Find volume of solid whose base is enclosed by circle and

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hank

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Sep 13, 2006
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Find the volume of the solid whose base is enclosed by the circle x^2 + y^2 = 1 and whose cross sections taken perpendicular to the x-axis are semicircles.

I'm just really stuck on the setup on this one.

I was thinking that I would need the semi-circle formula:

1/2 * pi * r^2

I would need to convert the circle into terms of x:

y = sqrt(1 - x^2)

And then take the integral with the limits of -1 to 1:

V = 1/2*pi S [-1 to 1] (sqrt(1 - x^2))^2 dx

But this evaluates to zero so can't be right.

Can someone show me how to set this one up?
 
Oh, I redid it and I got pi/3.

What I did was to find the area of one quadrant. Should I then double it, so I get both halves since it goes into two quadrants?

That would give me the correct answer of 2pi / 3.

Is this the correct reasoning?
 
You had it right the first time.

\(\displaystyle \L\\\frac{\pi}{2}\int_{-1}^{1}(1-x^{2})dx\)
 
Yeah, I went back and as I was working with it, I got it.

But now, a 2d part to that problem tells me to use a square instead of semi-circle.

Thus, I get:

A = s^2

Now, I want to integrate sqrt(1 - x^2)^2, I think for the limit of 0 to 1.

This gives me a result of 2/3 for one quadrant. Now, I'm thinking I have 4 quadrants so I should multiply that by 4 which gives me 8/3.

However, the book tells me the answer is 16/3.

There must be some wrinkle I'm missing here.

Here's my setup as outlined above:

V = 4 S [0 to 1] (sqrt(1 - x^2)^2) dx

What am I missing here?
 
You can't use just one quadrant on this one. Your squares aren't big enough. \(\displaystyle \sqrt{1-x^{2}}\) is only half the length of a side. I'm a little surprised you were not off by a factor of 4.
 
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