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Consider the function $$f(x,y,z)=x^2+4y^2+9z^2$$ find the value of $$\int \int \int_R e^{\sqrt{f(x,y,z)}}dxdydz$$ where $R$ is defined by $f(x,y,z) \leq 16$. First I did a change of variable and let $r=x, \; u=2y, \; v=3z$. After this I changed $f(r,u,v)$ to spherical coordinates and integrated. I got $\frac{2 \pi}{3}(10e^4 -2)$. Can someone confirm this of supply a good answer?

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I get the same result. –  Daniel Fischer Jan 8 '14 at 0:48
Well then I guess I was right, thanks! –  Wintermute Jan 8 '14 at 0:49

1 Answer 1

up vote 0 down vote accepted

If $f(x,y,z)=x^2+4y^2+9z^2$, then $f(u,v,w)=u^2 + v^2 + w^2$ with $u=x, v = 2y, w=3z.$


$$\iiint_R e^{\sqrt{f(x,y,z)}}dxdydz = \frac{1}{6} \iiint_R e^{\sqrt{f(u,v,w)}}dudvdw = \frac{2 \pi}{3}\int_{r=0}^4r^2e^rdr.$$

Integrating by parts twice gives

$$\frac{2 \pi}{3}\int_{r=0}^4r^2e^rdr = \frac{2 \pi}{3}\left.\left(r^2e^r - 2re^r + 2e^r\right)\right|_0^4 = \frac{2 \pi}{3}\left(10e^4-2\right).$$

So ... yup.

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