triple integral spherical coordinate

Question

I have a problem converting this question into a spherical form. $∫∫∫ z/√(x^2+y^2+z^2)dxdydz$ where R is the interior of a sphere $x^2+y^2+z^2 = 2z$

the limits of integration I found are: 0≤r≤2cosθ 0≤θ≤ π 0≤φ≤2 π

After converting this is my integrand $∫∫∫ rcosθr^2sinθ/√2rcosθ drdθdφ$ with limit given above. But this doesn't give me the right answer. Any help will be appreciated. Thanks in advance.

Answer 1

The equation of the sphere can be rewritten as $x^2+y^2+(z-1)^2=1$. Thus the sphere has center $(0,0,1)$ and radius $1$. In particular, it is not the sphere in the picture. Since the sphere lies above the $xy$-plane, you should have $$0\leq \theta \leq \frac{\pi}{2}.$$ Also the denominator $\sqrt{2 r \cos \theta}$ in the integrand looks wrong. The denominator is just $r$ since $r = \sqrt{x^2+y^2+z^2}$.

Answer 2

Maybe , instead of $$ z=r cos φ $$ $$ z=1+r cos φ $$

Then the boundaries for r would be from 0 to 1