# Thread: region between the curve

1. ## region between the curve

The region between the curve (y=1/x 2) and the x-axis from x=1/2 to x=2 is revolved about the y-axis to generate a solid. Find the volume of the solid.

I know the answer is pi ln16

i tried drawing it out but i am confused witht the "revolved about the Y-AXIS" i don't see it. i see x-axis though...

2. so you draw out the region, shade it, and then imagine it as being pasted onto a glass door, where the hinges of the door is the Y axis and spin the door.

The shape ends up looking a little like the red piece on the right side.

3. okay so i go to v= pi ∫ 1/x^4 dx

so i need to do du/u... and if so du= 3x^3 so do i add a negative on the outside?

4. yes because of power rule.

eg.

$\displaystyle \frac{d}{dx}(\frac{1}{x^2})$

is:

$\displaystyle -2*\frac{1}{x^3}$

5. so i got v= pi [-3x^3 (lnx^4)] from .5 to 2 (right?)
then do i include the .5 to 2 with the /3x^3?

6. Yes, when you do your substitution you substitute it for all the x's in your equation thing.

just to be clear your starting function is:

$\displaystyle y=\frac{1}{x^2}$ right?

7. Originally Posted by genlovesmusic09
okay so i go to v= pi ∫ 1/x^4 dx

so i need to do du/u... and if so du= 3x^3 so do i add a negative on the outside?
No, that is the wrong integral. That would be the volume if the region were rotated around the x axis. Since you are rotating around the y-axis, a small segment above the x-axis, with height y= $\displaystyle 1/x^2$ and thickness dx, rotated around the y-axis, will form a ring with $\displaystyle 2\pi x$, the circumference of the circle, height $\displaystyle 1/x^2$ and thickness dx and so volume $\displaystyle (2\pi x)(1/x^2)(dx)$. The entire volume is the "sum" of those which becomes the integral
$\displaystyle 2\pi\int_{x=1}^2 \frac{dx}{x}$.