Hello I am struggling to find the answer to converting:
$z = \sqrt{3x^2+3y^2}$ to spherical coordinates!
My answer I keep getting is...
$\rho\cos(φ) = \sqrt{3}\sqrt{(\rho\cos(\theta)\sin(\varphi))^2+(\rho\sin(\theta)\sin(\varphi))^2}$
But my submissions are wrong even after I simplify it... any help would be appreciated!
I don't know what to integrate, but maybe rewriting your identity as $\rho\cos{\varphi} = \sqrt{3}\sqrt{\rho^2\sin^2(\varphi)} = \sqrt{3}\rho\sin(\varphi)$ your problem becomes easier. I just used $\sin^2(\theta) + \cos^2(\theta) = 1$.
Well, it follows $\tan(\varphi) = 1/\sqrt{3}$ and then $\varphi = \pi/6$. Geometrically, you can see the constant value when you plot the graph:
Maybe it helps you to integrate whatever you want (the area of the surface, the volume that the cone contains, ...).