# surface area using double integral

• Jul 25th 2010, 04:09 PM
kientri123
surface area using double integral
my problem is like this: find the surface area of the portion of the cone $\displaystyle z=\sqrt{x^2+y^2}$ that lies inside the cylinder $\displaystyle x^2+y^2=2x$.

My solution:
$\displaystyle x^2+y^2=2x$
=> $\displaystyle (x-1)^2+y^2=1$
So my circle now has center at (1,0), radius = 1
=> $\displaystyle y=\sqrt{1-(x-1)^2}$

$\displaystyle A= \int_{0}^{2} \int_{-\sqrt{1-(x-1)^2}}^{\sqrt{1-(x-1)^2}} \sqrt {(\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2 +1} dydx$

$\displaystyle A= \int_{0}^{2} \int_{-\sqrt{1-(x-1)^2}}^{\sqrt{1-(x-1)^2}} \sqrt 2} dydx$
$\displaystyle A= \int_{0}^{2} 2\sqrt2(\sqrt{1-(x-1)^2}dx$

I stuck at this point, I dont know if I set up a right limit of integration, please help?
• Jul 26th 2010, 02:17 AM
Ackbeet
I agree with your steps so far. I would probably do a series of substitutions: u substitution to get rid of the x-1, and then a trig substitution for the remainder of the integral. Or, after the u substitution, you could simply recognize the remaining integral as the area under a well-known type of curve. That would enable you simply to write down the answer without further computation.
• Jul 31st 2010, 09:37 PM
kientri123
I got it. Thanks :)
• Aug 2nd 2010, 02:58 AM
Ackbeet
You're welcome. Have a good one!