Answer:
The area of the surface over R is [tex]\mathbf{ 2 \sqrt{17} + \dfrac{1}{2} \ In ( \sqrt{17} +4)}[/tex]
Step-by-step explanation:
From this study,
The surface area of f(x,y) is expressed by:
[tex]A = \iint_R \sqrt{1 +f^2_x+f^2_y}} \ dA[/tex]
where
[tex]0 \leq x \leq 2 \ and \ 0 \leq y \leq 2[/tex]
Given that:
f(x,y) = 9 - x²
Thus, [tex]f_x = -2x[/tex] and [tex]f_y = 0[/tex]
However, the area becomes:
[tex]A = \iint_R \sqrt{1+(-2x)^2+0^2} \ dxdy[/tex]
[tex]A = \iint_R \sqrt{1+4x^2} \ dxdy[/tex]
From above, replacing x with [tex]\dfrac{1}{2} \ tan \ u[/tex]
then [tex]dx = \dfrac{1}{2} sec^2 \ udu[/tex]
∴
[tex]\sqrt{1 + 4x^2 } \ dA= \dfrac{1}{2} \iint \sqrt{1 + 4u^2 } \ sec^2 \ udu[/tex]
[tex]\sqrt{1 + 4x^2 } \ dA= \dfrac{1}{2} \iint sec^2 \ udu[/tex]
[tex]\sqrt{1 + 4x^2 } \ dA= \dfrac{1}{2} ( \sqrt{4x^2 + 1x} + \dfrac{1}{2} In \sqrt{4x^2 + 1}+2x)[/tex]
Hence;
[tex]\mathsf{A = \int \limits ^2_0 \begin {bmatrix} \dfrac{1}{2}(\sqrt{4x^2 +1 x}+ \dfrac{1}{2} In ( \sqrt{4x^2+1}+2x \end {bmatrix}^2_0 \ dy}[/tex]
[tex]\mathbf{A = 2 \sqrt{17} + \dfrac{1}{2} \ In ( \sqrt{17} +4)}[/tex]
Therefore, the area of the surface over R is [tex]\mathbf{ 2 \sqrt{17} + \dfrac{1}{2} \ In ( \sqrt{17} +4)}[/tex]
