Express the area of the given surface as an interated double integral in polar coords

Express the area of the given surface as an interated double integral in polar coordinates, and then find the surface area.
A) the portion of the cone z=sqrt(x^{2} +y^{2} ) That lies inside the cylinder x^{2} +y^{2} =2x
First thing I did was transform x^{2} +y^{2} =2x into polar coords and got
r^{2} =2rcos(theta)
r=2cos(theta)
So I pretty much guessed that the integral goes from 0≤ r ≤ 2cos
and also guessed that theta went from 0≤theta ≤pi/2
Not really sure how to prove that is goes from theta to pi/2 just assumed it was in quardrant 1.
Now here is where I get confused,
I then got the partials of z^{2} = x^{2} +y^{2 } doing so I got
dz/dx=rcostheta
dz/dy=rsintheta
I plug it into the formula for surface area which is sqrt[(dz/dx)^{2} + (dz/dy)^{2} +1]da and im pretty sure this is incorrect.
Thanks for the help Re: Express the area of the given surface as an interated double integral in polar co

Hey mdhiggenz.
x^2 + y^2 = 2x gives
x^2 - 2x + y^2 = 0 or
(x-y)^2 = 0
I'm a bit unsure on how this is a cylinder: Usually a cylinder (axis-aligned) is x^2 + y^2 = r^2 with z varying if running along the z-axis.
I might be misinterpreting your post though. Re: Express the area of the given surface as an interated double integral in polar co

Quote:

Originally Posted by

chiro or
(x-y)^2 = 0

Whoops (Wink)
It's a circle radius one, centre (1,0). (So z can indeed vary giving a cylinder.)
Use
$\displaystyle \displaystyle{ S\ =\ \int \int _D \ \sqrt{\big{(}\frac{\partial z}{\partial x}\big{)^2} + \big{(}\frac{\partial z}{\partial y}\big{)^2} + 1}\ \ dA}$
and afterwards convert,
$\displaystyle \displaystyle{ \int \int _D \ f(x,y)\ dA \ = \ \int \int _P f(r \cos \theta, r \sin \theta)\ r \ dr \ d \theta$
So for the partial wrt x, http://www.ballooncalculus.org/draw/intMulti/sixa.png
... and similarly for y. Etc.
And theta traces the top half of the circle as it goes from 0 to pi/2 and the bottom half as it carries on from there up to pi. Spoiler :

$\displaystyle \displaystyle \begin{align*} \sqrt{(\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial x})^2 + 1}\ &=\ \sqrt{\frac{x^2}{x^2 + y^2}\ +\ \frac{x^2}{x^2 + y^2} + 1} \\ \\ &=\ \sqrt{2} \end{align*}$

and,

$\displaystyle \displaystyle \begin{align*} S\ &=\ \int \int _D \ \sqrt{(\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2 + 1}\ dA \\ \\ &=\ \int \int _D \ \sqrt{2}\ dA \\ \\ &=\ \int \int _P \ \sqrt{2}\ r\ dr\ d\theta \\ \\ &=\ \int _0^{\pi} \int _0^{2 \cos \theta} \ \sqrt{2}\ r\ dr\ d\theta \end{align*}$

http://www.ballooncalculus.org/draw/intMulti/sixb.png
... where (key in spoiler) ...

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