Calculating Volume: Rotating Curves About a Line

[tony873004]

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

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

 

[TD]

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?

 

[J77]

...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)

 

[HallsofIvy]

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. x^mxⁿ= x^m+n but (xⁿ)^m= x<sup>mn. (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.</sup>

 

[VietDao29]

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.

 

[tony873004]

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.

...(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.

 

[J77]

Nice LaTex style

For the limits, you can just put the numbers.