# Triple integral over a region.

I'm trying to find the volume of $D=\{(x,y,z): \displaystyle\frac{x^2}{a^2} + \displaystyle\frac{y^2}{b^2} + \displaystyle\frac{z^2}{c^2} \leq 1\}$.

I use the change of variables $x=ap\cos\theta\sin\phi$, $y=bp\sin\theta\sin\phi$, $z=cp\cos\phi$ and i have that the region of integration after changing variables should satisfy $p^2\cos^2\theta\sin^2\phi + p^2\sin^2\theta\sin^2\phi + p^2\cos^2\phi$ = $p^2(\sin^2\phi + \cos^2\theta)\leq 1$.

After that i'm stuck, how can i find the new of integration?. I know that the new region after the change of variables will be $E= \{p,\theta,\phi : 0\leq p \leq 1, 0 \leq \theta \leq \pi, 0 \leq \phi \leq 2\pi\}$ but i don't know "why" -and i use "why" because i can see that every point in $E$ satisfies $p^2(\sin^2\phi + \cos^2\theta)\leq 1$, what i don't know how you get the set $E$ in the first place-.

• Check your factorisation: $$p^2\cos^2\theta\sin^2\phi + p^2\sin^2\theta \sin^2\phi + p^2\cos^2\phi = p^2(\sin^2\phi(\cos^2\theta+\sin^2\theta)+\cos^2\phi))=p^2(\sin^2\phi + \cos^2\phi)=p^2$$ – James Jul 8 '13 at 20:20

I would use the change of variables $x = au$, $y = bv$, and $z = cw$. The transformation $\mathbf r(u,v,w) = (x,y,z)$ transforms the unit ball $B$ into $D$ and $dV = abc\, dudvdw$. Thus $$\mathrm{Vol} (D) = \iiint_D \, dxdydz = \iiint_B abc \, dudvdw = \frac{4abc\pi}{3}.$$
Hint: $$p^2\cos^2\theta\sin^2\phi + p^2\sin^2\theta\sin^2\phi + p^2\cos^2\phi = p^2 \sin^2\phi + p^2\cos^2\phi=p^2.$$