Calculus
posted by Hannah on .
We're learning disks, shells, and cylinders in school but we have a substitute and I've been trying to teach this to myself. Can you check them please? =) Thank you!
1) Find the volume of the solid formed when the region bounded by curves y=x^3 + 1, x= 1, and y=0 is rotated about the xaxis.
My answer is 23/14.
2) Find the volume of the solid of revolution obtained by revolving the region bounded by y=1/x and the lines x=pi/8 and x=pi/2 around the xaxis.
I got 6/pi for this one.
3) Find the area bounded by the curves f(x)= x^3 + x^2 and g(x)= 2x^2 + 2x.
My answer: 5/12
Can you tell me if I did them wrong, and if I did do them wrong what I Did wrong?

standard way to form volume if spun around xaxis
y = x^3 + 1 > this becomes the radius of your disk
we need the xintercept,
let x^3 + 1 = 0
x = 1 > this becomes our left boundary
Volume = π∫(x^3 + 1)^2 dx from x = 1 to x = 1
= π∫(x^6 + 2x^3 + 1) dx
= π [ x^7/7 + x^4/2 + x] from 1 to 1
= π( 1/7 + 1/2 + 1  ( 1/7) + 1/2  1)
= P π( 2/7 + 2)= 16π/7
for #2,
y = 1/x
so y^2 = 1/x^2
V = π∫1/x^2 dx from π/8 to π/2
= π [  1/x ] from π/8 to π/2
= π ( 2/π  (8/π)
= 2+8 = 6
#3, first you need the intersection:
x^3 + x^2 = 2x^2 + 2x
x^3  x^2  2x = 0
x(x^2  x  2) = 0
x(x2)(x+1) = 0
they intersect at x = 1, x=0 and x=2
So you must find the volume separately from
x = 1 to x = 0 , and then from x = 0 to x = 2
from 1 to 0 , the cubic is the upper curve, while from
0 to 2, the parabola is the upper curve
So you have
V = π∫( (x^3 + x^2)^2  (2x^2 + 2x)^2 ) dx from 1 to 0 + π∫( (2x^2 + 2x)^2  (x^3 + x^2)^2 ) dx from 0 to 2
= etc. 
Thank you very much Reiny. I understand it a bit better now.

I did it and I get 4653pi/105 but this is wrong. Have I calculated incorrectly?