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Because of symmetry,we need to compute the volume in the first octant only.Hence,if V is the total volume,then V=8\iint _{R}\int _{{0}}^{{c{\sqrt  {1-({\frac  {x^{2}}{a^{2}}}-{\frac  {y^{2}}{b^{2}}})}}}}dzdydx\, where R is the region bounded by the ellipse {\frac  {x^{2}}{a^{2}}}+{\frac  {y^{2}}{b^{2}}}=1\, in the first quadrant.Therefore,

V=8\int _{0}^{a}\int _{{0}}^{{b{\sqrt  {1-{\frac  {x^{2}}{a^{2}}}}}}}c{\sqrt  {1-{\frac  {x^{2}}{a^{2}}}-{\frac  {y^{2}}{b^{2}}}}}dydx\,

=8c\int _{0}^{a}\int _{{0}}^{{b{\sqrt  {1-{\frac  {x^{2}}{a^{2}}}}}}}{\sqrt  {1-{\frac  {x^{2}}{a^{2}}}-{\frac  {y^{2}}{b^{2}}}}}dydx\,

We shall first evaluate the inner integral,let I_{1}=\int _{{0}}^{{b{\sqrt  {p}}}}{\sqrt  {p-{\frac  {y^{2}}{b^{2}}}}}dy\, where p=1-{\frac  {x^{2}}{a^{2}}}\,

Setting {\frac  {y}{b}}={\sqrt  {p}}\sin \theta \,,we obtain dyb{\sqrt  {p}}\cos \theta d\theta \,,then

I_{1}=\int _{{0}}^{{{\frac  {\pi }{2}}}}({\sqrt  {p}}\cos \theta )b{\sqrt  {p}}\cos \theta d\theta \,

=bp\int _{{0}}^{{{\frac  {\pi }{2}}}}\cos ^{2}\theta d\theta \,

=bp\int _{{0}}^{{{\frac  {\pi }{2}}}}{\frac  {1+\cos 2\theta }{2}}d\theta \,

={\frac  {bp}{2}}[\theta +{\frac  {\sin 2\theta }{2}}]_{{0}}^{{{\frac  {\pi }{2}}}}\,

={\frac  {bp}{2}}{\frac  {\pi }{2}}={\frac  {bp\pi }{4}}\,

We therefore have

V=8c\int _{0}^{a}{\frac  {bp\pi }{4}}dx\,

=2cb\pi \int _{0}^{a}[1-{\frac  {x^{2}}{a^{2}}}]dx\,

=2cb\pi [x-{\frac  {1}{a^{2}}}{\frac  {x^{3}}{3}}]_{0}^{a}\,

=2cb\pi [a-{\frac  {a}{3}}]\,

={\frac  {4}{3}}cab\pi \,

={\frac  {4}{3}}\pi abc\,

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