Let $A=\{(x,y,z)\in \mathbb{R}^2 : x^2+y^2+z^2 \leq 1, 0\leq z \leq \frac{1}{2} \}$. Find the volume of $A$.
The volume I'm asked to find it's what is left of the unit semisphere minus the upper part cut by the plane $z=1/2$.
The volume of the semisphere is $\frac{1}{2} \frac{4}{3} \pi r^3 = \frac{2}{3} \pi$. Now I need to find the volume of the upper part of the semisphere above the plane $z=1/2$. To do that, I thought in changing to cylindral coordinates, so $\theta$ goes from $0$ to $2 \pi$ and $z$ goes from $1/2$ to $\sqrt{1-\rho^2}$. However, I'm not sure what is the domain of $\rho$ in this case. I think it goes from $0$ to $\frac{\sqrt{3}}{2}$ but my solution says otherwise.
If it is like I say, then the volume of that upper part is $\frac{5}{24}\pi$ and then
$$v(A)= \frac{2}{3}\pi - \frac{5}{24}\pi = \frac{11}{24}\pi$$
Am I correct? If not could you help me to understand from where to where does $\rho$ go?
Thanks in advance!
Use cylindrical polar coordinates to get the volume as the following integral $$ \int_0^{2\pi}\int_{0}^{0.5}\int_{0}^{\sqrt{1-z^2}}rdrdzd\theta=\pi\int_{0}^{0.5}(1-z^2)dz={11\over24}\pi $$