# Volume of a Solid

• October 4th 2008, 01:57 PM
veronicak5678
Volume of a Solid
I was hoping someone could double-check this for me:

Find the volume of a solid obtained by rotating the region bounded by y = x^2 + 1 , y = 9 - x ^2, about y = -1.

(Sorry I don't know how to use the symbols, so this may be hard to read)

pi integral from -5 to 5 ( ( 1+ 9-x^2)^2 - ( 1 + x^2 + 1 )^2) dx

After some algebra...

pi (96 x - (4/3) x ^3) from -5 to 5

626.66 pi

Not sure if I am doing this correctly, would appreciate some feedback.
• October 4th 2008, 03:16 PM
skeeter
recheck your limits of integration ...

$x^2 + 1 = 9 - x^2$

$2x^2 - 8 = 0$

$2(x+2)(x-2) = 0$

using symmetry ...

$V = 2\pi \int_0^2 [(9 - x^2) + 1]^2 - [(x^2 + 1) + 1]^2 \, dx = 256\pi
$
• October 4th 2008, 03:24 PM
veronicak5678
I see what I did wrong. Thsnks for your help.

There's one more thing I need help with...

The base of a solid is a square with vertices (1,0), (0,1), (0, -1), and (-1, 0). Each cross-section perpendicular to the x-axis is a semi-circle. Find the volume.

I was attempting to find a quarter of the volume and just multiply by 4. I found the line above the x-axis to be y = 1-x, so attempting to use that, i came up with

8pi * Integral (1-x^2)(1-x) dx
• October 4th 2008, 03:52 PM
skeeter
I would do half the volume and double.

on the positive side of x ...

region is bounded above by y = 1-x and below by y = x-1

diameter = (1 - x) - (x - 1) = 2(1 - x)

cross-sectional area = $\frac{\pi}{2}(1-x)^2$

$V = 2 \int_0^1 \frac{\pi}{2}(1-x)^2 \, dx$

you should be able to finish from here.
• October 4th 2008, 03:58 PM
veronicak5678
OK, this wasn't finished after all...

On the first one:

Using the new limits of integration, I come up with 362.666 pi, not 256. Here are my last few steps:

2pi Integral from 0 to 2 96 -4x^2

Integrated to 96x -(4x^3)/3 from 0 to 2

192-10.66 * 2pi

Why do we have different answers?
• October 4th 2008, 05:59 PM
skeeter
$V = 2\pi \int_0^2 [(9 - x^2) + 1]^2 - [(x^2 + 1) + 1]^2 \, dx$

$V = 2\pi \int_0^2 (10 - x^2)^2 - (x^2 + 2)^2 \, dx$

$V = 2\pi \int_0^2 [(100 - 20x^2 + x^4) - (x^4 + 4x^2 + 4)^2 \, dx$

$V = 2\pi \int_0^2 (96 - 24x^2) \, dx$

$2\pi\left[96x - 8x^3\right]_0^2 = 2\pi(192 - 64) = 2\pi(128) = 256\pi$
• October 4th 2008, 07:16 PM
veronicak5678
I just dropped a 0 for no apparent reason. Too much time studying. Thanks again!