Question: The section of a certain solid cut by any plane perpendicular to the x-axis is a circle with the ends of a diameter lying on the parabolas y^2 = 4x and x^2 = 4y. Find its volume.
Attempted Answer:
I drew the following diagram: 
I assumed the diameter of the circle would be in the region bounded by the parabolas, with $ y^2 = 4x $ being the higher edge and $ x^2 = 4y $ being the lower edge.
The diameter for each circle/disk would be the difference between the two y's, So, $ y^2 = 4x $ becomes $ y =2\sqrt{x} $ and $ x^2 = 4y $ becomes $ y = \frac {x^2}{4} $. The difference is $ 2\sqrt{x} - \frac {x^2}{4} $. I take this to be the diameter of each circle, and so the area of each circle is $ \pi(\frac{1}{4})(2\sqrt{x} - \frac {x^2}{4})^2 $.
I expand and get the area of each circle as $ \frac{\pi}{4} * (4x - x^\frac{3}{2} + \frac{x^4}{16}) $
I then integrated the circles from x = 0 to x = 4.
$ \frac{\pi}{4} \int_0^4 (4x - x^\frac{3}{2} + \frac{x^4}{16}) dx = 8\pi $
My answer is wrong. I suspect I have misread the question. Some help would be appreciated.