Calculate the volume of solid consisting of the cylinder $x^2+y^2\leq 4, 0 \leq z \leq 2$ and by cone $x^2+y^2\leq z^2, 2\leq z \leq 5.$
I tried to draw the figure on geogebra and I'm trying to use cylinder coordenate but I could not.
My attempt: put $x=r\cos(\theta), y=r \sin(\theta)$ and $z=z.$
We have $r^2\leq z^2,$ then $r\leq z$. Futhermore, $0\leq r\leq 2$ and $0\leq \theta \leq 2\pi.$
Therefore the volume is $\int\int\int 1dzdxdy =\int_{0}^{2}\int_{0}^{2\pi}\int_{r}^{5}r dzdrd\theta = \dfrac{44\pi}{3} $
The answer is $47\pi$ but if I do as above, I can't get it.
Can anyone help me, please?
Yes, the answer is $47\pi$. This is so because the volume of the cylinder is$$\int_0^{2\pi}\int_0^2\int_0^2\rho\,\mathrm d\rho\,\mathrm dz\,\mathrm d\theta=8\pi,$$whereas the volume of the cone is$$\int_0^{2\pi}\int_2^5\int_0^z\rho\,\mathrm d\rho\,\mathrm dz\,\mathrm d\theta=37\pi.$$