Triple integral over a half hemisphere?

Question

Calculate the integral $\displaystyle \iiint_K (x + 2) d{x} d{y} d{z}$ when $K$ is given by the inequalities $x^2 + y^2 + z^2 ≤ 1$ and $z ≥ 0$.

But the correct answer is $\frac{4\pi}3$ and they gave a clue which is to use symmetric properties but I don't know how.

Answer 1

• Small mistake: since $z\ge0,$ $\phi\in[0,\pi/2],$ not $[0,\pi].$
• Big mistake: $$\iiint \sin\phi\cdot(\dots)dr d\theta d\phi\ne\int_0^\pi\sin\phi d\phi\iint\left(\int_0^1\dots dr\right)d\theta d\phi.$$
• Easy solution: $\displaystyle \iiint_Kxdxdydz=0$ by symmetry, and $\displaystyle \iiint 2dxdydz=$ twice the volume of half the unit sphere$=\dfrac{4π}3.$