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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.

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Be sure you have a clear visual of what it is that you are computing. Hopefully this will help some:

Mathematica graphics

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How did you plot this region? – EsX_Raptor Dec 15 '12 at 1:43
    
Using Mathematica. – JohnD Dec 15 '12 at 2:05

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.

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