# Volume of revolution problem

• Jul 16th 2008, 05:12 PM
Zolthas
Volume of revolution problem
This is the only problem on my homework that I can't seem to figure out.

Find the volume of the solid formed by revolving the region bounded by the graphs of y=x^2+1, y=0, x=0, and x=1 about the y-axis using the washer method.

Any help would be greatly appreciated.
• Jul 16th 2008, 06:17 PM
meymathis
With these it is crucial to draw the picture.

You have $y=x^2+1$ is a parabola at the top. See the attached image.

So method by washers: the washers are perpendicular to the axis of rotation. So they are horizontal in this case, and integration is parallel to the axis of rotation (so we need x's as a function of y).

With washers you need to figure you the two radii (inner and outer).
Since the axis of rotation is the line x=0, we need the distances from this line.

From y=0 to y=1, inner radius is 0, outer is 1 (note that the
From y=1 to y=2, inner radius is ?, outer is 1

Let's figure out the inner radius: we need to write the inner radius is a function of y. What we have is $y = x^2+1$
$y-1 = x^2$
$\pm\sqrt{y-1} = x$

but this is the positive one (the curve is for positive x) so
From y=1 to y=2, inner radius is $\sqrt{y-1}$, outer is 1

Now (informally) the volume of the washer is $\pi(R^2-r^2)dy$ where R is the outer and r is the inner radii.

So

$\int_0^1\pi(1^2-0^2)dy + \int_1^2 \pi(1^2-(\sqrt{y-1})^2)dy$

Make sense?
• Jul 16th 2008, 06:43 PM
Zolthas
Ok so after integrating I got:
$[\pi - 0] + [ 2\pi - 3\pi/2]
$

$
\pi + \pi/2 = 3\pi/2
$

I hope I did this right, I have been making little dumb mistakes after working on this stuff for hours.
• Jul 16th 2008, 06:57 PM
ticbol
Quote:

Originally Posted by Zolthas
This is the only problem on my homework that I can't seem to figure out.

Find the volume of the solid formed by revolving the region bounded by the graphs of y=x^2+1, y=0, x=0, and x=1 about the y-axis using the washer method.

Any help would be greatly appreciated.

( I am late, as usual. My typing is just a little bit faster than the speed of a starfish walking. See if you can use my solution anyway.)

By washer method?

It is done easier by shell method, where dV = (2pi*x)[(x^2 +1)*dx] and the integration is from x=0 to x=1.

If it should be by washer method, then divide the solid into two solids.

One solid will be a disc that is below the y=1 horizontal line. It has a radius of 1 unit and a height of 1 unit also. Use disc method here.
dV1 = pi[1^2]dy
dV1 = pi dy

The other solid is the one above the y=1 horizontal line. And here you can use the washer method.
dV2 = pi[(1^2 -x^2)*dy]
dV2 = pi[1 -(y-1)]dy
dV2 = pi[2 -y]dy

So for the original solid,
V = (pi)INT.(0 to 1)[dy] +(pi)INT.(1 to 2)[2 -y]dy

-------
Either way, by shell method or by disc+washer method, you should get the volume to be 1.5pi cubic units.
• Jul 16th 2008, 07:02 PM
meymathis
Looks good to me.(Rock)
• Jul 16th 2008, 07:08 PM
Zolthas
Quote:

Originally Posted by ticbol
By washer method?

It is done easier by shell method, where dV = (2pi*x)[(x^2 +1)*dx] and the integration is from x=0 to x=1.

I agree that the shell method would be easier, but the problem makes you use the washer method.

Quote:

Either way, by shell method or by disc+washer method, you should get the volume to be 1.5pi cubic units.
So I got the right answer then (3pi/2).

Thanks a ton for the help guys, I really appreciate it.