Science:Math Exam Resources/Courses/MATH101/April 2015/Question 10 (b)/Solution 2

From part (a), we have the volume of solid $S$ as

$V=8\pi \int _{-1}^{1}{\sqrt {1-x^{2}}}dx.$

Let $C_{1}:x^{2}+y^{2}=1$ , be the circle centered at the origin with the radius 1, then for $y>0$, the integrand $y={\sqrt {1-x^{2}}}$ is the semi-circle in the upper half-plane. Therefore, we can interpret the integral $\int _{-1}^{1}{\sqrt {1-x^{2}}}dx$ as the area between the upper part of the circle and x-axis, so that

$\int _{-1}^{1}{\sqrt {1-x^{2}}}dx={\frac {1}{2}}\cdot ({\text{area of the circle }}C_{1})={\frac {1}{2}}\pi .$

Plugging this into the volume formula, we have $V=\color {blue}4\pi ^{2}$.