91Ó°ÊÓ

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

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 $r^2+z^2=4$

$⇒z=Â\pm \sqrt{4-r^2}$

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

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

$⇒x^2+y^2=2y$

And

$r^2=2r\sin \mathrm{θ}$ (Cylindrical coordinates)

$⇒r=0$or $r=2\sin \mathrm{θ}$

The limits are

$-\sqrt{4-r^2}≤z≤\sqrt{4-r^2},0≤r≤2\sin \mathrm{θ},0≤\mathrm{θ}≤\mathrm{π}$

Hence, volume of solid is given by

$V=∫∫{∫}_{E}dV$

$=∫∫{∫}_{E}rdrd\mathrm{θ}dz$

role="math" localid="1652295002803" $={∫}_0^{\mathrm{π}}{∫}_0^{2\sin \mathrm{θ}}{∫}_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}}rdrd\mathrm{θ}dz$

role="math" localid="1652295023673" $={∫}_0^{\mathrm{π}}{∫}_0^{2\sin \mathrm{θ}}{\left[z\right]}_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}}rdrd\mathrm{θ}$

Putting limits, we get

$={∫}_0^{\mathrm{π}}{∫}_0^{2\sin \mathrm{θ}}r\left(2\sqrt{4-r^2}\right)drd\mathrm{θ}$

Solving second integral

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

role="math" localid="1652295190867" $=\left(-1\right){∫}_0^{\mathrm{π}}{∫}_0^{2\sin \mathrm{θ}}\left(-2r\right){\left(4-r^2\right)}^{\frac{1}{2}}drd\mathrm{θ}$

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

Putting limits, we get

$=\left(-\frac{2}{3}\right){∫}_0^{\mathrm{π}}\left(8{\cos }^3\mathrm{θ}-8\right)d\mathrm{θ}$

$=\left(\frac{16}{3}\right){∫}_0^{\mathrm{π}}\left(1-{\cos }^3\mathrm{θ}\right)d\mathrm{θ}$

$=\left(\frac{32}{3}\right){∫}_0^{\frac{\mathrm{π}}{2}}\left(1-{\cos }^3\mathrm{θ}\right)d\mathrm{θ}$

Simplifying

$=\left(\frac{32}{3}\right){∫}_0^{\frac{\mathrm{π}}{2}}\left(1-\left(1-{\sin }^2\mathrm{θ}\right)\cos \mathrm{θ}\right)d\mathrm{θ}$

$=\left(\frac{32}{3}\right){∫}_0^{\frac{\mathrm{π}}{2}}\left(1-\cos \mathrm{θ}+{\sin }^2\mathrm{θ}\cos \mathrm{θ}\right)d\mathrm{θ}$

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

Solving

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

$=\frac{48\mathrm{Ï€}-64}{9}$