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I'm asked in an exercise to calculate the mass of the ball $x^2+y^2+z^2=1$ with a density of $e^x+e^y+e^z$ at a given point. We've only learned triple integration with Cartesian coordinates so far so I'm trying to set up a triple integral using those. But I get sort of stuck in figuring out how I want to set up the integral.

My first thought was, I should have one coordinate, say z, go from $-1$ to $1$, y from $-\sqrt{1-z^2}$ to $\sqrt{1-z^2}$ and x from $-\sqrt{1-y^2-z^2}$ to $\sqrt{1-y^2-z^2}$. But the resulting integral turned out to be hard to calculate and the answer seems wrong.

Any tips would be appreciated :). Thanks!

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I think you mean $\le1$ rather than $=1$, if you indeed want a ball. – anon May 21 '12 at 22:09
Haven't you learned yet how to transform an integral from one coordinate system to another using the Jacobian of the transformation? – Américo Tavares May 21 '12 at 22:10
That's /probably/ what I meant, but the exercise itself actually types it as $x^2+y^2+z^2=1$ for whichever reason. I'm guessing it's a typo since they're still calling it a ball though. – ro44 May 21 '12 at 22:11
@AméricoTavares: Nope, not yet. We've only just started studying triple integration. – ro44 May 21 '12 at 22:11
up vote 8 down vote accepted

One way to simplify matters is to note that integration is linear and that the region is symmetric under permutations of $x$, $y$, $z$, so the answer will be $3$ times the integral of $e^x$ over the ball. Slice the ball at a given $x$ and you get a disk of radius $\sqrt{1-x^2}$. So this reduces the computation to $$ 3 \int_{-1}^1 \pi (1-x^2) e^{x}\ dx$$

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Thanks for your assistance! – ro44 May 21 '12 at 22:22

$$M = \int_{V} (\exp(x) + \exp(y) + \exp(z)) dx dy dz = 3 \int_V \exp(x) dxdydz$$ $$\int_V \exp(x) dxdydz = \int_{x=-1}^{x=1} \int_{y=-r(x)}^{y = r(x)} \int_{z=-\sqrt{r(x)^2-y^2}}^{z = \sqrt{r(x)^2 - y^2}} \exp(x) dz dy dx$$ where $r(x) = \sqrt{1-x^2}$ $$\int_V \exp(x) dxdydz = \int_{x=-1}^{x=1} \int_{y=-r(x)}^{y = r(x)} 2 \sqrt{r(x)^2-y^2} \exp(x) dy dx = \int_{x=-1}^{x=1}\exp(x) \pi r(x)^2 dx\\ = \pi\int_{x=-1}^{x=1}\exp(x) (1-x^2) dx = \frac{4 \pi}{e}$$ Hence, the mass is $$\dfrac{12 \pi}{e}$$

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Thanks for your assistance! – ro44 May 21 '12 at 22:22

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