# Thread: Surface Area in Polar Coordinates

1. ## Surface Area in Polar Coordinates

Find the surface area of the surface z=cosh(sqrt(x^2+y^2)) above the region in the xy plane given in polar coordinates:
r is between 0 and theta
theta is between 2 and 4

Ok. I used the formula:
Surface area equals the square root of the partial derivative of x squared plus the partial derivative of y squared plus 1. After doing it out, the (x^2+y^2) turned into 1, leaving me with the square root of 1 + (sinhsqrt(x^2+y^2))^2.
I took the integral of this from 0 to theta by switching the (x^2+y^2) into just r (polar coordinates). I got that this integral equaled cosh(r)tanh(r) and in turn cosh(theta)tanh(theta). Taking this integral from 2 to 4, I got cosh(4)-cosh(2). I thought I did everything correct, but this is not the answer. What did I do wrong?

Thanks for any help. I hope you can understand my thought process.

2. Originally Posted by MikeWakefield32
r is between 0 and theta
You mean,
$\displaystyle 0\leq r \leq \theta$?

That is not possible, cuz the region is unbounded!
(The curve produces an Archimedean Spiral).

3. 1theta it says. I didn't think that mattered.

4. My fault the region is bounded. Because you gave theta a limitation. Sorry.
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Then the integration is,
$\displaystyle \int_A \int \sqrt{1+f_x^2+f_y^2} dA$
Where, $\displaystyle A$ is the region you described. Since $\displaystyle A$ is simpler expressed in polar coordinates we will use polar integration.
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The surface is,
$\displaystyle f(x,y)=\cosh \sqrt{x^2+y^2}$
Then,
$\displaystyle f_x=\frac{x}{\sqrt{x^2+y^2}} \sinh \sqrt{x^2+y^2}$
Then,
$\displaystyle f_y=\frac{y}{\sqrt{x^2+y^2}} \sinh \sqrt{x^2+y^2}$
The next part is two square each one,
$\displaystyle f_x^2=\frac{x^2}{x^2+y^2}\sinh^2 \sqrt{x^2+y^2}$
$\displaystyle f_y^2=\frac{y^2}{x^2+y^2}\sinh^2 \sqrt{x^2+y^2}$
$\displaystyle f_x^2+f_y^2=\frac{x^2+y^2}{x^2+y^2}\sinh^2 \sqrt{x^2+y^2}=\sinh^2 \sqrt{x^2+y^2}$
$\displaystyle 1+f_x^2+f_y^2=1+\sinh^2 \sqrt{x^2+y^2}$
But,
$\displaystyle \cosh^2 u=1+\sinh^2 u$
Therefore,
$\displaystyle 1+f_x^2+f_y^2=\cosh^2 \sqrt{x^2+y^2}$
Thus, the integration is,
$\displaystyle \int_A \int \sqrt{\cosh^2 \sqrt{x^2+y^2}} dA$
This simplifies as,
$\displaystyle \int_A \int |\cosh \sqrt{x^2+y^2}| dA$
But, $\displaystyle \cosh u >0$ thus,
$\displaystyle \int_A \int \cosh \sqrt{x^2+y^2} dA$
Now, finally you can use polar substituion,
$\displaystyle \sqrt{x^2+y^2}\rightarrow r$
$\displaystyle \int_2^4 \int_0^{\theta} r \cosh r dr, d\theta$
Can you solve it from here?

5. Originally Posted by ThePerfectHacker
Can you solve it from here?
Let me continue.

To solve this integral the problem is to find,
$\displaystyle \int r \cosh r dr$
You gonna use integration by parts.
Let, $\displaystyle u=r$ and $\displaystyle v'=\cosh r$
Then, $\displaystyle u'=1$ and $\displaystyle v=\sinh r$
Using parts,
$\displaystyle r\sinh r -\int \sinh r dr$
Thus,
$\displaystyle r\sinh r - \cosh r$
Thus,
$\displaystyle \int_2^4 \int_0^{\theta} r\cosh r dr\, d\theta$ is,
$\displaystyle \int_2^4 r\sinh r-\cosh r \big|_0^\theta d\theta$
Thus,
$\displaystyle \int_2^4 \theta \sinh \theta - \cosh \theta +1 d\theta$
The problem now is to find,
$\displaystyle \int \theta \sinh \theta d\theta$
Let,
$\displaystyle u=\theta$ and $\displaystyle v'=\sinh \theta$
Then,
$\displaystyle u'=1$ and $\displaystyle v=\cosh \theta$
Thus, by parts,
$\displaystyle \theta \cosh \theta -\int \cosh \theta d\theta$
Thus,
$\displaystyle \theta \cosh \theta - \sinh \theta$
Thus, you have, (entire integral)
$\displaystyle \theta \cosh \theta -\sinh \theta -\sinh \theta +\theta\big|_2^4$