Answer:
Use the formula [tex]Area(S)=\iint_{S} 1 dS= \iint_{D} \lVert r_{u}\times r_{v} \rVert dudv[/tex]
Step-by-step explanation:
Let [tex]r(x,y)=(x,y,5-3x^2-2y^2)[/tex] be the explicit parametrization of the paraboid. The intersection of this paraboid with the xy plane is the ellipse given by
[tex]\dfrac{x^2}{\frac{5}{3}}+\dfrac{y^{2}}{\frac{5}{2}}=1[/tex]
The partial derivatives of the parametrization are:
[tex]\begin{array}{c}r_{x}=(1,0,-6x)\\r_{y}=(0,1,-4y)\end{array}[/tex]
and computing the cross product we have
[tex]r_{x}\times r_{y}=(6x,4y,1)[/tex]. Then
[tex]\lVert r_{x}\times r_{y}\rVert =\sqrt{1+36x^{2}+16y^{2}}[/tex]
Then, if [tex]R[/tex] is the interior region of the ellipse the superficial area located above of the xy is given by the double integral
[tex]\iint_{R}\sqrt{1+36x^2+16y^2}dxdy=\int_{-\sqrt{5/3}}^{\sqrt{5/3}}\int_{-\sqrt{5/2}\sqrt{1-\frac{x^2}{5/3}}}^{\sqrt{5/2}\sqrt{1-\frac{x^2}{5/3}}}\sqrt{1+36x^2+16y^2}dy dx=30.985[/tex]
The last integral is not easy to calculate because it is an elliptic integral, but with any software of mathematics you can obtain this value.
