Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 3 / 10B (Vector Calculus)

Table of contents
- a 10B
  - Solution a
- b 10B
  - Solution b
- c 10B
  - Solution c

a

Solution

Let

I = \int_{0}^{\infty} e^{- a x^{2}} d x .

Then

I^{2} = \int_{0}^{\infty} \int_{0}^{\infty} e^{- a (x^{2} + y^{2})} d x d y .

Using polar coordinates on the first quadrant,

I^{2} = \int_{0}^{π /2} \int_{0}^{\infty} e^{- a r^{2}} r d r d θ = \frac{π}{4 a} .

Since

I > 0

, the Gaussian integral is

\int_{0}^{\infty} e^{- a x^{2}} d x = \frac{π}{4 a} .

b

Solution

For fixed

x > 0

,

x \neq = 1

,

\int_{0}^{1} x^{y} d y = \frac{x - 1}{l o g x},

with limiting value one at

x = 1

. Integrating over

0 < x < 1

in the opposite order gives

\int_{0}^{1} \int_{0}^{1} x^{y} d y d x = \int_{0}^{1} \frac{d y}{y + 1} = lo g 2.

On the other hand, after setting

x = e^{- u}

,

\int_{0}^{1} \frac{x - 1}{l o g x} d x = \int_{0}^{\infty} \frac{e ^{- u} - e ^{- 2 u}}{u} d u .

Consequently

\int_{0}^{\infty} \frac{e ^{- u} - e ^{- 2 u}}{u} d u = lo g 2 .

c

Solution

In polar coordinates, the circle is

r = 2

, the line

y = 3 x

is

θ = π /3

, and

y = 1

is

r sin θ = 1

. The inequalities select

\frac{π}{6} < θ < \frac{π}{3}, csc θ < r < 2 .

The surface height is

z = \frac{x y}{x ^{2} + y ^{2}} = sin θ cos θ .

Thus the required volume is

V = \int_{π /6}^{π /3} \int_{c s c θ}^{2} sin θ cos θ r d r d θ = \int_{π /6}^{π /3} (2 sin θ cos θ - \frac{1}{2} cot θ) d θ = \frac{1}{2} - \frac{1}{4} lo g 3 .

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