I am trying to use spherical coordinates to find the volume of 1/2 of a cylinder with height = 1 and radius = 1.
Area i wanna find is marked in blue spun around the z-axis
I wanna find a triple integral on the form: $$ \int{\int{\int{\rho^2sin(\phi)\ d\rho\ d\phi\ d\theta}}} $$
The integral in terms of $\theta$ is from $0$ to $2\pi$
The integral in terms of $\phi$ is from $tan^{-1}(1)=\frac{\pi}{4}$ to $\frac{\pi}{2}$
The integral in terms of $\rho$ is from $0$ to the line that goes from point C to y=1 which we can find using pytagoras: $$ \rho = \sqrt{z^2 + r^2} $$ We know that $r$ always is 1 and $z = \rho cos(\phi)$ This gives us: $$ \rho = \sqrt{\rho^2cos^2(\phi) + 1} $$ Solving this for $\rho$ gives us: $$ \rho = \sqrt{\frac{1}{1-cos^2(\phi)}} $$
If we plug in the limits and solve the triple integral we get: $$ \int_0^{2\pi}{\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{\int_{0}^{\sqrt{\frac{1}{1-cos^2(\phi)}}}{\rho^2sin(\phi)\ d\rho\ d\phi\ d\theta}}} = \frac{2\pi}{3} $$
This volume will be 1/2 of a cylinder with height = 1 and radius = 1, the answer i expected was $ \frac{\pi}{2} $
I am not really sure where i am doing something wrong..