# Calculating the area of a paraboloid inside a cylinder

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

• 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. Nov 21 at 0:29
• 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. Nov 21 at 8:05