cylindrical polar coordinates

(a)

Consider $y=0$ , then $x=r$ , so

$\begin{aligned} \frac{z}{a} &= 1 - \left( \frac{r}{b} \right)^2 \\ \frac{r}{b} &= \sqrt{1-\frac{z}{a}} \end{aligned}$

The polar angle is $\alpha$ , thus the cylindrical polar coordinates are

$\left( b\sqrt{1-\frac{z}{a}} , \alpha, z \right)$

(b)

Let

$r_z = b\sqrt{1-\frac{z}{a}}$

Then

$\begin{aligned} V &=\int dV \\ &= \int_{z=0}^{a} \int_{\theta=0}^{\alpha} \int_{r=0}^{r_z} r dr d\theta dz \\ &= \int_{\theta=0}^{\alpha}d\theta \int_{z=0}^{a} \left[ r^2/2 \right]_0^{r_z} dz \\ &= \frac{\alpha}{2} b^2 \int_{z=0}^{a} 1-\frac{z}{a} dz \\ &=\frac{1}{2}\alpha b^2 \left[ z - \frac{z^2}{2a} \right]_0^{a} \\ &= \frac{1}{4} \, \alpha \, a \, b^2 \end{aligned}$

NST First Year Maths

Some unofficial solutions to Cambridge NST IA maths papers

2016 Paper 2 Question 12