Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2020 / Paper 1 / 12 / b / i

Table of contents
- Solution i

Solution

z > 0

, the elliptic paraboloid has a circular cross-section

0 \leq r \leq z

, of area

π z

. The cylindrical coordinates volume element therefore reduces the volume integral to

K = \int_{0}^{\infty} \int_{0}^{2 π} \int_{0}^{z} f^{'} (z) r d r d θ d z = π \int_{0}^{\infty} z f^{'} (z) d z .

For the specified

f

, the derivative is

e^{- z} (z^{- 1/2} + \frac{1}{2} z^{- 3/2})

. Thus

z f^{'} (z)

is integrable at zero and decays exponentially at infinity. Apply integration by parts first on a truncated interval and then take the endpoint limits:

K = π [z f (z)]_{0}^{\infty} - π \int_{0}^{\infty} f (z) d z = π \int_{0}^{\infty} \frac{e ^{- z}}{z} d z .

Both boundary terms vanish because

z f (z) = - z e^{- z}

. With

z = u^{2}

, integration by substitution gives the Gaussian integral:

K = 2 π \int_{0}^{\infty} e^{- u^{2}} d u = π^{3/2} .

The integrand is singular at the apex but the resulting improper integral is convergent.

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