Science:Math Exam Resources/Courses/MATH101/April 2017/Question 07 (a)/Solution 1

The region $R$ and the reservoir are given on $x$-$y$ plane as follows:

Slicing the reservoir horizontally, for $y$ between $-4$ and $0$, we have a circle cross section at $y$ whose area is is $\pi x^{2}=\pi {\sqrt {y+4}}$.

Combining with the height $dy$ of each slice, we get the volume of the slice at $y$ as $dV=\pi x^{2}dy=\pi {\sqrt {y+4}}dy$.

Summing them up for $y\in (-4,0)$, we obtain the volume $V$ of the reservoir;

${\begin{aligned}V=\int _{-4}^{0}dV=\int _{-4}^{0}\pi {\sqrt {y+4}}dy=\pi \int _{0}^{4}{\sqrt {z}}dz={\frac {2}{3}}\pi \left[({\sqrt {z}})^{3}\right]_{0}^{4}={\frac {2}{3}}\pi \cdot 8={\frac {16}{3}}\pi .\end{aligned}}$

Here, the third equality follows from the substitution $z=y+4$.

Answer: $\color {blue}{\frac {16}{3}}\pi$