Being that some people just taught me something interesting i would like to apply it.

Use cylindrical coordinates to find the volume of the region bounded above by the plane $z = 2x$ and below by the paraboloid $z = x^{2} + y^{2}$.

it may appear that i messed this up but it should actually be fine i switched my upper and lower bounds twice in the setup below sorry for the confusion

I would like to convert into polar and try and be clever by integrating from $\pi /3$ to $4\pi /3$ but i cant seems to figure out where i want to integrate r over perhaps

$$\int^{4\pi /3}_{\pi /3}\int^{2 \cos \theta}_{0} \big(r^{3} -2r^{2} \cos \theta\big) dr d\theta$$

Does this make any sense to anyone?

ok none of this makes any sense basically i want to use the fact that z=2x and its the the lower bound this is like looking at a line in the x-z plane that cuts it in half u can view this as a triangle of hyp=2 and x=1 so if we convert this to an angle ( $\pi /3$) we should be able to integrate over the angle $\pi /3$ to $\pi + \pi /3$ as this will give us the angles we cover in the x-z plane if we can integrate those angles in the paraboloid is should give us what we are looking for?

$$\int^{4\pi /3}_{\pi /3}\int^{2 \cos \theta}_{0} \big(r^{3} -2r^{2} \cos \theta\big) dr d\theta=$$ $$\int^{4\pi /3}_{\pi /3}\big(\cos^{4} \theta -\frac{2}{3} \cos^{4} \theta\big)d\theta= \frac{1}{3}\int^{4\pi /3}_{\pi /3} \cos^{4} \theta d\theta$$

Could someone perhaps inegrate this diffrently so i could compare answers i think im completly out to lunch? is there some software i could use to verify what the answer was?

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