} + y^{2} \leq a^{2}

, on which

x \geq 0

. The vertical thickness is

(p - q) x

. Using the symmetry of the disk about

x = a

, or integrating with

x = a + r cos ϕ

, gives

V = (p - q) \iint_{D} x d A = (p - q) a π a^{2}, V = π a^{3} (p - q) .

For the curved cylindrical surface use

x = a (1 + cos ϕ)

y = a sin ϕ

, with

0 \leq ϕ \leq 2 π

and

q x \leq z \leq p x

a d ϕ d z

. Its surface area is therefore

A = \int_{0}^{2 π} a (p - q) a (1 + cos ϕ) d ϕ, A = 2 π a^{2} (p - q) .

The sloping planar faces are excluded from this curved surface area.

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