MvCalc49

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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}}}}\,

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