Volumes By Cylindrical Shells (2 questions)

  • Thread starter Jet1045
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  • #1
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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.

Homework Equations







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 :)
 

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Answers and Replies

  • #2
tiny-tim
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Hey Jet1045! :smile:
(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. :wink:

Try again! :smile:
 
  • #3
Dick
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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.
 
  • #4
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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?!
 
  • #5
Dick
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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?
 
  • #6
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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 :)
 
  • #7
Dick
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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.
 
  • #8
49
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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.
 

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