The prompt is to find the volume of the solid made by cutting out the ball $x^2 + y^2 + z^2 \le 1$ by a half cone $z = \sqrt{x^2 + y^2}.$
We know that in cylindrical coordinate system, $x = r\cos\theta$, $y = r\sin\theta$, $z = z$ and $x^2 + y^2 = r^2$
Following this, we can find value of r and z from the 2 equations, $$x^2 + y^2 + z^2 = 1$$ $$r^2 + z^2 = 1$$ equation 1 $$z = \sqrt{x^2 + y^2}$$ $$z^2 = r^2$$ equation 2
From equations 1 and 2, we can tell $\frac{-1}{2} < r < \frac{1}{2}$ and $\frac{-1}{2} < z < \frac{1}{2}$
Using these limits, we can construct the iterated integral as $$\int\int_{\frac{-1}{2}}^{\frac{1}{2}}\int_{\frac{-1}{2}}^{\frac{-1}{2}}r\,dz\,dr\,d\theta$$ but I'm not sure how to find limits for $\theta$. Is this the correct way to solvethis problem?
Intersection sphere/cone: $$r^2 + z^2 = 1, r^2 = z^2\implies r^2 = 1/2\implies r = \sqrt 2/2.$$ The correct limits are: $$V = \int_0^{2\pi}\int_0^{\sqrt 2/2}\int_r^{\sqrt{1 - r^2}}r\,dzdrd\theta.$$