Substitute \({f_x}{\rm{ for }}\frac{{\partial z}}{{\partial x}}{\rm{ and }}{f_y}{\rm{ for }}\frac{{\partial z}}{{\partial y}}\) in equation (3),
\(A(S) = \iint_D {\sqrt {1 + {{\left( {{f_x}} \right)}^2} + {{\left( {{f_y}} \right)}^2}} }dA\) …… (5)
Rewrite the expression in equation (1) as follows,
\(0 \le {f_x} \le 1\)
Take square and rewrite the expression as follows,
\(\begin{aligned}{}{(0)^2} \le {\left( {{f_x}} \right)^2} \le {(1)^2}\\0 \le {\left( {{f_x}} \right)^2} \le 1\end{aligned}\) …… (6)
Rewrite the expression in equation (2) as follows,
Take square and rewrite the expression as follows,
\(\begin{aligned}{}{(0)^2} \le {\left( {{f_y}} \right)^2} \le {(1)^2}\\0 \le {\left( {{f_y}} \right)^2} \le 1\end{aligned}\) …… (7)
Add equations (6) and (7) as follows,
\(\begin{aligned}{}(0 + 0) \le {\left( {{f_x}} \right)^2} + {\left( {{f_y}} \right)^2} \le 1 + 1\\0 \le {\left( {{f_x}} \right)^2} + {\left( {{f_y}} \right)^2} \le 2\end{aligned}\)
Add \(1\) and rewrite the expression as follows,
\(\begin{aligned}{}0 + 1 \le 1 + {\left( {{f_x}} \right)^2} + {\left( {{f_y}} \right)^2} \le 2 + 1\\1 \le 1 + {\left( {{f_x}} \right)^2} + {\left( {{f_y}} \right)^2} \le 3\end{aligned}\)
Take square roots and rewrite the expression as follows,
\(\begin{aligned}{}\sqrt 1 \le \sqrt {1 + {{\left( {{f_x}} \right)}^2} + {{\left( {{f_y}} \right)}^2}} \le \sqrt 3 \\1 \le \sqrt {1 + {{\left( {{f_x}} \right)}^2} + {{\left( {{f_y}} \right)}^2}} \le \sqrt 3 \end{aligned}\) …… (8)
Use the following property and find the range of area of the surface of the plane.
If \(m \le f(x,y) \le M\) for all\((x,y){\rm{ in }}D\), then \(mA(D) \leqslant \iint_D f(x,y)dA \leqslant MA(D){\text{. }}\)
Use the property and rewrite the expression in equation (8) as follows,
\(\iint_D {(1)}dA \leqslant \iint_D {\sqrt {1 + {{\left( {{f_x}} \right)}^2} + {{\left( {{f_y}} \right)}^2}} }dA \leqslant \iint_D {(\sqrt 3 )}dA\)
For equation (5), substitute \(A(S){\text{ for }}\iint_D {\sqrt {1 + {{\left( {{f_x}} \right)}^2} + {{\left( {{f_y}} \right)}^2}} }dA\) and simplify the expression as follows,
\(A(D) \le A(S) \le \sqrt 3 A(D)\) …… (9)
As \(D\)is the region of circle with equation \({x^2} + {y^2} \le {R^2},A(D)\) is the area of the circle with radius \(R.\)
From equation (4), the area of circle with radius \(R{\rm{ is }}\pi {R^2}{\rm{. }}\)
Substitute \(\pi {R^2}{\rm{ for }}A(D)\) in equation (9),
\(\pi {R^2} \le A(S) \le \sqrt 3 \pi {R^2}\)
Thus, the range of the area of the surface of the equation\(z = f(x,y)\) is \(\pi {R^2} \le A(S) \le \sqrt 3 \pi {R^2}.\)
