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. 
 A: 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 $I = \frac{(2-\sqrt{2}) \pi}{32}$.
A: 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
$$\int_0^1\int_0^{\pi/2}\int_{\pi/4}^{\pi/2}p^3\sin^2\varphi\cos\theta\,\mathrm d\theta\,\mathrm d\varphi\,\mathrm dp=\frac14\cdot\frac\pi4\left(1-\frac1{\sqrt2}\right)=\frac\pi2\left(\frac18-\frac{\sqrt2}{16}\right)\;.$$
