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Solids of revolution 
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#1
Mar1010, 03:56 AM

P: 184

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 3^2y^2 = \pi \int_0^1 9xe^{2x^2}[/tex] [tex]\frac{9\pi}{4}\cdot (e^{2x^2}1)\bigg_0^1[/tex] But this is all wrong, why? 


#2
Mar1010, 06:55 AM

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P: 26,148

Hi James889!
First, are you trying to find the area of a surface of revolution, or the volume of a solid of revolution? Anyway, how did you get 9/4 out of that? Do the two parts separately (you seem to be suffering from a sort of humanfly syndrome ). 


#3
Mar1010, 08:16 AM

P: 184

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
Mar1010, 08:36 AM

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Solids of revolution
Anyway … But you're using ∫πy^{2}dx, which is a volume. For the correct formula, see http://en.wikipedia.org/wiki/Surface_of_revolution. 


#5
Mar1010, 08:45 AM

P: 184

I am so bad at this



#6
Mar1010, 10:55 AM

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P: 26,148

Have you got it now?
If not, show us what you have so far. 


#7
Mar1010, 11:54 AM

P: 184

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{36e^2}{4}  (0\frac{e}{4})[/tex] 


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