Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2011 / Paper 2 / 18

Table of contents
- a 18
  - Solution a
- b 18
  - Solution b
- c 18
  - Solution c

a

Solution

The integrand separates into a radial and an angular factor, so

\int_{0}^{π /2} \int_{0}^{R} r^{2} cos θ sin θ d r d θ = [\frac{r ^{3}}{3}]_{0}^{R} [\frac{s i n ^{2} θ}{2}]_{0}^{π /2} = \frac{R ^{3}}{6} .

The differential printed in the PDF is already

d r d θ

. No further polar Jacobian is to be inserted into this given iterated integral.

b

Solution

The intersections satisfy

6 x - x^{2} = 2 x + 3

, or

(x - 1) (x - 3) = 0

, so the bounded region has

1 \leq x \leq 3

. Over that interval the parabola is above the line, and their height difference is

(x - 1) (3 - x)

. Thus its area is

A = \int_{1}^{3} (4 x - x^{2} - 3) d x = [2 x^{2} - \frac{x ^{3}}{3} - 3 x]_{1}^{3} = \frac{4}{3} .

Using upper minus lower ensures a positive geometric area.

c

Solution

The projection onto the

x

-

y

plane is the disk

x^{2} + y^{2} \leq 4

. In polar coordinates, the height is

4 - r^{2}

and the polar area element is

r d r d θ

. The volume is therefore

V = \int_{0}^{2 π} \int_{0}^{2} (4 - r^{2}) r d r d θ = 2 π [2 r^{2} - \frac{r ^{4}}{4}]_{0}^{2} = 8 π .

The radial factor here is required because this integral is being constructed from the geometric volume, unlike the already specified integral in (a).

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