Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2007 / Paper 1 / Section B / 7

Table of contents
- i 7
  - Solution i
- ii 7
  - Solution ii

i

Solution

The region is

0 \leq y \leq 1

,

2 y \leq x \leq 2

. Reversing the order describes the same triangle as

0 \leq x \leq 2

,

0 \leq y \leq x /2

. Since the integrand is continuous on this compact region, Fubini's theorem permits

\int_{0}^{1} \int_{2 y}^{2} e^{x^{2}} d x d y = \int_{0}^{2} \int_{0}^{x /2} e^{x^{2}} d y d x = \frac{1}{2} \int_{0}^{2} x e^{x^{2}} d x = \frac{e ^{4} - 1}{4} .

The new inner integral introduces the factor

x

needed for the change of variables

u = x^{2}

.

ii

Solution

In cylindrical coordinates the spherical cap has

1 < z < 2

,

0 < r < 2 - z^{2}

and

0 < θ < 2 π

. Its mass density is

k z

and the volume element is

r d r d θ d z

. Therefore

M = \int_{1}^{2} \int_{0}^{2 π} \int_{0}^{2 - z^{2}} k zr d r d θ d z = πk \int_{1}^{2} z (2 - z^{2}) d z = \frac{πk}{4} .

This is the mass of a spherical cap with height-proportional density with

R = 2

and

c = 1

.

For its curved surface area, write

z = 2 - r^{2}

with

0 \leq r \leq 1

. The surface integral of one becomes

A = \int_{0}^{2 π} \int_{0}^{1} 1 + (\frac{d z}{d r})^{2} r d r d θ = 2 π 2 \int_{0}^{1} \frac{r}{2 - r ^{2}} d r = 2 π (2 - 2) .

It agrees with the curved area of a spherical cap,

2 π R h

, for

h = 2 - 1

. The planar closing disk is not part of the requested curved surface.

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