Surface Integration Problem: How to Compute \int\int_{D}zdS?

[rshalloo]

Homework Statement

Let $D=\{(x,y,z)| z^{2}=1+x^{2}+y^{2} , 1<z<3\}$ Compute$\int\int_{D}zdS$

Homework Equations

From lectures I know;
$\int\int_{D}\delta dS=\int\int_{D}\delta\sqrt{(\frac{\partial f}{\partial x})^{2}+(\frac{\partial f}{\partial y})^{2}+1}dxdy$

The Attempt at a Solution

I'm not sure what I'm doing is correct my answer seems wrong;

$z^{2}=1+x^{2}+y^{2}$
so

$z=\sqrt{1+x^{2}+y^{2}}$

Taking partial derivatives of x and y and substituting into equation from 2 I get

$\int\int_{D}\sqrt{5x^{2}+5y^{2}+1}dxdy$

making change of variables to cylindrical coordinates

$\int\int_{D}\sqrt{5r^{2}+1}r drdt$

making substitution

$u=5r^{2}+1$

I get

$\int\int_{D}\sqrt{u} dudt$

I brought the limits through as well to have limits for u of 0 and 41 and limits of t of 0 and 2$\pi$

giving me a final answer of 1099.675108 which seems completely wrong;

Any suggestions?

 

[hamsterman]

I got 2 instead of 5. Really, $\frac {\partial z}{\partial x} = \frac{x}{z}$ and $\frac {\partial z}{\partial y} = \frac{y}{z}$
$z \sqrt{1+\frac{x^2}{z^2} + \frac{y^2}{z^2}} = z \sqrt{ \frac{z^2 + x^2 + y^2}{z^2} } = \sqrt{1+2x^2+2y^2}$

I then got 73.39 which is a lot more likely.