Given the following function: $$f(x,y,z)=x+y+z$$
And domain:
$$D(f)=\{(x,y,z): {\sqrt {x^2 + y^2}} \leq z \leq {\sqrt {4-x^2 - y^2}}\}$$
Solve the following integral:
$$\iiint_D f $$
I tried the following solution by using spherical coordinates system:
$$ x=r\cdot cos(\theta)\cdot sin(\omega) $$
$$ y=r\cdot sin(\theta)\cdot sin(\omega) $$
$$ z=r\cdot cos(\omega) $$
And the determinant of the Jacobian matrix:
$$ J=r^2\cdot sin(\omega) $$
I started with finding limits for $\theta$,$\omega$,$r$: $$0\leq\theta\leq 2 \pi$$ $$0\leq\omega\leq 0.25 \pi$$ $$0\leq r\leq 2$$
Then we have following integral: $$\int_0^{0.25\pi} \int_0^{2\pi} \int_0^2 f(r,\theta,\omega) \cdot J dr d\theta d\omega$$
Is this the right way to solve this? I'm not sure about the limits and the coordinate system I used. Please advise.
When you convert the conditions$$\sqrt{x^2+y^2}\leqslant z\leqslant\sqrt{4-x^2-y^2},$$what we get is$$r\lvert\sin\omega\rvert\leqslant r\cos\omega\leqslant\sqrt{4-r^2\sin^2\omega}.$$It follows from the first inequality that $0\leqslant\omega\leqslant\frac\pi4$ and from the second one that $r\leqslant2$. So, your limits of integration are right and the integral that you should compute is$$\int_0^{\pi/4}\int_0^{2\pi}\int_0^2\bigl(r\cos(\theta)\sin(\omega)+r\sin(\theta)\sin(\omega)+r\cos(\omega)\bigr)r^2\sin(\omega)\,\mathrm dr\,\mathrm d\theta\,\mathrm d\omega.$$It turns out that it is equal to $2\pi$.