1. ## Integral volume

I have the following problem. I need to find the volume of a solid that is obtained when the region under the curve y = 1-(x^2/16) is revolved around the x axis between x=-4 and x=4

So far i have: pi (integral 4,-4) (1-(x^2/16))^2 dx

Any help would be appreciated

2. Originally Posted by menco
I have the following problem. I need to find the volume of a solid that is obtained when the region under the curve y = 1-(x^2/16) is revolved around the x axis between x=-4 and x=4

So far i have: pi (integral 4,-4) (1-(x^2/16))^2 dx

Any help would be appreciated
Dear menco,

$\displaystyle V=\displaystyle\pi\int_{-4}^{4}\left(1-\frac{x^2}{16}\right)^2dx$

All you have to do is expand the brackets and integrate.

$\displaystyle V=\displaystyle\pi\int_{-4}^{4}\left(1-\frac{x^2}{8}+\frac{x^4}{256}\right)dx$

$\displaystyle V=\displaystyle\pi\left(\int^4_{-4}dx-\int^4_{-4}\frac{x^2}{8}dx+\int^4_{-4}\frac{x^4}{256}dx\right)$

Hope you can continue.

3. I went through it and as far as i know I find the derivative of the integral then substitube 4 for x and minus -4 for x

I get a final result of 15.56919154 units

4. Originally Posted by menco
I went through it and as far as i know I find the derivative of the integral then substitube 4 for x and minus -4 for x

I get a final result of 15.56919154 units
The correct answer should be, $\displaystyle \dfrac{64\pi}{15}=13.40412865531645$

5. Well i figured it out however i couldnt work out how you got (64pi/15) I just got (pi x 4.26')

Thanks for the help

6. Originally Posted by menco
Well i figured it out however i couldnt work out how you got (64pi/15) I just got (pi x 4.26')

Thanks for the help
Used a computer algebra system. You could do it by hand but it would take a lot of time because you have to deal with huge fractions.