Solids of revolution question !

Mark44

Yes, so you should be integrating from 1 to 2, not from 0 to 2.

lionely

Oh... so that's how I should get 1/2pi thanks

Mark44

That's not what I'm getting. How did you get that?

iRaid

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.

lionely

But I thought I should integrate 1/x^2

then evaluate [-1/x] from x= 1 to x =2?

Mark44

That's not the whole integrand.
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.

lionely

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.

Mark44

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.

lionely

the 1 is from y=1

and the other is y=1/x

also the answer in my text book is (1/2)∏

Mark44

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...