Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. –  Julian Kuelshammer Dec 14 '12 at 23:53
add comment

2 Answers

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

Mathematica graphics

share|improve this answer
    
How did you plot this region? –  Josué Molina Dec 15 '12 at 1:43
    
Using Mathematica. –  JohnD Dec 15 '12 at 2:05
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.