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

The total volume is therefore

$\int _{0}^{4}\pi ({\sqrt {y}})^{2}\,dy+\int _{4}^{6}\pi (6-y)^{2}\,dy={\color {blue}\pi \int _{0}^{4}y\,dy+\pi \int _{4}^{6}(6-y)^{2}\,dy}.$