# Triple Integral with difficult limits.

I don't understand how to count integral bounds from the inequality.

$$\iiint\limits_{V}(x-2)dV$$ where

$$V=\left\{(x, y, z):\frac{(x-2)^2}{9}+\frac{(y+3)^2}{25}+\frac{(z+1)^2}{16} < 1\right\}$$

• Hint: $V$ is symmetric about $x=2$. – user99914 Jul 19 '15 at 17:36

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!!

• @anichka Please let me know how I can improve my answer. I really just want to give you the best answer I can. – Mark Viola Jul 22 '15 at 15:24