# Thread: Volume of revolution problem

1. ## 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.

2. 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?

3. 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.

4. 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.

5. Looks good to me.

6. 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.

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.