Integration of graph $z = x^2-y^2$

Question

How would I set up the integral using the characterization of $S$?

Answer 1

First write the equation of the surface as: $\Phi(x,y,z)=z-x^2+y^2=0$. Taking the gradient of $\Phi$, one gets a vector $\vec n$ normal to the surface:

$$\vec n = \vec\nabla\Phi(x,y,z)=(-2x,2y,1)\,.$$

This vector forms an angle $\theta$ with the $z$-axis whose cosine is just the third component of this vector divided by its magnitude:

$$ \cos(\theta) = \frac{1}{\sqrt{4x^2+4y^2+1}}\,.$$

The element of area for this surface is $dxdy/\cos(\theta)$, so the area $A$ of the portion of the surface with projection $\Omega$ in to the $xy$ plane is given by:

$$A=\iint_\Omega\frac{dxdy}{\cos(\theta)}=\iint_\Omega\sqrt{4x^2+4y^2+1}\;dxdy\,.$$

In this case, the region $\Omega$ of the $xy$ plane is the triangle bounded by the lines: $y=x,y=-x$ and $x=2$, i.e. to the inside of the triangle with vertices at $(0,0),(2,-2),(2,2)$. Therefore, doing first the integral in $y$, and then the the integral in $x$ one gets:

$$A=\int_0^2dx\int_{-x}^x\sqrt{4x^2+4y^2+1}\;dy$$

Answer 2

The function is positive in the shaded region

The two lines are where $z=0$, or $y=x$ and $y=-x$

Therefore, the domain is $-x \le y \le x, \ 0 \le x \le 2$ and the surface integral is given by $$ \int_{0}^{2}\int_{-x}^{x} \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} dy\ dx $$