Calculate the area of $z=\frac{x^2}{2}+\frac{y^2}{2}\;$ that is enclosed by $x^2+\frac{y^2}{4}=1$

Question

The exercise is the text in the title. I'm studying surface integrals. To start I thougt to make a change to cartesian coordinates with $z$ as a function of $x$ and $y$, that is, $z=\frac{x^2+y^2}{2}$ . With this I make the following parameterization: $O(x,y)=[x/2,y,z=\frac{x^2+y^2}{2}]$ Then I calculated the norm of the normal vector of $O(x,y)$ that is $(-x,\frac{-y}{2}, 1/2)$. Is it okay what I did? How did I continue the exercise?

Answer 1

According to the definition of surface area you have to evaluate $$A=\iint_S\sqrt{1+f_x^2+f_y^2}dxdy=\iint_S\sqrt{1+x^2+y^2}dxdy$$ where $z=f(x,y)=\frac{x^2+y^2}{2}$ (a paraboloid) and $S=\{(x,y):x^2+\frac{y^2}{4}\leq 1\}$ (the interior of an ellipse). After letting $X=x$, and $Y=y/2$ you should be able to use the polar coordinates. Unfortunately in the final result elliptic integrals are involved and there is no closed form.