Question

Find the surface area of the part of the paraboloid z=5-3x^2-2y^2 located above the xy plane. (10 points) z

Answer 1

Answer:

Use the formula $Area(S)=\iint_{S} 1 dS= \iint_{D} \lVert r_{u}\times r_{v} \rVert dudv$

Step-by-step explanation:

Let $r(x,y)=(x,y,5-3x^2-2y^2)$ be the explicit parametrization of the paraboid. The intersection of this paraboid with the xy plane is the ellipse given by

$\dfrac{x^2}{\frac{5}{3}}+\dfrac{y^{2}}{\frac{5}{2}}=1$

The partial derivatives of the parametrization are:

$\begin{array}{c}r_{x}=(1,0,-6x)\\r_{y}=(0,1,-4y)\end{array}$

and computing the cross product we have

$r_{x}\times r_{y}=(6x,4y,1)$. Then

$\lVert r_{x}\times r_{y}\rVert =\sqrt{1+36x^{2}+16y^{2}}$

Then,  if $R$ is the interior region of the ellipse the superficial area located above of the xy  is given by the double integral

$\iint_{R}\sqrt{1+36x^2+16y^2}dxdy=\int_{-\sqrt{5/3}}^{\sqrt{5/3}}\int_{-\sqrt{5/2}\sqrt{1-\frac{x^2}{5/3}}}^{\sqrt{5/2}\sqrt{1-\frac{x^2}{5/3}}}\sqrt{1+36x^2+16y^2}dy dx=30.985$

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.