Respuesta :
The area of the surface is [tex]\rm \dfrac{1}{54}(53\sqrt{53} -17\sqrt{17} )[/tex] and this can be determined by doing the integration.
Given :
The part of the surface [tex]\rm z = 9+5x+2y^2[/tex] that lies above the triangle with vertices (0, 0), (0, 1), (2, 1).
The formula of the surface area is given by:
[tex]\rm A(S) = \int \int_D\sqrt{1+\left(\dfrac{\delta z}{\delta x}\right)^2 +\left(\dfrac{\delta z}{\delta y}\right)^2 }[/tex]
Now, the values of D varies from:
D = {(x,y): 0 [tex]\leq[/tex] y [tex]\leq[/tex] 1 , 0 [tex]\leq[/tex] x [tex]\leq[/tex] 2y}
Now, the area of the surface is given by:
[tex]\rm A(S) = \int \int_D\sqrt{1+4^2+(6y)^2 }\;dA[/tex]
[tex]\rm A(S) = \int^1_0 \int^{2y}_0\sqrt{17+36y^2 }\;dx\;dy[/tex]
[tex]\rm A(S) = \int^1_02y\sqrt{17+36y^2 }\;dy[/tex]
Now, let [tex]u=17+36y^2[/tex]
[tex]du = 72y dy[/tex]
[tex]\rm \dfrac{du}{36}=2ydy[/tex]
Now, the integral becomes:
[tex]\rm A(S)= \dfrac{1}{36}\sqrt{u} \; du[/tex]
[tex]\rm A(S) = \dfrac{1}{54}\times u^{3/2}+C[/tex]
Now, substitute the value of u in the above expression.
[tex]\rm A(S) = \dfrac{1}{54}\times \left[\left[\sqrt{17+36y^2} \right]^{3/2}\right]^1_0[/tex]
Now, simplify the above expression.
[tex]\rm A(S) = \dfrac{1}{54}(53\sqrt{53} -17\sqrt{17} )[/tex]
For more information, refer to the link given below:
https://brainly.com/question/18125359
