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.