Calculating Volume: Rotating Curves About a Line

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Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line.
y=1/x, x=1, x=2, y=0; about the x-axis

About halfway down, I diverge into two methods. The one on the left I use the method from FrogPad's post in my other thread (or at least my effort to understand the method he describes, btw, thank you). But it gives me 3pi/2. The one on the right gives me the same answer as the back of the book, pi/2. Maybe neither is right, and I coincidentlly got the right answer using the method on the right :bugeye:

Also, am I using the right symbols, etc.?

vol2.gif
 
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Why do you start off with a summation? Also: the limits are unclear (what is the variable?).

The volume of the solid you obtain by rotating f(x) about the x-axis between x = a and x = b is given by:

\pi \int\limits_a^b {f\left( x \right)^2 dx}

In your specific case, with f(x) = 1/x and the limits given; this comes down to:

\pi \int\limits_1^2 {\frac{1}{{x^2 }}dx}

Can you follow? Can you now compute this integral?
 
...and

(x^{-1})^2\ne x

but

(x^{-1})^2= x^{-2}

with the limits, your left-hand column would be right (assuming the integrand is right, as above or in prev. post)
 
You started off fine. I don't know why you "diverged" as you put it.
(It would also, by the way,be better to write more than just math symbols in your work: explain what you are doing and what each formula represents.)

Using the "disk" method you can imagine the volume divided into many thin disk with center on the x-axis, thickness \Delta x, and radius equal to the y value= \frac{1}{x}. The area of each disk is \pi r^2= \pi \frac{1}{x^2} and so the volume is \pi \frac{1}{x^2} \Delta x. The approximate volume of the whole figure then is the Riemann Sum \Sigma \pi \frac{1}{x^2}\Delta x.
Now, for some reason, you have \Sigma \pi x \Delta x.
Surely you know that 1/x2 is not x!
(Are you subtracting exponents- that's for multiplication, not repeated powers. xmxn= xm+n but (xn)m= xmn. (x-1)2= x-2.

Now take the limit as the number of disks goes to infinity: the Riemann Sum \Sigma \pi \frac{1}{x^2}\Delta x becomes the integral
\pi \int_1^2 \frac{1}{x^2}dx= \pi \int_0^1 x^{-2} dx[/itex]<br /> That should be easy to integrate.
 
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Others have pointed out the errors in your OP. If you are about to integrate:
\pi \mathop {\int} \limits_{1} ^ 2 x \ dx, the way in the left is correct.
You should note that:
\mathop {\int} \limits_{a} ^ b x \ dx = F(b) - F(a), where F is the antiderivative of f.
And in this case, you'll have: F(x) = \frac{1}{2} x ^ 2 + C.
 
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TD said:
Why do you start off with a summation? Also: the limits are unclear (what is the variable?).
The first few problems from the notes were done that way. I guess to show us why the integral is set up that way. After the first few examples, the teacher skips that step.

HallsofIvy said:
...(It would also, by the way, to more than just math symbols in your work: explain what you are doing and what each formula represents.)...
Point well taken, especially when I give up for the night and try to pick it up in the morning.

I re-did it and got the correct answer. I'll post my solution anyway, just in case I did something wrong and coincidentally got the right answer.

Thanks for all your explanations. This is starting to become clear. :smile:
vol3.gif
 
Nice LaTex style :smile:

For the limits, you can just put the numbers.
 
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