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

Use the formula
`Area(S)=\iint_(S) 1 dS= \iint_(D) \lVert r_(u)* r_(v) \rVert dudv`

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

`(x^2)/((5)/(3))+(y^(2))/((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)* r_(y)=(6x,4y,1)`. Then

`\lVert r_(x)* 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)√(1+36x^2+16y^2)dxdy=\int_(-√(5/3))^(√(5/3))\int_{-√(5/2)\sqrt{1-(x^2)/(5/3)}}^{√(5/2)\sqrt{1-(x^2)/(5/3)}}√(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.