$$\iiint (xyz) dx\,dy\,dz$$ I am asked to evaluate this integral over the region $$D:=\left \{ (x,y,z) \in\mathbb{R}^3 :x^2+y^2+z^2 \leq 4 \wedge y > x\right \}$$ Here is the integral I set up: $$\iint_{Pr_{y,x}(D)}\int_{-\sqrt{4-y^2-x^2}}^{\sqrt{4-y^2-x^2}}(xyz) dx dy dz$$=$$\iint_{Pr_{y,x}(D)}(xy)dx dy\int_{-\sqrt{4-y^2-x^2}}^{\sqrt{4-y^2-x^2}}(z) dz$$ which integral of z is
$$\int_{-\sqrt{4-y^2-x^2}}^{\sqrt{4-y^2-x^2}}(z) dz=(\frac{4-y^2-x^2-4+y^2+x^2}{2})=0$$
and then we get:
$$\iint_{Pr_{y,x}(D)}\int_{-\sqrt{4-y^2-x^2}}^{\sqrt{4-y^2-x^2}}(xyz) dx dy dz=0$$
Is this correct?
Yes, this is correct. In fact, we can conclude this immediately without doing any calculations since the domain of integration is symmetric in $z$, while the function $xyz$ is odd in $z$, so the integral must be zero.