Evaluating the volume of $4z^2=x^2+y^2$

Question

I know the volume of the cone $$4z^2=x^2+y^{2}$$ for $0\leq z\leq3$ is equal to $36\pi$ since its height is 3 and its base radius is 6. So, In the first octant, the volume should be $$36\pi\div4=9\pi$$ But I can't find this value using integrals. I tried to write using spherical coordinates: it's clear that $$0\leq\rho\leq6$$ and $$0\leq\theta\leq\frac{\pi}{2}$$ and on the plane $yOz$ to find the interval for $\phi$, I realized that, if $x=0$, then $$4z^2=y^2\Rightarrow z=\frac{y}{2}$$ so, for $z=3,y=6$. So, $$\tan(\phi)=\frac{6}{3}=2\Rightarrow 0\leq\phi\leq\arctan(2)$$ However, Wolfram says the integral $$\int_{0}^{\pi/2}\int_{0}^{\arctan(2)}\int_{0}^{6}\rho\sin^{2}(\phi)d\rho d\phi d\theta=\frac{9}{10}\pi(5\arctan(2)-2)$$ What am I doing wrong?

Answer 1

The top of your cone is flat, yet this isn't reflected in your integral. Constant bounds on all the variables gives you regions shaped like spheres. That's why they are called spherical coordinates.

You will need $\rho$ to go from $0$ to $\dfrac{3}{\cos\phi}$.

Also, it's $\iiint\rho^2\sin\phi\,d\rho\, d\phi\, d\theta$, not $\iiint\rho\sin^2\phi\,d\rho \,d\phi \,d\theta$.

Answer 2

Don't use spherical coordinates, but still use radial in $(x,y)$.

The area of the segment of a circle radius: $$A = \frac{r^2\theta}{2}$$

Radius squared: $$r^2 = 4z^2$$

Using area.

$$\int_0^3\int_0^{\pi/2} 2z^2 d\theta dz$$

$$\int_0^3 \pi z^2 dz = 9\pi $$