Triple integral - wedge shaped solid

Question

Find the volume of of the wedge shaped solid that lies above the xy plane, below the $z=x$ plane and within the cylinder $x^2+y^2 = 4$.

I'm having serious trouble picturing this. I think the z bounds are [0,x], but can't figure out any of the other ones.

Answer 1

looks like this should be the required vol $$2\int_{0}^{\pi/2} \int_{0}^{2} \int_{0}^{rcos\theta} rdzdrd\theta $$

Answer 2

I believe that you are describing a variation of the hoof of Archimedes (see figure below). If that's the case, I can give you a detailed solution and show you how to reduce the triple integral for volume to a single integration. From what you describe, the height would be $R$, rather than the $2R$ shown in the figure.

Answer 3

Planes $x={\rm const.}$ intersect this solid $B$ in rectangles of width $2\sqrt{4-x^2}$ and height $x$ (see the figure in Cye Waldman's answer). One therefore obtains $${\rm vol}(B)=\int_0^2 2\sqrt{4-x^2}\>x\>dx=-{2\over3}(4-x^2)^{3/2}\biggr|_0^2={16\over3}\ .$$