# Solids of revolution

1. Mar 10, 2010

### 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?

Last edited: Mar 10, 2010
2. Mar 10, 2010

### 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 ).

3. Mar 10, 2010

### James889

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

4. Mar 10, 2010

### tiny-tim

Yes, but how did they get toegther?

Anyway
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.

5. Mar 10, 2010

### James889

I am so bad at this :tongue:

6. Mar 10, 2010

### tiny-tim

Have you got it now?

If not, show us what you have so far.

7. Mar 10, 2010

### James889

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

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

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

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

8. Mar 10, 2010

### tiny-tim

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)