Solids of revolution

  • Thread starter James889
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  • #1
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Hi,

I have the area [tex]D(x,y): \sqrt{x}e^{x^2}\leq y \leq 3,~~ 0\leq x \leq 1[/tex]

That is rotated about the x axis, and i need to calculate the area

[tex]\pi \int_0^1 3^2-y^2 = \pi \int_0^1 9-xe^{2x^2}[/tex]

[tex]\frac{-9\pi}{4}\cdot (e^{2x^2}-1)\bigg|_0^1[/tex]

But this is all wrong, why?
 
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Answers and Replies

  • #2
tiny-tim
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Hi James889! :smile:

First, are you trying to find the area of a surface of revolution, or the volume of a solid of revolution? :confused:

Anyway, how did you get 9/4 out of that?

Do the two parts separately (you seem to be suffering from a sort of human-fly syndrome :rolleyes:).
 
  • #3
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Hi James889! :smile:

First, are you trying to find the area of a surface of revolution, or the volume of a solid of revolution? :confused:

Anyway, how did you get 9/4 out of that?

Do the two parts separately (you seem to be suffering from a sort of human-fly syndrome :rolleyes:).

Hi Tim,

Im trying to find the area of a solid.
I just factored out the 9 from the integral, the [tex]1/4[/tex] is from integrating [tex]xe^{2x^2}[/tex]
 
  • #4
tiny-tim
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I just factored out the 9 from the integral, the [tex]1/4[/tex] is from integrating [tex]xe^{2x^2}[/tex]

Yes, but how did they get toegther? :confused:

Anyway
Im trying to find the area of a solid.

You mean the surface area?

But you're using ∫πy2dx, which is a volume.

For the correct formula, see http://en.wikipedia.org/wiki/Surface_of_revolution" [Broken].
 
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  • #5
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I am so bad at this :tongue:
 
  • #6
tiny-tim
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Have you got it now?

If not, show us what you have so far. :smile:
 
  • #7
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Turned out i had misread the question, they did ask for the volume of the solid
[tex]
D(x,y): \sqrt{x}e^{x^2}\leq y \leq 3,~~ 0\leq x \leq 1
[/tex]

Same as before, baby steps.
[tex]\pi\int_0^1 9-(\sqrt{x}e^{x^2})^2[/tex]

[tex]9x-\frac{e^{2x^2}}{4}\bigg|_0^1[/tex]

[tex]\pi\cdot\frac{36-e^2}{4} - (0-\frac{e}{4})[/tex]
 
  • #8
tiny-tim
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Fine, except the last term should be e0/4, = 1/4. :wink:

(and use more brackets, to show you have the π in the right place)
 

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