Science:Math Exam Resources/Courses/MATH101/April 2016/Question 08 (b)/Solution 1

We solve for $x$ in both equations, yielding $x=\sqrt{y}$ and $x=6-y$.

The two curves intersect at $y=4$, and so two integrals are needed:

for $y\le 4$ the radius of the disc is $x=\sqrt{y}$, while for $y\ge 4$ the radius of the disc is $x=6-y$.

The total volume is therefore

$\underset{0}{\overset{4}{\int }}\pi (\sqrt{y}{)}^{2}dy+\underset{4}{\overset{6}{\int }}\pi (6-y{)}^{2}dy=\pi \underset{0}{\overset{4}{\int }}ydy+\pi \underset{4}{\overset{6}{\int }}(6-y{)}^{2}dy.$