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<record version="12" id="6672">
 <title>example of integration with respect to surface area of a paraboloid</title>
 <name>Example7OfIntegrationWithRespectToSurfaceArea</name>
 <created>2005-01-27 16:35:41</created>
 <modified>2006-06-14 14:42:12</modified>
 <type>Example</type>
<parent id="6660">surface integration with respect to area</parent>
 <creator id="2760" name="yark"/>
 <author id="2760" name="yark"/>
 <author id="6075" name="rspuzio"/>
 <classification>
	<category scheme="msc" code="28A75"/>
 </classification>
 <preamble></preamble>
 <content>In this example we examine the paraboloid given by the equation $z = x^2 + 3 y^2$.  Putting $g(x,y) = x^2 + 3 y^2$, we have
 $$\sqrt{1 + \left( \frac{\partial g}{\partial x} \right)^{\!2} + \left( \frac{\partial g}{\partial y} \right)^{\!2}}
= \sqrt{1 + \left( 2 x \right)^2 + \left( 6 y \right)^2}
= \sqrt{1 + 4 x^2 + 36 y^2 }$$
and hence
 $$\int_S f(x,y) \, d^2 A = \int f(x,y) \sqrt{ 1 + 4 x^2 + 36 y^2 } \, dx \, dy.$$

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