91Ó°ÊÓ

Let givenRbe a simply connected region in the xy-plane.

${∫}_{s}1dS=∫{∫}_R\sqrt{{\left(\frac{∂z}{∂x}\right)}^2+{\left(\frac{∂z}{∂y}\right)}^2+1dA}...........\left(1\right)$

$$\begin{gathered} \frac{∂z}{∂x}=\frac{∂}{∂x}(x^2+y^2) \\ =2x \end{gathered}$$

and,

$$\begin{gathered} \frac{∂z}{∂y}=\frac{∂}{∂y}(x^2+y^2) \\ =2y \end{gathered}$$

Then,

$$\begin{gathered} dS=\left(\sqrt{{\left(\frac{∂z}{∂x}\right)}^2+{\left(\frac{∂z}{∂y}\right)}^2+1}\right)dA \\ =\left(\sqrt{{\left(2x\right)}^2+{\left(2y\right)}^2+1}\right)dA \\ =\left(\sqrt{4x^2+4y^2+1}\right)dA \end{gathered}$$

Then,by(1),the surface area of the paraboloid surface I will be,

$$\begin{gathered} {∫}_{s}1dS=∫{∫}_R\sqrt{{\left(\frac{∂z}{∂x}\right)}^2+{\left(\frac{∂z}{∂y}\right)}^2+1dA} \\ =∫{∫}_R\left(\sqrt{4x^2+4y^2+1}\right)dA \end{gathered}$$

Here, the surface S is the portion of the following paraboloid; $z=x^2-y^2$

Now, first find $\frac{∂z}{∂x}$and $\frac{∂z}{∂x}$. The first partial derivatives of z are,

$$\begin{gathered} \frac{∂z}{∂x}=\frac{∂}{∂x}(x^2-y^2) \\ =2x \end{gathered}$$

And,

$$\begin{gathered} \frac{∂z}{∂y}=\frac{∂}{∂y}(x^2-y^2) \\ =-2y \end{gathered}$$

Then,

$$\begin{gathered} dS=\left(\sqrt{{\left(\frac{∂z}{∂x}\right)}^2+{\left(\frac{∂z}{∂y}\right)}^2}+1\right)dA \\ =\left(\sqrt{{\left(2x\right)}^2+{\left(-2y\right)}^2+1}\right)dA \\ =\left(\sqrt{4x^2+4y^2+1}\right)dA \end{gathered}$$

Then,by(1),the surface area of the paraboloid surface II will be,

$$\begin{gathered} {∫}_{s}1dS=∫{∫}_R\sqrt{{\left(\frac{∂z}{∂x}\right)}^2+{\left(\frac{∂z}{∂y}\right)}^2+1dA} \\ =∫{∫}_R\left(\sqrt{4x^2+4y^2+}\right)dA \end{gathered}$$

For each surface,the surface area is,

${∫}_{s}1dS=∫{∫}_R\left(\sqrt{4x^2+4y^2+1}\right)dA$

Where,the limits determined by R,

Thus,it is proved that the areas of both surface are equal

Hence proved