# Finding the volume of a solid made with base as a parabola.

The base of the solid is the region between the $x$-axis and the parabola $y=4-x^2$. The vertical cross sections of the solid perpendicular to the $y$-axis are semicircles. Compute the volume of the solid.

I know the area formula for a semicircle is $\frac{(\pi*r^2)}{2}$, but would you use $r=4 - x^2$ even though the question only says the base is made from this parabola?

Each semicircle has a radius $x$, so the solid has volume
$$\frac{\pi}{2} \int_0^2 dy \, x^2 = \frac{\pi}{2} \int_0^2 dy \, (4-y) = 3 \pi$$
• @mathtastic: "The base of the solid is the region between the x-axis and the parabola $y=4-x^2$..." – Ron Gordon Jan 16 '15 at 0:56
• @ Bryce - you made the same mistake I did and just assumed he used $dx$. Instead, he uses the differential $dy$ and notice $y=4-x^2$ becomes $x=\sqrt{4-y}$ so that, when plugged into the formula you indicated in your post, you obtain Ron's integral. – 123 Jan 16 '15 at 1:03