Q 47. - Calculus

Question

Use a triple integral with either cylindrical or spherical coordinates to find the volumes of the solids described below:

The region inside both the sphere with equation $x^2+y^2+z^2=4$ and the cylinder with equation $x^2+{\left(y-1\right)}^2=1$.

Step-by-Step Solution

Answer

The volume of solid is $\frac{48\mathrm{\pi }-64}{9}$units.

Step 1: Given Information

The region inside both the sphere is determined by equation $x^2+y^2+z^2=4$ and the cylinder with equation $x^2+{\left(y-1\right)}^2=1$.

Step 2: Simplification and evaluation of limits

The term $x^2+y^2$ appears in both the given equations, therefore use cylindrical coordinates.

In rectangular coordinates, equation of sphere is $x^2+y^2+z^2=4$

In cylindrical coordinates, equation of sphere is $r^2+z^2=4$

$\Rightarrow z=\pm \sqrt{4-r^2}$

In $xy$ plane, above which the lies the surface is given by equation

$x^2+{\left(y-1\right)}^2=1$  (Rectangular coordinates)

$\Rightarrow x^2+y^2=2y$

And

$r^2=2r\sin \theta$   (Cylindrical coordinates)

$\Rightarrow r=0$or $r=2\sin \theta$

The limits are

$-\sqrt{4-r^2}\le z\le \sqrt{4-r^2}, 0\le r\le 2\sin \theta , 0\le \theta \le \mathrm{\pi }$

Step 3: Evaluating the volume

Hence, volume of solid is given by

$V=\int \int {\int }_{E}dV$

$=\int \int {\int }_{E}rdrd\theta dz$

$={\int }_0^{\mathrm{\pi }}{\int }_0^{2\sin \theta }{\int }_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}}rdrd\theta dz$

$={\int }_0^{\mathrm{\pi }}{\int }_0^{2\sin \theta }{\left[z\right]}_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}}rdrd\theta$

Putting limits, we get

$={\int }_0^{\mathrm{\pi }}{\int }_0^{2\sin \theta }r\left(2\sqrt{4-r^2}\right)drd\theta$

Solving second integral

As $\frac{d}{dr}\left(4-r^2\right)=-2r$

$=\left(-1\right){\int }_0^{\mathrm{\pi }}{\int }_0^{2\sin \theta }\left(-2r\right){\left(4-r^2\right)}^{\frac{1}{2}}drd\theta$

$=\left(-1\right){\int }_0^{\mathrm{\pi }}{\left[\frac{{\left(4-r^2\right)}^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}\right]}_0^{2\sin \theta }d\theta$

Putting limits, we get

$=\left(-\frac{2}{3}\right){\int }_0^{\mathrm{\pi }}\left(8{\cos }^3\theta -8\right)d\theta$

$=\left(\frac{16}{3}\right){\int }_0^{\mathrm{\pi }}\left(1-{\cos }^3\theta \right)d\theta$

$=\left(\frac{32}{3}\right){\int }_0^{\frac{\mathrm{\pi }}{2}}\left(1-{\cos }^3\theta \right)d\theta$

Simplifying

$=\left(\frac{32}{3}\right){\int }_0^{\frac{\mathrm{\pi }}{2}}\left(1-\left(1-{\sin }^2\theta \right)\cos \theta \right)d\theta$

$=\left(\frac{32}{3}\right){\int }_0^{\frac{\mathrm{\pi }}{2}}\left(1-\cos \theta +{\sin }^2\theta \cos \theta \right)d\theta$

$=\left(\frac{32}{3}\right){\left(\theta -\sin \theta +\frac{{\sin }^3\theta }{3}\right)}_0^{\frac{\mathrm{\pi }}{2}}$

Solving

$=\frac{32}{3}\left(\frac{\mathrm{\pi }}{2}-\frac{2}{3}\right)$

$=\frac{48\mathrm{\pi }-64}{9}$