VC4.36

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We shall obtain first volume V in the positive quadrant.Clearly,the required volume integral is 8V.We integrate the element of volume dx dy dz first w.r.t z from xy-plane to the surface of the cylinder x^{2}+y^{2}=a^{2}\,,then integrate w.r.t y from x-axis to the circle x^{2}+y^{2}=a^{2}\, in the xy-plane(the limits of y are from 0 to {\sqrt  {(a^{2}-x^{2})}}\,) and finally integrate w.r.t x from 0 to a.

Therefore,the required volume=8\int _{{x=0}}^{{a}}\int _{{y=0}}^{{{\sqrt  {(a^{2}-x^{2})}}}}\int _{{z=0}}^{{{\sqrt  {(a^{2}-x^{2})}}}}dxdydz\,

=8\int _{{x=0}}^{{a}}\int _{{y=0}}^{{{\sqrt  {(a^{2}-x^{2})}}}}[z]_{{0}}^{{{\sqrt  {(a^{2}-x^{2})}}}}dxdy\,

=8\int _{{x=0}}^{{a}}\int _{{y=0}}^{{{\sqrt  {(a^{2}-x^{2})}}}}{\sqrt  {(a^{2}-x^{2})}}dxdy\,

=8\int _{{x=0}}^{{a}}{\sqrt  {(a^{2}-x^{2})}}[y]_{{0}}^{{{\sqrt  {(a^{2}-x^{2})}}}}dx\,

=8\int _{0}^{a}(a^{2}-x^{2})dx=8[a^{2}x-{\frac  {x^{3}}{3}}]_{0}^{a}=8(a^{3}-{\frac  {a^{3}}{3}})={\frac  {16a^{3}}{3}}\,


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