 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-v-resit/question-3/c/solution

Start with the solid between the elliptic paraboloid

z = x^{2} + y^{2}

and the upper plane

z = 10 - 2 y

D : x^{2} + (y + 1)^{2} \leq 11,

since their vertical separation is

11 - x^{2} - (y + 1)^{2}

. The second lower bound,

z \geq 2 - 2 x

, removes the part between the elliptic paraboloid and this lower plane wherever

2 - 2 x > x^{2} + y^{2} ⟺ (x + 1)^{2} + y^{2} < 3.

E

. Its centre is distance

2

from the centre of

D

, and

2 + 3 < 11

because

5 + 26 < 11

. Thus

E

lies wholly inside

D

. In particular the lower plane never exceeds the upper plane on

E

, so the subtraction describes exactly the requested solid.

Vol (R) = \iint_{D} (11 - x^{2} - (y + 1)^{2}) d x d y - \iint_{E} (3 - (x + 1)^{2} - y^{2}) d x d y .

Translate polar coordinates to the centre of each disk. A general integral of this form is

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

Using

A = 11

and

A = 3

gives

Vol (R) = \frac{π}{2} (1 1^{2} - 3^{2}) = 56 π .

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