# Value of $\int z$ over half sphere.

What is the value of $\int_Az$ where $A=\{(x,y,z)\in\mathbb{R}^3|x^2+y^2+z^2\leq a^2,z>0\}$?

I used the change of variables $g(r,\phi,\theta)=(r\sin(\phi)\cos(\theta),r\sin(\phi)\sin(\theta),r\cos(\phi))$ and got the new integral $\int_Br\cos(\phi)\cdot r^2\sin(\phi)$ where $B=[0,a]\times[0,\pi]\times[0,\pi]$. I obtained these bounds by reasoning that the half-sphere in the positive z-axis was simply the half circle of radius $a$ rotated over the positive z-axis (in both cases the rotation was $\pi$ radians).

Now to evaluate:

$$\int_Br^3\sin(\phi)\cos(\phi)\,d\phi\,d\theta\,dr$$ $$=\int_0^a\int_0^\pi\int_0^\pi r^3\sin(\phi)\cos(\phi)\,d\phi\,d\theta\,dr$$ $$=\int_0^a\int_0^\pi -\frac{r^3}{2}\cos^2(\phi)|_0^\pi\,d\theta\,dr=0$$

But the function we are integrating (i.e. $f(x,y,z)=z$) is positive over the area of integration, so how can the integral be $0$? What am I missing?

• I suggest you do the calculations again to get the correct result. – Marko Jan 21 '16 at 14:25

Boundaries for one of your integrals are not correct. Since $z>0$, the angle $\phi$ varies from $0$ to $\frac{\pi}{2}.$ Also, the angle $\theta$ varies from $0$ to $2\pi$. Here is the problem.
• Thanks for catching my mistake. I've got $\frac{a^4\pi}{4}$. – Zelzy Jan 21 '16 at 14:33