I need to find the volume of solid enclosed by the cone $z =\sqrt{x^2+ y^2}$ between the planes $z =1$ and $z =2$
Now using Spherical Coordinates I can set up the integral as:
$\displaystyle \int_{0}^{2\pi}\int_{0}^{\pi/4}\int_{sec\phi}^{2sec\phi} {\rho}^2sin\phi\text{ } d{\rho}\text{ } d{\phi}\text{ } d\theta$
Just for my practice I also want to find this via cylindrical Coordinates. But, I don't understand how should I express the region in terms of $dz$ $dr$
Can anyone please explain to me step by step how should I express this in cylindrical coordinates ?
Thank You.
Best Answer
If you insist in cylindrical coordinates you need two integrals $$ \int _0^{2\pi} \int _0^1\int_1^2dzrdrd\theta +\int _0^{2\pi} \int _1^2\int_r^2dzrdrd\theta=$$
$$\pi + 4\pi/3 = 7\pi/3$$
where the first integral evaluates the middle cylinder and the second one evaluates the rest of the volume.
However the easier way is to use the solid of revolution formula.
The volume is found by the integral $$\int _1^2 \pi r^2 dz $$ where $r^2 = z^2$
Therefore the answer is $$\int _1^2 \pi r^2 dz =\int _1^2 \pi z^2 dz =\frac {7\pi }{3} $$