Mentor

## Solids of revolution question !

Yes, so you should be integrating from 1 to 2, not from 0 to 2.
 Oh... so that's how I should get 1/2pi thanks
 Mentor That's not what I'm getting. How did you get that?
 lionely, you have to understand the concept of this. Believe me, if you just learn the formulas and not HOW this stuff works, you'll fail during the exam. The next thing you're going to be learning is the shell method which builds upon this even more. When your teacher starts throwing in rotations around lines that aren't the x and y axis, you'll be completely screwed.
 But I thought I should integrate 1/x^2 then evaluate [-1/x] from x= 1 to x =2?

Mentor
 Quote by lionely But I thought I should integrate 1/x^2
That's not the whole integrand.
 Quote by lionely then evaluate [-1/x] from x= 1 to x =2?
These are the right integration limits, but it looks like you don't have the right integrand.

I don't think you have a drawing (or maybe a good drawing) of the region that's being rotated. The region being rotated is the roughly triangular region above the curve y = 1/x, below the line y = 1, and to the left of the line x = 2.

Each area element is a vertical strip of width Δx. Each area element, when rotated, is in the shape of a washer whose volume is the area of the washer times its width, Δx.
 OKAY SIR I THINK I got it if not I truly hate integration , xD. I got 1/2pi from evaluating [-1/x] NOW I now find the volume under y=1 for the same interval I get pi NOW I then subtract the volumes ( Pi - 1/2pi) = 1/2pi cu. units.
 Mentor Edit: I made a mistake in evaluating the antiderivative. No, that's the same as you got earlier, and it's still not right. What does your integral look like? $$\pi \int_1^2 (f(x)) dx$$ If you "Quote" my post, you can use my LaTeX above and fill in the integrand.

 Quote by Mark44 No, that's the same as you got earlier, and it's still not right. What does your integral look like? $$\pi \int_1^2 (1-(1/x^2)) dx$$ If you "Quote" my post, you can use my LaTeX above and fill in the integrand.
the 1 is from y=1

and the other is y=1/x

also the answer in my text book is (1/2)∏
 Mentor Perfect. The answer is as you show, ##\pi/2##. I subbed in a wrong value in my work. I'm sorry if I caused you any grief...