# Surface Area in Polar Coordinates

• August 2nd 2006, 09:08 AM
MikeWakefield32
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.
• August 2nd 2006, 09:16 AM
ThePerfectHacker
Quote:

Originally Posted by MikeWakefield32
r is between 0 and theta

You mean,
$0\leq r \leq \theta$? :confused:

That is not possible, cuz the region is unbounded!
(The curve produces an Archimedean Spiral).
• August 2nd 2006, 09:18 AM
MikeWakefield32
1theta it says. I didn't think that mattered.
• August 2nd 2006, 09:46 AM
ThePerfectHacker
My fault the region is bounded. Because you gave theta a limitation. Sorry.
---
Then the integration is,
$\int_A \int \sqrt{1+f_x^2+f_y^2} dA$
Where, $A$ is the region you described. Since $A$ is simpler expressed in polar coordinates we will use polar integration.
---
The surface is,
$f(x,y)=\cosh \sqrt{x^2+y^2}$
Then,
$f_x=\frac{x}{\sqrt{x^2+y^2}} \sinh \sqrt{x^2+y^2}$
Then,
$f_y=\frac{y}{\sqrt{x^2+y^2}} \sinh \sqrt{x^2+y^2}$
The next part is two square each one,
$f_x^2=\frac{x^2}{x^2+y^2}\sinh^2 \sqrt{x^2+y^2}$
$f_y^2=\frac{y^2}{x^2+y^2}\sinh^2 \sqrt{x^2+y^2}$
$f_x^2+f_y^2=\frac{x^2+y^2}{x^2+y^2}\sinh^2 \sqrt{x^2+y^2}=\sinh^2 \sqrt{x^2+y^2}$
$1+f_x^2+f_y^2=1+\sinh^2 \sqrt{x^2+y^2}$
But,
$\cosh^2 u=1+\sinh^2 u$
Therefore,
$1+f_x^2+f_y^2=\cosh^2 \sqrt{x^2+y^2}$
Thus, the integration is,
$\int_A \int \sqrt{\cosh^2 \sqrt{x^2+y^2}} dA$
This simplifies as,
$\int_A \int |\cosh \sqrt{x^2+y^2}| dA$
But, $\cosh u >0$ thus,
$\int_A \int \cosh \sqrt{x^2+y^2} dA$
Now, finally you can use polar substituion,
$\sqrt{x^2+y^2}\rightarrow r$
$\int_2^4 \int_0^{\theta} r \cosh r dr, d\theta$
Can you solve it from here?
• August 2nd 2006, 10:14 AM
ThePerfectHacker
Quote:

Originally Posted by ThePerfectHacker
Can you solve it from here?

Let me continue.

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