Answer to Question #167688 in Calculus for Poojashree K

Question

Find the volume of the solid inside the surface r

2 + z

2 = 4 and outside the surface r = 2 cos θ.

Answer 1

Solution.

In cylindrical coordinates, we can express the cylinder as

$r=2\cos{\theta}, 0\leq\theta\leq\pi$

and the sphere as

$r^2+z^2=4.$

Then, $z=\pm \sqrt{4-r^2}.$

So,

$V=\int\limits_0^π\int\limits_0^{2\cos{\theta}}\int\limits_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}}rdzdrd\theta= \newline =\int\limits_0^π\int\limits_0^{2\cos{\theta}}2\int\limits_{0}^{\sqrt{4-r^2}}rdzdrd\theta= \newline =\int\limits_0^π\int\limits_0^{2\cos{\theta}}2r\sqrt{4-r^2}drd\theta= \newline$ $=\int\limits_0^π(-\frac{2}{3}(4-r^2)^{\frac{3}{2}})|_0^{2\cos{\theta}}d\theta= \newline$ $=\frac{2}{3}\int\limits_0^π(8-8\sin^3\theta )d\theta= \newline$ $=\frac{16}{3}\int\limits_0^π(1-\sin^3\theta )d\theta=$ $=\frac{16}{3}\cdot \frac{1}{4}\int\limits_0^π(4-3\sin\theta+\sin{3\theta} )d\theta=$ $=\frac{4}{3}(4\theta+3\cos\theta-\frac{1}{3}\cos{3\theta} )|_0^π=$ $=\frac{4}{3}(4π-3+\frac{1}{3}-3+\frac{1}{3})= \newline =\frac{4}{3}(4π-\frac{16}{3})= \newline =\frac{16π}{3}-\frac{64}{9}.$

Answer. $\frac{16π}{3}-\frac{64}{9}.$