Past exam of the mathematical methods II course of the University of Oxford / 2022 / 7

Table of contents
- Solution 7

Solution

The two planes bound

z - 1 \leq y \leq 1 - z

, and the elliptical cross-section has

- 1 \leq z \leq 1

. Under

x = 2 r cos θ

,

z = r sin θ

, the absolute Jacobian determinant for

(x, z)

is

2 r

. Therefore

V = \int_{0}^{2 π} \int_{0}^{1} \int_{r s i n θ - 1}^{1 - r s i n θ} 2 r d y d r d θ = \int_{0}^{2 π} \int_{0}^{1} 4 r (1 - r sin θ) d r d θ = 4 π .

As a check, the cross-sectional ellipse has area

2 π

, while the part of the thickness proportional to

z

integrates to zero by reflection. The mean thickness is therefore two.

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