Integrating function over half-ball 
Let $B:=\{(x,y,z)\in\mathbb{R}^3:x^2+y^2+z^2\le 4, z\ge 0\}$. Calculate $\int_B \dfrac{(x^2+y^2)z}{(x^2+y^2+z^2)^2}d\lambda_3(x,y,z)$.

I see that the boundary of $B$ is $D:=\{(x,y,z)\in\mathbb{R}^3:(x^2+y^2\le4,z=0)$ or $x^2+y^2+z^2=4\}$. I feel like I need to find a parametrization of $B$ and then transform it. Any help would be greatly appreciated.

In spherical coordinates, you have $$\int_B\frac{(r^2-r^2\cos^2\varphi)r\cos\varphi}{r^4}\,r\sin\varphi\, dr\, d\theta\, d\varphi$$
with $r$ fromt $0$ to $2$, $\theta$ from $0$ to $2\pi$, and $\varphi$ from $0$ to $\pi/2$. Can you simplify and take it from here?
With $r$ and $\theta$ disappearing from the calculation, you have $$4\pi\int_0^{\pi/2}\sin^3(\varphi)\cos(\varphi)\,d\varphi$$ Note that symmetry makes this equal to $$4\pi\int_0^{\pi/2}\sin(\varphi)\cos^3(\varphi)\,d\varphi$$ and that summing these equals (with the Pythagorean identity):$$4\pi\int_0^{\pi/2}\sin(\varphi)\cos(\varphi)\,d\varphi$$ So $$2I=4\pi\int_0^{\pi/2}\sin(\varphi)\cos(\varphi)\,d\varphi=4\pi\int_0^{\pi/2}\frac12\sin(2\varphi)\,d\varphi$$
(My scratch work earlier had left out that $2$ on the left.) The last integral is one lump of a sine wave, contracted both horizontally and vertically by factors $2$. Since one regular sine lump has area $2$, the last integral is $\frac12$, and therefore $I=\pi$.
