How to calculate the surface area of the elliptical paraboloid given by $z=x^2 + 4y^2$

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Also, $z$ should lie in the closed interval of $[0,4]$.

I know the general method to this - to find the function $\langle x, y, z(x,y)\rangle$ differentiated with respect to $x$, then with respect to $y$, finally taking their cross-product and double integrating. However, I would really appreciate a step-by-step solution as not only do I end up with an integral I don't exactly know how to integrate, but I also feel like I am making small errors and would just like to make sure I'm on the right path.

Thank you so much.

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$dS = \|(-2x,-8y,1)\|\ dy\ dx\\ \sqrt{4x^2 + 64y^2 + 1} \ dy\ dx$

$\iint \sqrt{4x^2 + 64y^2 + 1} \ dy\ dx$

Limits: $z=4$

$4 = x^2 + 4y^2$

We can stay in cartesian in which case:

$x = 2\sqrt {1-y^2}\\y=1$

Or, we could flip to an elliptical version of polar coordinates

$x = 2r\cos t\\y = r\sin t\\ dy\ dx = 2r \ dr\ d\theta.$

Up to you.

At $z=0,x,y=0$

Lets stay in cartesian, as I don’t know if you are comfortable with coordinate transformations.

$\int_0^1\int_0^{2\sqrt{1-y^2}} \sqrt{4x^2 + 64y^2 + 1}\ dx \ dy$