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

User FPK
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Answer:

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.

Find the surface area of the part of the paraboloid z=5-3x^2-2y^2 located above the-example-1
User JustinHui
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7.9k points