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

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November 4th 2012, 10:30 AM
mdhiggenz

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(x²+y²) That lies inside the cylinder x²+y²=2x

First thing I did was transform x²+y²=2x into polar coords and got

r²=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 z²=x²+y²doing 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
November 4th 2012, 06:34 PM
chiro

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.
November 5th 2012, 04:54 AM
tom@ballooncalculus

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

$\displaystyle{ S\ =\ \int \int _D \ \sqrt{\big{(}\frac{\partial z}{\partial x}\big{)^2} + \big{(}\frac{\partial z}{\partial y}\big{)^2} + 1}\ \ dA}$

and afterwards convert,

$\displaystyle{ \int \int _D \ f(x,y)\ dA \ = \ \int \int _P f(r \cos \theta, r \sin \theta)\ r \ dr \ d \theta$

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:

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

and,

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

http://www.ballooncalculus.org/draw/intMulti/sixb.png

... where (key in spoiler) ...

Spoiler:

http://www.ballooncalculus.org/asy/chain.png

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable or as indicated on the line, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

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