# Volumes By Cylindrical Shells (2 questions)

## Homework Statement

HEY EVERYONE! Sorry for putting up another quesiton, but i'm almost done my assignment for the week. I understanding volumes by cylindrical shells, but i get confused when rotating about a line thats not the x or y axis. I have uploaded the two questions.

## The Attempt at a Solution

For the first question here is the integral i set up:

(2pi) integral from 0 to 4 of x(32-x^2-x^2)dx

For the second integral i set up:

(2pi) integral from y(8- cuberoot(64y)) dy

I am probably doing the same thing wrong in both questions since both are rotating about a line which the shaded region is touching. If someone could tell me where i'm going wrong in my integrals i'd appreciate it :)
THANKS :)

#### Attachments

tiny-tim
Homework Helper
Hey Jet1045! (2pi) integral from 0 to 4 of x(32-x^2-x^2)dx

Your cylinder must (like any cylinder) have a height a radius and a thickness.

Your thickness (dx) and your height are correct, but your radius is wrong. Try again! Dick
Homework Helper
You should picture the problem as finding the volume by integrating over the area of cylinders. The area of a cylinder is 2pi*r*h where h is the height of the cylinder and r is the radius of the cylinder from it's axis. Take the first first one, I'll agree that h=32-x^2-x^2 but I don't think r=x. r should be the distance of the cylinder from x=4. What's that? And think about the limits again too.

haha oh my god, how do you guys figure this stuff out so fast.... i feel so stupid lol.

OHHH so for the first question the radius (distance from the line x=4) would be 4-x then?

And for the second question with respect to y, the radius would be 8-y.

so my integrals are:
(2pi) integral from 0 to 4 of (4-x)(32-x^2-x^2)dx

AND

(2pi) integral from 0 to 8 of (8-y)(8- cuberoot(64y))dy

or am i still way off?! hahaha and are my limits of integration correct?!

Dick
Homework Helper
haha oh my god, how do you guys figure this stuff out so fast.... i feel so stupid lol.

OHHH so for the first question the radius (distance from the line x=4) would be 4-x then?

And for the second question with respect to y, the radius would be 8-y.

so my integrals are:
(2pi) integral from 0 to 4 of (4-x)(32-x^2-x^2)dx

AND

(2pi) integral from 0 to 8 of (8-y)(8- cuberoot(64y))dy

or am i still way off?! hahaha and are my limits of integration correct?!

Looks ok. But I still don't like the limits on the first one. Remember the cylinder are rotated around x=4. Do any of them cover the part of your graph with x<0?

Looks ok. But I still don't like the limits on the first one. Remember the cylinder are rotated around x=4. Do any of them cover the part of your graph with x<0?

GOOD CATCH, for some reason, im so used to having regions bounded only in one quadrant, but this does go into x<0. So the limits would be -4 to 4. So since it goes into x<0 this won't affect my radius of 4-x right?

btw the second question i just submitted and got right, so thank you :)

Dick
Homework Helper
GOOD CATCH, for some reason, im so used to having regions bounded only in one quadrant, but this does go into x<0. So the limits would be -4 to 4. So since it goes into x<0 this won't affect my radius of 4-x right?

btw the second question i just submitted and got right, so thank you :)

Right, 4-x works for the radius in both the positive and negative parts.

perfect, i just submitted my answer and it was right!
Thanks so much, for some reason, i just assumed that because the shaded region is touching the line which it rotates about, that i wouldnt have to subtract anything.