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Revision difference : example of integration with respect to surface area of a paraboloid
Version 3 Version 2
In this example, we examine the paraboloid given by the equation $z = x^2 + 3 y^2$. We have In this example, we examine the paraboloid given by the equation $z = x^2 + 3 y^3$. We have
$$\sqrt{1 + \left( \frac{\partial g}{\partial x} \right)^2 + \left( \frac{\partial g}{\partial y} \right)^2} =$$ $$\sqrt{1 + \left( \frac{\partial g}{\partial x} \right)^2 + \left( \frac{\partial g}{\partial y} \right)^2} =$$
$$\sqrt{1 + \left( 2 x \right)^2 + \left( 6 y \right)^2} = \sqrt{ 4 x^2 + 36 y^2 }$$ $$\sqrt{1 + \left( 2 x \right)^2 + \left( 6 y \right)^2} = \sqrt{ 4 x^2 + 36 y^2 }$$
and, hence and, hence
$$\int_S f(x,y) \, d^2 A = \int f(x,y) \sqrt{ 4 x^2 + 36 y^2 } \, dx \, dy.$$ $$\int_S f(x,y) \, d^2 A = \int f(x,y) \sqrt{ 4 x^2 + 36 y^2 } \, dx \, dy.$$
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