surface area using double integral

**kientri123** · July 25th 2010, 04:09 PM

my problem is like this: find the surface area of the portion of the cone $z=\sqrt{x^2+y^2}$ that lies inside the cylinder $x^2+y^2=2x$ .

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

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

$A= \int_{0}^{2} \int_{-\sqrt{1-(x-1)^2}}^{\sqrt{1-(x-1)^2}} <br /> \sqrt 2}<br /> dydx$
$A= \int_{0}^{2} 2\sqrt2(\sqrt{1-(x-1)^2}dx$

I stuck at this point, I dont know if I set up a right limit of integration, please help?

**Ackbeet** · July 26th 2010, 02:17 AM

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.

**kientri123** · July 31st 2010, 09:37 PM

I got it. Thanks

**Ackbeet** · August 2nd 2010, 02:58 AM

You're welcome. Have a good one!

Thread: surface area using double integral

LinkBack

Thread Tools

Display

surface area using double integral

Similar Math Help Forum Discussions

Surface area double integrals

Double integral surface area with polar coordinates

Computing surface area with double integrals help

surface area ( double integral)

Find the area of a surface (double integral)

Search Tags