MHB How to Find the Surface Area of a Pringle Using Double Integration?

[wonder_koo]

does anyone know how to solve this/can lead me on a direction to where I will get an answer that actually makes sense lol? I keep getting a negative answer/0. For context, I'm tryna find the surface area of a pringle and all the sources I've visited always estimate the projected 2D region as a circle + rounded off the y and x coefficients to the same number which allows them to simplify to polar coordinates w/ just r in the integration and then solve it BUT this gives only an estimated area of the pringle

 

[pasmith]

The question is:

Find the surface area of the surface z = \frac{y^2}{4} - \frac{x^2}{4.507} lying within the ellipse <br /> \frac{x^2 }{6.76} + \frac{y^2}{4} = 1.

You can parametrize the interior of the ellipse (x/a)^2 + (y/b)^2 = 1 as <br /> \begin{split}<br /> x &amp;= ar\cos \theta \\<br /> y &amp;= br\sin \theta<br /> \end{split} where 0 \leq r \leq 1 and 0 \leq \theta \leq 2\pi. Then the surface area of z = f(r,\theta) is given by <br /> S = \int_0^{2\pi} \int_0^1 \left\| \frac{\partial \mathbf{r}}{\partial r} \times \frac{\partial \mathbf{r}}{\partial \theta}\right\|\,dr\,d\theta where \left\| \frac{\partial \mathbf{r}}{\partial r} \times \frac{\partial \mathbf{r}}{\partial \theta}\right\|^2 = <br /> b^2\left(\frac{\partial f}{\partial \theta} \sin \theta - \frac{\partial f}{\partial r}r\cos \theta \right)^2 + a^2\left(<br /> \frac{\partial f}{\partial r}r\sin \theta + \frac{\partial f}{\partial \theta}\cos\theta<br /> \right)^2 + a^2b^2r^2.