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So I have to calculate the area of a paraboloid $2z=\frac{x^2}{a}+\frac{y^2}{b}$ inside a cylinder $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. So I plugged in my formula: $$\iint_R\sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2}dA$$ The terms in square root simplify to $\sqrt2$ if I have did math correctly, But I am having difficulty in limits and subsequent solving. I thought of letting $x$ vary from $-a\sqrt{1-\left(\frac{y}{b}\right)^2}$ to $a\sqrt{1-\left(\frac{y}{b}\right)^2}$ and later $y$ as $-b$ to $b$.

I thought of using polar coordinates but I dont know how to apply it in case of ellipse where $r$ is variable so i would like an answer solving and explain by polar coordinates. Nevertheless, I am getting the wrong answer, so I really would appreciate some other method besides this formula. Please help and thanks in advance

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    $\begingroup$ For starters, the integrand is certainly not $\sqrt 2$. For example, what is it at the origin? If you know about change of variables in double integrals, suitably stretched polar coordinates would be ideal. $\endgroup$
    – Ted Shifrin
    Nov 21 at 0:29
  • $\begingroup$ Hint: Let $u=\frac{x}{a}$ and $v=\frac{y}{b}$. This leads to $ab\iint_R\sqrt{1+u^2+v^2}\mathrm du \mathrm dv$. Now you can proceed with polar coordinates. $\endgroup$
    – Anton Vrdoljak
    Nov 21 at 8:05

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