# MvCalc49

Let the center of the sphere be at the origin and let the axis of the hole be along the z-axis. The volume V of the sphere is ${\frac {4}{3}}\pi a^{3}\,$ and that of the circular hole is obtained as follows.
Volume of the upper half of the hole=$\iint _{R}f(x,y)dxdy=\iint _{R}zdxdy\,$ where z is obtained from the equation $x^{2}+y^{2}+z^{2}=a^{2}\,$ and R is the circle in the XY-plane,that is $x^{2}+y^{2}=b^{2}\,$.Hence,the volume V1 of the circular hole is
$V_{1}=2\iint _{R}{\sqrt {a^{2}-x^{2}-y^{2}}}dxdy\,$ where R is given by $x^{2}+y^{2}=b^{2}\,$. Changing into polar coordinates,we obtain,
$v_{1}=2\int _{{0}}^{{2\pi }}\int _{0}^{b}{\sqrt {a^{2}-r^{2}}}rdrd\theta ={\frac {4\pi }{3}}[a^{3}-(a^{2}-b^{2})^{{{\frac {3}{2}}}}]\,$
Hence the remaining portion =$V-V_{1}={\frac {4\pi }{3}}a^{3}-{\frac {4\pi }{3}}[a^{3}-(a^{2}-b^{2})^{{{\frac {3}{2}}}}]={\frac {4\pi }{3}}(a^{2}-b^{2})^{{{\frac {3}{2}}}}\,$