Question

Can someone check my answer for the following problem?
"Determine the volume of the solid generated by revolving the area enclosed by the following curves about the y-axis: y=1, x=1, x^2 + y^2 = 1."

I used the Washer Method, with upper and lower limits of 1 and 0 respectively.
I got 2pi/3, but my book has pi/3. Did I do something wrong? If so, can someone help me solve this correctly?

Any and all help is greatly appreciated! :)

Answer 1

The sphere has a volume of 4/3pir^3 where r is 1, 4/3pi. The cube which encases it has a volume of 2*2*2=8. I'm getting 4-2pi/3 as the area of that enclosed part so I'm confused. I know that the half of the sphere would be 2pi/3, if that helps.

Answer 2

Sorry, perhaps I didn't word my question clear enough– the volume that is to be revolved is the three-sided chunk BETWEEN the outside of the "circle" formed by x^2 + y^2 = 1 and the square formed by x=1 & y=1. The sliver in the corner is what needs revolved, so it would be an upside-down, flat-bottomed bowl that is the solid figure, with the sphere being the empty space, not included in the shape.
(My book has a graph) :)

Answer 3

The geometric outlook is that we have an outer bound of a cylinder and we remove the inner upper-half sphere. The volume of the cylinder is V=pi r^2 h (r=1,h=1) = pi. The volume of the upper hemisphere is V = 1/2 (4/3 pi r^3) (r=1) = 2/3 pi. So the total volume is the difference:
pi - 2/3 pi = pi/3.

What did you write down for the whole integral where you set it up?

Answer 4

If we chose Washer method, we have an outer radius and an inner radius and we are revolving around the y-axis (x=0), we have this set up:
\( \displaystyle \pi \int_{a}^{b} R^2 - r^2 \ dy \) R = 1, r = sqrt(1 - y^2), a=0, b=1
\( \displaystyle \pi \int_{0}^{1} 1^2 - \left( \sqrt{1 - y^2} \right)^2 \ dy \)
Can you compare this to your integral and see what went wrong?

Answer 5

Those are the limits, and radii I used... :/ I wrote the inner radius as (1 - y)^2, because when solving the x^2 + y^2 = 1, I took the square root of both sides, and I simplified (1 - y^2)^(1/2) to be (1 - y). Is that okay?

Answer 6

That would not work. Because:
\( \sqrt{1 + a^2} \color{red} \ne \sqrt{1} + \sqrt{a^2} = 1 + a \)
The square root does not distribute through addition, it only distributes through multiplication (same as rules of exponents).
This would be true:
\( \sqrt{ (1 + a)^2 } = 1 + a \) The square root and square cancel out.

Answer 7

The way you get that 1/3 coefficient out is because you simply cancel the square root and squared parts:
\( \left( \sqrt{1 + y^2} \right)^2 = 1 + y^2 \)
and then your integrand simplifies into:
\( 1 - (1 + y^2) = y^2 \)
When you integrate \(y^2\), you get \(\dfrac{y^3}{3} \) and taking this from 0 to 1 the numerator is 1/3 - 0/3 = 1/3.

Answer 8

I got it, and now understand what I did wrong! Thank you very much!

Answer 9

Glad to help! :)