# Express inequlities in spherical coordinates

The problem is following: Assume that we have a area, lets call it $K$ that is limited by a cone $z=-\sqrt{x^2+y^2}$ and by a sphere $x^2+y^2+z^2 = 16$

Now, in this problem they want me to describe the area K using inequlities. More specifically in spherical coordinates

My attempt:

Spherical coordinates are given by:

$\left\{x = r \sin\theta \cos\phi \\y=r \sin\theta \sin\phi \\z=r \cos\theta\right\}$

where $0 \leq \theta \leq \pi, \space 0\leq \phi \leq 2\pi$

So this gives us: $x^2+y^2 = r^2 \sin\theta$

After that I'm basically a bit lost, the only restriction we have, for the variables $x$, $y$ and $z$ are that they range between $\pm4$.

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Do you mean $K$ is a volume, that is, the three dimensional solid that is bordered by the cone and sphere? It sounds like it.
You lost a square of the $\sin$ in your expression for $x^2+y^2$. It equals $r^2\sin^2 \theta$. Can you do a similar calculation of $x^2+y^2+z^2$?
Then you need to decide which side of the cone and sphere form the region of interest. This will be shown by changing the equal signs into inequalities. For example, you probably want the interior of the sphere. In Cartesian coordinates, this would be $x^2+y^2+z^2 \le 16$ (or $x^2+y^2+z^2 \lt 16$ if you want an open ball). Similarly, for the cone, you probably want the volume around the negative $z$ axis, so you want $z \le -\sqrt{x^2+y^2}$. If you substitute your conversions to spherical coordinates into these you are home.