# Evaluating the integral $\int x\,dV$ where $V$ is the region bounded the surface $x^2+y^2+z^2=1$ and the planes $x = 0$, $y = x$, $z=0$

Evaluate the integral $\int x\,dV$ inside domain $V$, where $V$ is bounded by the planes $x=0$, $y=x$, $z=0$, and the surface $x^2+y^2+z^2=1$.

Answer given: $\dfrac{1}{8} - \dfrac{\sqrt{2}}{16}$

Uh, so I did it in spherical coordinates, which equals

$$\iiint p^2 \sin φ \;dp dφ dθ$$

$∫dp$ runs from $0$ to $1$

$∫dφ$ runs from $0$ to $\frac{\pi}{2}$ (right??)

$∫dθ$ runs from $-\frac{\pi}{2}$ to $\frac{\pi}{4}$ (because of the line $y = x$ in the $xy$ plane)

I do not get the given answer though.

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Hmm, on second thought, the question seems ambiguous. There are two possible interpretations of the region $V$: the one I was thinking of and the one you were thinking of (the other six are all equal in volume, though their $x$-values might be different in sign). Does the question specify which one is intended? –  Zev Chonoles Apr 12 '12 at 3:19
Don't forget the integrand. We have $x = r \cos \theta = p \sin \varphi \cos\theta$ so the integrand should be $(p\sin \varphi \cos \theta)(p^2\sin\varphi)$. –  nullUser Apr 12 '12 at 3:32
The question doesn't specify. In either case, the integral of dφ would run from zero to pi/2, right? I worked it your way and almost got the answer, but then the denominators changed to 32.. –  Anon Apr 12 '12 at 3:45
Yes, $\varphi$ should be going from $0$ to $\pi/2$. –  Zev Chonoles Apr 12 '12 at 3:47
@ZevChonoles I think you mean phi goes from 0 to pi/2. Theta should be between 0 and pi/4 correct? –  gsingh2011 Apr 12 '12 at 3:51

As Zev pointed out, the question is ill-posed since there's more than one region bounded by these surfaces. From the answer, it seems that the region $0\le p\le1$, $0\le\varphi\le\pi/2$, $\pi/4\le\theta\le\pi/2$ was intended, but even then the given answer is missing a factor of $\pi/2$. So I think the main conclusion from this exercise should be not to put too much stock in its source :-)

As has been pointed out in comments, your integrand is just the Jacobian and you forgot to include the original integrand $x=p\sin\varphi\cos\theta$. The required integral is

$$\int_0^1\int_0^{\pi/2}\int_{\pi/4}^{\pi/2}p^3\sin^2\varphi\cos\theta\,\mathrm d\theta\,\mathrm d\varphi\,\mathrm dp=\frac14\cdot\frac\pi4\left(1-\frac1{\sqrt2}\right)=\frac\pi2\left(\frac18-\frac{\sqrt2}{16}\right)\;.$$

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I may have made a mistake, but it seems easier to use rectilinear coordinates here. I get:

$$I = \int_{z=0}^1 \int_{y=0}^{\sqrt{1-z^2}} \int_{x=0}^{\min(y,\sqrt{1-y^2-z^2})} x \; dx dy dz$$

The min can be removed by splitting the integral into:

$$I = \int_{z=0}^1 \int_{y=0}^{\frac{1}{\sqrt{2}}\sqrt{1-z^2}} \int_{x=0}^{y} x \; dx dy dz + \int_{z=0}^1 \int_{y=\frac{1}{\sqrt{2}}\sqrt{1-z^2}}^{\sqrt{1-z^2}} \int_{x=0}^{\sqrt{1-y^2-z^2}} x \; dx dy dz$$

However, when I integrate this (a little tedious, but not particularly difficult), I get $I = \frac{(2-\sqrt{2}) \pi}{32}$.

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You can use double dollar signs to get displayed equations, which look nicer and are easier to read. –  joriki Apr 12 '12 at 8:18
@joriki: I added the double dollars; I think it just indents more? –  copper.hat Apr 12 '12 at 8:23
Which browser are you using? It might make sense to report this as a bug at meta.math.stackexchange.com. The displayed equations should look quite noticably different, less condensed, with much larger integral signs. (Also they're centred rather than indented.) –  joriki Apr 12 '12 at 8:25
The splitting point is $\sqrt{(1-z^2)/2}$, not $\sqrt{1-z^2}/2$. With that fixed, this gives the right result (Wolfram|Alpha computations for the first and second integral). –  joriki Apr 12 '12 at 10:35
@joriki: good catch! –  copper.hat Apr 12 '12 at 17:05