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
Not sure if I am doing this correctly, would appreciate some feedback.
recheck your limits of integration ...
using symmetry ...
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
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)
radius = (1 - x)
cross-sectional area =
you should be able to finish from here.
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?
I just dropped a 0 for no apparent reason. Too much time studying. Thanks again!