I want to calculate the volume of a cone having base radius $3$ units and height $6$ units by setting up a triple integral in cylindrical coordinates. I must get answer as $18π$ cubic units. My cone has its vertex at the point $(0,0,6)$ and base circle is in $xy$ plane. I first came up with the equation of the cone which is as follows:
$36-z^2= 4(x^2+y^2)$ which can be written as $z=\sqrt{36-4r^2}$ in cylindrical coordinates. Then I set up the triple integral as: $$\int_0^3 \int_0^{2\pi} \int_0^{\sqrt{36-4r^2}} r dz d\vartheta dr$$
But this integral gives me result $36\pi$, double of the actual volume. Why is it so? What am I doing wrong here? Please suggest.
Please note that the equation of cone you have come up with is not in line with your problem statement. $36-z^2= 4(x^2+y^2) \implies 4x^2 + 4y^2 + z^2 = 36$ is equation of an an ellipsoid surface.
The correct equation of the cone is,
$(6-z)^2= 4(x^2+y^2) = 4r^2$
And if you are going in order $dz$ and then $dr$, the integral to find its volume between $0 \leq z \leq 6$ should be,
$\displaystyle \int_0^{2\pi} \int_0^3 \int_0^{6-2r} r \ dz \ dr \ d\theta$