Volume integration help A volume sits above the figure in the $xy$-plane bounded by the equations $y = \sqrt{x}$, $y = −x$ for $0 ≤ x ≤ 1$. Each $x$ cross section is a half-circle, with diameter touching the ends of the curves. What is the volume?
a) Sketch the region in the $xy$ plane.
b) What is the area of a cross-section at $x$?
c) Write an integral for the volume.
d) Find the value of the integral.

The question has been essentially fully answered by JohnD: The picture does it all.
The cross section at $x$ has diameter $AB$, where $A$ is the point where the vertical line "at" $x$ meets the curve $y=\sqrt{x}$, and $B$ is the point where the vertical line at $x$ meets the line $y=-x$.
So the distance $AB$ is $\sqrt{x}-(-x)$, that is, $\sqrt{x}+x$. So the radius at $x$ is $\dfrac{\sqrt{x}+x}{2}$. The area of the half-circle with this radius is $A(x)$, where
$$A(x)=\frac{\pi}{2}\left(\frac{\sqrt{x}+x}{2}\right)^2.$$
The required volume is
$$\int_0^1 A(x)\,dx.$$
Once the setup has been done, the rest is just computation. We want to integrate $\dfrac{\pi}{8}(\sqrt{x}+x)^2$. Expand the square, and integrate term by term. 
