91Ó°ÊÓ

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

The cylindrical and rectangular coordinates are related as

$r=\sqrt{x^2+y^2},\tan \mathrm{θ}=\frac{y}{x},z=z$

and $x=r\cos \mathrm{θ},y=r\sin \mathrm{θ},z=z$

Rectangular coordinates are $x^2+y^2+z^2=4$and $x^2+(y-1{)}^2=1$

Cylindrical coordinates are $z=Â\pm \sqrt{4-r^2}$and $r=2\sin \mathrm{θ}$

Hence, cylindrical limits are

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

At every point, density is proportional to the square of point distance from $z$axis.

$⇒\mathrm{ÒÏ}=k\left(x^2+y^2\right)=k{r}^2$

Required Mass,

$m={∭}_E\mathrm{ÒÏ}dxdydz$

$m={∭}_{E}k\left(x^2+y^2\right)dxdydz$

$m={∫}_{\mathrm{θ}=0}^{\mathrm{π}}{∫}_{r=0}^{2\sin \mathrm{θ}}{∫}_{z=-\sqrt{4-r^2}}^{\sqrt{4-r^2}}\left(k{r}^2\right)rdzdrd\mathrm{θ}$

$m={∫}_{\mathrm{θ}=0}^{\mathrm{π}}{∫}_{r=0}^{2\sin \mathrm{θ}}\left(k{r}^3\right)\left(2\sqrt{4-r^2}\right)drd\mathrm{θ}$

localid="1653041357266" $m=2k{∫}_{\mathrm{θ}=0}^{\mathrm{π}}{∫}_{r=0}^{2\sin \mathrm{θ}}{r}^3\sqrt{4-r^2}drd\mathrm{θ}$

Simplify further

$m={\left.2k{∫}_{\mathrm{θ}=0}^{\mathrm{π}}\left(\frac{1}{2}\left(\frac{2}{5}{\left(4-r^2\right)}^{5/2}-\frac{8}{3}{\left(4-r^2\right)}^{3/2}\right)\right)\right|}_{r=0}^{2\sin \mathrm{θ}}d\mathrm{θ}$

$m=2k{∫}_{\mathrm{θ}=0}^{\mathrm{π}}\left(\frac{32}{15}\right)\left(\left(3{\cos }^2\mathrm{θ}-5\right)\left|{\cos }^3\mathrm{θ}\right|+2\right)d\mathrm{θ}$

For splitting integral, we can use

$\left|\cos \mathrm{θ}\right|=\left\{\begin{array}{cc}\cos \mathrm{θ}& 0≤\mathrm{θ}≤\mathrm{π}/2\\ -\cos \mathrm{θ}& \mathrm{π}/2≤\mathrm{θ}≤\mathrm{π}\end{array}\right.$

$m=2k\left\{{∫}_{\mathrm{θ}=0}^{\mathrm{π}/2}\left(\frac{32}{15}\right)\left(\left(3{\cos }^2\mathrm{θ}-5\right){\cos }^3\mathrm{θ}+2\right)d\mathrm{θ}+{∫}_{\mathrm{θ}=\mathrm{π}/2}^{\mathrm{π}}\left(\frac{32}{15}\right)\left(\left(-3{\cos }^2\mathrm{θ}+5\right){\cos }^3\mathrm{θ}+2\right)d\mathrm{θ}\right\}$$=2k\left\{\left(\frac{32}{225}\right)(15\mathrm{π}-26)+\left(\frac{32}{225}\right)(15\mathrm{π}-26)\right\}$

$=2k\left\{\frac{960\mathrm{Ï€}-1664}{225}\right\}$

Hence,$m=k\left(\frac{1920\mathrm{Ï€}-3328}{225}\right)$