Volume of a Solid of Revolution

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  • #1
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Homework Statement


Find the volume generated by rotating the area bounded by [itex] y^2 = 8x, x = 2 [/itex] and the x-axis about the y-axis.


Homework Equations


Volume of cylinder, volume of disk.


The Attempt at a Solution


I think this can be solved by subtracting the 'empty' volume from the volume of a cylinder of radius 2:

[tex] \pi r^2h - \int_0^4 \frac{\pi y^2}{8}\,dy [/tex]

I get an answer of [itex] \pi 40/3 [/itex] but the solution provided is [itex] 128/5\pi [/itex].

I'm sure my arithmetic is not that far off :-)

What did I do wrong?

Thanks,
Sheldon
 

Answers and Replies

  • #2
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Homework Statement


Find the volume generated by rotating the area bounded by [itex] y^2 = 8x, x = 2 [/itex] and the x-axis about the y-axis.


Homework Equations


Volume of cylinder, volume of disk.


The Attempt at a Solution


I think this can be solved by subtracting the 'empty' volume from the volume of a cylinder of radius 2:

[tex] \pi r^2h - \int_0^4 \frac{\pi y^2}{8}\,dy [/tex]

I get an answer of [itex] \pi 40/3 [/itex] but the solution provided is [itex] 128/5\pi [/itex].

I'm sure my arithmetic is not that far off :-)

What did I do wrong?

Thanks,
Sheldon
I see an error, but I still don't get the provided answer. I believe the following is correct:

[tex] \pi r^2h - \int_0^4 \frac{\pi y^4}{64}\,dy [/tex]

However I get the answer: [itex] 64\pi/5 [/itex]

Thanks!
Sheldon
 
  • #3
HallsofIvy
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I see an error, but I still don't get the provided answer. I believe the following is correct:

[tex] \pi r^2h - \int_0^4 \frac{\pi y^4}{64}\,dy [/tex]

However I get the answer: [itex] 64\pi/5 [/itex]

Thanks!
Sheldon
That integral is the correct one. I don't see how you could get [itex]64\pi/5[/itex] out of it! Show exactly how you integrated.
 
  • #4
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That integral is the correct one. I don't see how you could get [itex]64\pi/5[/itex] out of it! Show exactly how you integrated.
When x = 2, y = 4, so that's the height of the cylinder:

[tex] \pi(2^2)4 - \int_0^4 \frac{\pi y^4}{64}\,dy = 16\pi -\pi\left[\frac{y^5}{(5)(64)}\right]_0^4 [/tex]


[tex] 16\pi - \frac{16\pi}{5} = \frac{80\pi}{5} - \frac{16\pi}{5} = \frac{64\pi}{5} [/tex]

Am I making some dumb mistake here?

Thanks for your help!
Sheldon
 
  • #5
Your integration is actually perfect.

Just notice that your answer is 1/2 the book's answer. That should lead you in the right direction.
 
  • #6
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Your integration is actually perfect.

Just notice that your answer is 1/2 the book's answer. That should lead you in the right direction.
Hi Wolf,

I noticed that, but I don't understand it. I am starting to wonder if the book answer is wrong, but this is a new topic, and I am not feeling very confident.

If I multiply by two (or integrate from -4 to 4), then I get the book answer. But the problem states that the area is bounded by the X-axis. Multiplying by 2 would give the volume for the object both above and below the X-axis. Isn't that right?

Thanks for your help,
Sheldon
 

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