# Solids of revolution

James889
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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Homework Helper
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 ).

James889
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 $$xe^{2x^2}$$

Homework Helper
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" [Broken].

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James889
I am so bad at this :tongue:

Homework Helper
Have you got it now?

If not, show us what you have so far.

James889
$$D(x,y): \sqrt{x}e^{x^2}\leq y \leq 3,~~ 0\leq x \leq 1$$

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

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

$$\pi\cdot\frac{36-e^2}{4} - (0-\frac{e}{4})$$