# surface integral

• Feb 26th 2013, 12:38 PM
DonnieDarko
surface integral
Find surface area given when rotating $\displaystyle y=(x+5)^{\frac{1}{2}}$ and $\displaystyle 5^{\frac{1}{2}} \leqslant x \leqslant 3$ around y asis.
How do i do this? Any help :)
• Feb 26th 2013, 01:30 PM
HallsofIvy
Re: surface integral
Where did you get this problem? If you are taking a Calculus class in which you are given a problem like this, you should have learned several ways to do problems like this. The simplest, for this problem, is the "disk method". For a specific y value, $\displaystyle x= y^2- 5$ and the radius of the circle, when rotated around the y-axis, is x. The area of that circle is $\displaystyle \pi x^2= \pi(y^2- 5)^2= \pi(y^4- 10y^2+ 25)$. Taking "dy" as the thickness of the disk, the volume is $\displaystyle \pi(y^2- 10y+ 25)$. Integrate that. (I suspect that your limits are intended to be $\displaystyle 5^{1/2}\le y\le 3$. Check it.)
• Feb 26th 2013, 01:50 PM
DonnieDarko
Re: surface integral
Yep, i see now. Thank you very much , now i realize it. We had almost same problem also.