Find the surface area of the hyperbolic paraboloid $z = 13 + x^2 – y^2$ that lies above the closed disk $x^2 + y^2 \le 4$.
I know that the surface area is equal to: $\iint$$\sqrt{1+[f_x(x,y)]^2 + [f_y(x,y)]^2}\, dA$ over some region $R$.
$$f_x(x,y) = 2x$$
$$f_y(x,y) = -2y$$
It's now equal to:
$\iint$$\sqrt{1+4x^2 + 4y^2}$ $dA$ for some region $R$
I'm stuck at this point. Can someone help me finish this? Thank you!
Best Answer
Square roots in integrals generally make problems difficult, but here it's a clue that the problem can be simplified by making a change of variable to polar coordinates. Let $x = r \cos \theta$, $y=r\sin\theta$.
To get you started, we write the region of integration in polar coordinates. The closed disk $x^2+y^2\leq 4$ is given in polar coordinates by $\{(r,\theta)|0\leq r\leq 2, 0 \leq \theta < 2\pi\}$. This gives you very easy limits of integration.
You've already found the integrand in terms of $x$ and $y$. Use the polar coordinate transformation to express it in terms of $r$ and $\theta$, and use trigonometric identities to make the integrand very simple.
Finally, remember that whenever you make a change in variable you must multiply by the appropriate Jacobian determinant. (It's easy to forget this step. Remember that $dA=dxdy$ in our original coordinates, but $dA\neq drd\theta$ in our new coordinates - something's missing.)