## Solids of revolution

Hi,

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

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

$$\pi \int_0^1 3^2-y^2 = \pi \int_0^1 9-xe^{2x^2}$$

$$\frac{-9\pi}{4}\cdot (e^{2x^2}-1)\bigg|_0^1$$

But this is all wrong, why?
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 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor 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 human-fly syndrome ).

 Quote by tiny-tim 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 human-fly syndrome ).
Hi Tim,

Im trying to find the area of a solid.
I just factored out the 9 from the integral, the $$1/4$$ is from integrating $$xe^{2x^2}$$

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## Solids of revolution

 Quote by James889 I just factored out the 9 from the integral, the $$1/4$$ is from integrating $$xe^{2x^2}$$
Yes, but how did they get toegther?

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.
 I am so bad at this
 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor Have you got it now? If not, show us what you have so far.
 Turned out i had misread the question, they did ask for the volume of the solid $$D(x,y): \sqrt{x}e^{x^2}\leq y \leq 3,~~ 0\leq x \leq 1$$ Same as before, baby steps. $$\pi\int_0^1 9-(\sqrt{x}e^{x^2})^2$$ $$9x-\frac{e^{2x^2}}{4}\bigg|_0^1$$ $$\pi\cdot\frac{36-e^2}{4} - (0-\frac{e}{4})$$
 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor Fine, except the last term should be e0/4, = 1/4. (and use more brackets, to show you have the π in the right place)