Gaussian Curvature: Calculating for g(x,y)=xy\frac{1}{(x^2+y^2+1)^2}

[foxjwill]

Homework Statement

Is the gaussian curvature at a point on the surface $$g(x,y)=xy$$

$$\frac{1}{(x^2+y^2+1)^2}?$$​

Homework Equations

shape operator:

$$S(\textbf{x})=-D_\textbf{x}\hat{\textbf{n}}=\frac{\partial (n_x, n_y)}{\partial (x,y)}$$​

Gaussian Curvature = $$|S(\textbf{x})|$$

$$\hat{\textbf{n}}=\frac{\nabla g}{\|\nabla g\|}$$

The Attempt at a Solution

I basically plugged stuff into the above equations. I'm not sure if they're all correct.

 

[lippyka]

\begin{align}\nabla g &= (y\frac{1}{(x^2+y^2+1)^2},x\frac{1}{(x^2+y^2+1)^2}) \\\|\nabla g\| &= \sqrt{y^2\frac{1}{(x^2+y^2+1)^4} + x^2\frac{1}{(x^2+y^2+1)^4}} \\&= \frac{\sqrt{x^2+y^2}}{(x^2+y^2+1)^2} \\\hat{\textbf{n}} &= \frac{(y\frac{1}{(x^2+y^2+1)^2},x\frac{1}{(x^2+y^2+1)^2})}{\frac{\sqrt{x^2+y^2}}{(x^2+y^2+1)^2}} \\&= (\frac{y(x^2+y^2+1)^2}{(x^2+y^2)^{3/2}}, \frac{x(x^2+y^2+1)^2}{(x^2+y^2)^{3/2}}) \\D_\textbf{x}\hat{\textbf{n}} &= \frac{\partial (\frac{y(x^2+y^2+1)^2}{(x^2+y^2)^{3/2}}, \frac{x(x^2+y^2+1)^2}{(x^2+y^2)^{3/2}})}{\partial (x,y)} \\&= (\frac{(x^2+y^2+1)^2(-3yx)+(x^2+y^2)^{\frac{3}{2}}(2xy)}{(x^2+y^2)^{\frac{5}{2}}}, \frac{(x^2+y^2+1)^2(-3xy)+(x^2+y^2)