Calculating Area of 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?

 

[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 ).

 

[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}

 

[tiny-tim]

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" .

 

[James889]

I am so bad at this

 

[tiny-tim]

Have you got it now?

If not, show us what you have so far.

 

[James889]

Turned out i had misread the question, they did ask for the volume of the solid
<br /> D(x,y): \sqrt{x}e^{x^2}\leq y \leq 3,~~ 0\leq x \leq 1<br />

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})

 

[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)