Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2010 / Paper 2 / 18

Split the integrand into its three coordinate contributions. Each integrates independently over the rectangular box:

I = (\frac{2}{3}) (4) (6) + (2) (\frac{16}{3}) (6) + (2) (4) (18) = 16 + 64 + 144 = 224 .

For example, the last factor is

\int_{- 3}^{3} z^{2} d z = 18

. No spherical or cylindrical Jacobian is needed for this Cartesian integral.

b

The region is the triangle

0 \leq y \leq x \leq 1

. Reversing the integration order makes the inner integral elementary:

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

The integrand is continuous on this compact triangle, so Fubini theorem justifies exchanging the order.

9 - r^{2} = r^{2}

r = 3/ 2

and

z = 9/2

. The solid has

0 \leq r \leq 3/ 2

0 \leq θ \leq 2 π

, and

r^{2} \leq z \leq 9 - r^{2}

r d z d r d θ

V = 2 π \int_{0}^{3/ 2} (9 - 2 r^{2}) r d r = 2 π [\frac{9}{2} r^{2} - \frac{1}{2} r^{4}]_{0}^{3/ 2} = \frac{81 π}{4} .

The upper-minus-lower height is nonnegative throughout that radial interval.

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