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y = 2.

Using washers, carefully find the volume of the half-torus obtained by revolving the part of the unit

disk above the x-axis around y = 2.

2. First of all, am I even setting this up properly?

I am using the washer method to solve. This is how I thought to do it...

But when I put the integral problem I get into Wolfram Alpha, it comes up with a different answer. Is my integration really wrong?

R= 2

r= 2 - √(1 - x^2)

∫-1 to 1 of [∏(2)^2 - ∏(2 - √(1 - x^2))^2]dx

=∫ ∏[-4(1 - x^2)^(1/2) + 1 - x^2]

= ∏[-4 * 2/3(1 - x^2)^3/2 * -1/2x + x - ((x^3)/3) |-1 to 1

= Substituting... ∏(1 - 1/3) - ∏(-1 + 1/3)

= 4∏/3 <-----That's my answer... But when I put my integral into Wolfram Alpha, it says the answer is 15.5504...

Ugh. Would you please tell me what I'm doing wrong?

Thank you! :)