# Another volume problem

1. Sep 14, 2006

### 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.?

2. Sep 14, 2006

### 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?

3. Sep 14, 2006

### 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)

4. Sep 14, 2006

### HallsofIvy

Staff Emeritus
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
$$\pi \int_1^2 \frac{1}{x^2}dx= \pi \int_0^1 x^{-2} dx[/itex] That should be easy to integrate. Last edited: Sep 15, 2006 5. Sep 14, 2006 ### VietDao29 Others have pointed out the errors in your OP. If you are about to integrate: [tex]\pi \mathop {\int} \limits_{1} ^ 2 x \ dx$$, 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: $$F(x) = \frac{1}{2} x ^ 2 + C$$.

Last edited: Sep 14, 2006
6. Sep 14, 2006

### tony873004

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.

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.

7. Sep 15, 2006

### J77

Nice LaTex style

For the limits, you can just put the numbers.