Triple Integral with difficult limits. I don't understand how to count integral bounds from the inequality.
$$\iiint\limits_{V}(x-2)dV $$ where
$$V=\left\{(x, y, z):\frac{(x-2)^2}{9}+\frac{(y+3)^2}{25}+\frac{(z+1)^2}{16} < 1\right\}$$
 A: As @JohnMa hinted, we infer from the symmetry about $x-2=0$ that the volume integral is zero.  If one wishes to forgo using the symmetry argument, then we can proceed to evaluate the volume integral directly.  

METHOD 1:
We use Cartesian coordinates and take as the inner integral, the integration over $x$ and write
$$\begin{align}
\int_{2-3\sqrt{1-\left(\frac{y+3}{5}\right)^2-\left(\frac{z+1}{4}\right)^2}}^{2-3\sqrt{1-\left(\frac{y+3}{5}\right)^2+\left(\frac{z+1}{4}\right)^2}} (x-2)\,dx&=\int_{-\sqrt{1-\left(\frac{y+3}{5}\right)^2-\left(\frac{z+1}{4}\right)^2}}^{\sqrt{1-\left(\frac{y+3}{5}\right)^2-\left(\frac{z+1}{4}\right)^2}}9x\,dx\\\\
&=\frac92 \left.x^2\right|_{-\sqrt{1-\left(\frac{y+3}{5}\right)^2-\left(\frac{z+1}{4}\right)^2}}^{\sqrt{1-\left(\frac{y+3}{5}\right)^2-\left(\frac{z+1}{4}\right)^2}}\\\\
&=\frac92\left(1-\left(\frac{y+3}{5}\right)^2-\left(\frac{z+1}{4}\right)^2\right)\\\\
&-\frac92\left(1-\left(\frac{y+3}{5}\right)^2-\left(\frac{z+1}{4}\right)^2\right)\\\\
&=0
\end{align}$$
Thus, the volume integral is zero as expected!

METHOD 2:
We first change variables with 
$$\frac{x-2}{3}=u$$ 
$$\frac{y+3}{5}=v$$ 
$$\frac{z+1}{4}=w$$ 
The Jacobian for the transformation is trivially $60$.  Thus, 
$$\int_V (x-2) dx\,dy\,dz=\int_{V'}(3u)\,60du\,dv\,dw$$
where $V'$ is the spherical region defined by $u^2+v^2+w^2\le1$.  Now, we change coordinates again, this time using a spherical coordinate system with 
$$\begin{align}
u&=R\sin t\,\cos s\\\\
v&=R\sin t\,\sin s\\\\
w&=R\cos t
\end{align}$$
The Jacobian here is $R^2\,\sin t$ and thus
$$\begin{align}
\int_{V'}(3u)\,60du\,dv\,dw&=180\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{1}R \sin t\cos s R^2 \sin t dR\, dt\,ds\\\\
&=\frac{45\pi}{2}\int_{0}^{2\pi}\cos s\,ds\\\\
&=0
\end{align}$$
as expected again!!
