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

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

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

  • $\begingroup$ Hint: $V$ is symmetric about $x=2$. $\endgroup$ – 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.


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!


We first change variables with




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!!

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

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