example of integration with respect to surface area of a paraboloid

In this example we examine the paraboloid given by the equation $z=x^{2}+3y^{2}$ . Putting $g(x,y)=x^{2}+3y^{2}$ , we have

\\sqrt{1+\\left(\\frac{\\partial g}{\\partial x}\\right)^{\\!2}+\\left(\\frac{\\partial g%}{\\partial y}\\right)^{\\!2}}=\\sqrt{1+\\left(2x\\right)^{2}+\\left(6y\\right)^{2}}=%\\sqrt{1+4x^{2}+36y^{2}}

and hence

\\int_{S}f(x,y)\\,d^{2}A=\\int f(x,y)\\sqrt{1+4x^{2}+36y^{2}}\\,dx\\,dy.