spheresvolume
I'm having some problems finding the volume inside a sphere of radius 2 and above the cone $\sqrt3z = \sqrt{x^2 + y^2}$.
I tried integrating using spherical coordinates and got $\int_0^{2\pi}\int_0^{\pi/2}\int_1^2 \rho^2 sin\phi$ $d\rho$ $d\phi$ $d\theta$, which gave me $14/3\pi$, but the correct answer is $8/3 \pi$, anyone knows what might have gone wrong or what is the correct way to solve this problem?
Much appreciated!
The parameterized (polar) curve $$r(\theta) = 2 \sin \theta$$ can be written as the parameterized (Cartesian) curve $$(x(\theta), y(\theta)) = (2 \sin \theta \cos \theta, 2 \sin^2 \theta) = (\sin 2 \theta, \cos 2 \theta + 1);$$ since both of the components in the last expression have period $\pi$, the parameterization traces out the circular boundary twice over the interval $[0, 2 \pi]$; we only want to integrate over each direction $\theta$ once, so we should replace the interval over which we integrate $\theta$ by $[0, \pi]$.
We can see too why the triple integral you wrote evaluates to zero: The upper limit on the middle integral, is $2 \sin \theta$, and the symmetry of that function (together with the fact that $\theta$ appears nowhere else in the integral) ensures that the integral over $[\pi, 2\pi]$ has the same magnitude but the opposite sign as the desired integral over $[0, \pi]$.
Your answer gives you the volume of a cylinder with height $2$.
See the picture:
$$\frac{2}{\rho}=\cos \phi\implies \rho=\frac{2}{\cos\phi}=2\sec\phi$$
Best Answer
It looks like you have an incorrect bound on the angle $\phi$. The cone in question is the upper half of $3z^2 = r^2$, so at height $z = 1$, for example, the radius of the cone is $3$. Drawing an appropriate right triangle will allow you to find the angle between your cone's side and the $z$-axis. This will be the upper bound on $\phi$ in your integral.
Happy calculating!