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Given $f(x,y,z) = \sqrt{1+(x^2+y^2+z^2)^{\frac{3}{2}}}$ and $D=\{(x,y,z) : x^2+y^2+z^2 \leq r^2\}$, evaluate $\int\int\int_D f(x,y,z)dxdydz$.

I've thought that spherical coordinates would be the best way to go, then $x=\rho \cos\theta \sin\varphi$, $y=\rho\sin\theta\sin\varphi$ and $z=\rho\cos\varphi$.

Now because of $x^2+y^2+z^2 \leq r^2$ follows $\rho^2 \cos^2\theta\sin^2\varphi + \rho^2\sin^2\theta\sin^2\varphi + \rho^2\cos^2\varphi = \rho^2\sin^2\varphi(\cos^2\theta+\sin^2\theta)+\rho^2\cos^2\varphi = \rho^2 \leq r^2 \implies -r \leq \rho \leq r$.

If both 'dissapear' from the equation above does that mean that $0\leq \theta \leq 2\pi$ and and $0\leq \varphi \leq 2\pi$? Otherwise, how can i get the limits for $\theta$ and $\varphi$?

And the first question doesn't seem likely since if i put the former limits i'd have $\displaystyle\int_{0}^r\int_0^{2\pi}\int_0^{2\pi} \sqrt{1+\rho^\frac{3}{2}}\;\rho^2\sin\varphi\;d\varphi d\theta d\rho = -\displaystyle\int_{0}^r\int_0^{2\pi}\bigg(\sqrt{1+\rho^\frac{3}{2}}\;\rho^2\cos 2\pi - \sqrt{1+\rho^\frac{3}{2}}\;\rho^2\cos 0\bigg) d\theta d\rho = 0$

Another question: does it matter if i evaulate $\int\int\int_D f(\varphi,\theta,d\rho)d\varphi d\theta d\rho$ or $\int\int\int_D f(\varphi,\theta,d\rho)d\theta d\varphi d\rho$?

Cure
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1 Answers1

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I'll just start by going through the steps, rather than going straight to correcting your work.

We'd like to find (I think) $$ \int\int\int_D \sqrt{1+(x^2+y^2+z^2)^{3/2}}dx\,dy\,dz $$ Changing coordinates (using the jacobian $\rho^2\sin\varphi$) gives us $$ \int\int\int_D \sqrt{1+\rho^{3}}\rho^2\sin\varphi\,d\rho\,d\theta\,d\varphi $$ Now, to describe $D$: remember that we can describe all points using some $\rho>0$, $0<\theta<2\pi$, and $0<\varphi<\pi$. Since we want all points such that $\rho<r$, this integral can be written as $$ \int_0^{\pi}\int_0^{2\pi}\int_0^r \rho^2\sin\varphi \sqrt{1+\rho^{3}}\,d\rho\,d\theta\,d\varphi $$ Notice that $\varphi$ goes only from $0$ to $\pi$.

Taking that integral will give you the answer

To your second question, you could change the order of integration and get the same answer (see Fubini's theorem). So,

$$ \int_0^r\int_0^{2\pi}\int_0^{\pi} \rho^2\sin\varphi \sqrt{1+\rho^{3}}\,d\varphi\,d\theta\,d\rho $$

Would give you the same answer.

Ben Grossmann
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  • (Nevermind, I'm an idiot that mentally multiplied by -1 one too many times) – Emily Jul 24 '13 at 23:46
  • @Arkamis I'll put it this way: any point in the top half of the cartesian plane can be described using $0<\varphi<\pi/2$, $0<\theta<2\pi$, and $\rho>0$. Any point in the lower half can be described using $\pi/2<\varphi<\pi$, $0<\theta<2\pi$, and $\rho>0$. If we take $\varphi$ all the way up to $2\pi$, then we take every point twice, doubling the resulting integral. – Ben Grossmann Jul 24 '13 at 23:46
  • Yeah, I was recalling the parametric description in my head, and was thinking about the $\sin \phi \cos \theta$ term, assuming that we'd always be getting $-1 \cdot -1$ there. It's been a long day. – Emily Jul 24 '13 at 23:47