Surface integral of an paraboloid cut by a plane.

Question

I'm trying to calculate the surface integral of the portion of the paraboloid $x^2+y^2=2ay$ cut by the plane $y=a$. My attempt was to substitute the plane on the equation for the paraboloid, and that would give the following integral, using polar coordinates, $$A = \iint_T r\ dr\ d\theta,$$ where $T : \{\frac{x^2}{2} + \frac{y^2}{2} \leq a^2\},$ so that means that I'm calculating the area of that ellipse, which results $A = 2 \pi ,$ but the right answer is $A = 2 \pi a^2 \frac{(3\sqrt{3} - 1)}{3}.$ I know I'm missing a lot, but I don't get what I am doing wrong.

Answer 1

Somthing is wrong with "right answer" or with the problem statement. I solve in cartesian coordinates as follow.

$x^2+y^2-2ay=0 \rightarrow y=a \pm \sqrt{a^2-x^2}$

Substituting $y$ for get the $x$ limits. $x^2+a^2-2a^2=x^2-a^2=0 \rightarrow x=\pm \lvert a \rvert$

Assuming $a>0$, $$\begin{equation}\begin{split} &\int_{-a}^{a}\int_{y=a \pm \sqrt{a^2-x^2}}^{y=a} dy \,dx \\ &=\int_{-a}^{a}a-a \pm \sqrt{a^2-x^2} \,dx \\ &=\int_{-a}^{a}\pm \sqrt{a^2-x^2} \,dx \\ &=\pm \dfrac{\pi a^2}{2} \end{split} \end{equation}$$