MvCalc52

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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})}}dz dy dx\, 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}}dy dx\,

=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}}dy dx\,

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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