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Surface Area of 3D Surface z=f(x,y) By Integration Example

The surface area of a surface  
\[S\]
 is
\[A= \int_S \sqrt{(\frac{\partial z}{\partial x})^2 +(\frac{\partial z}{\partial y})^2 +1} dS \]

A cone has equation  
\[z=\sqrt{x^2 +y^2} \]
.
The surface area of that part of the cone bounded by  
\[0 \leq x \leq 4, 1 \leq y \leq 6\]
 is
\[\begin{equation} \begin{aligned} A &= \int^4_0 \int^6_1 \sqrt{(\frac{\partial z}{\partial x})^2 +(\frac{\partial z}{\partial y})^2 +1} dy dx \\ &= \int^4_0 \int^6_1 \sqrt{ (\frac{x}{\sqrt{x^2+y^2}})^2 +(\frac{y}{\sqrt{x^2+y^2}})^2 +1} dy dx \\ &= \int^4_0 \int^6_1 \sqrt{2} dy dx \\ &= \int^4_0 [y \sqrt{2}]^6_1 dx \\ &= \int^6_1 5 \sqrt{2} dx \\ &= [5x \sqrt{2}]^4_0 \\ &= 20 \sqrt{2} \end{aligned} \end{equation}\]