Respuesta :
The area of the surface is 5/3 [tex]\sqrt{14}[/tex][tex]\pi[/tex]
Consider that the part of the plane x+2y +3z=1 that lies inside the cylinder [tex]x^{2}[/tex] + [tex]y^{2}[/tex] =5
The surface area of an equation z= g(x,y) is given by the following formula:
[tex]\int\limits^d_d \sqrt {\alpha z/\alpha x]^{2} . [\alpha z/\alpha y]^{2} +1da[/tex]
where the D is the domain of integration.
Since the plane lies inside the cylinder[tex]x^{2}[/tex] + [tex]y^{2}[/tex] = 5, we have the domain of
D= ( (r,0) |0 [tex]\leq[/tex]r [tex]\leq[/tex][tex]\sqrt{5}[/tex],0[tex]\leq[/tex]θ[tex]\leq[/tex]2[tex]\pi[/tex])
The partial derivatives of z = 1/3 ( 1 - x - 2y) are
[tex]\alpha z/ \alpha x[/tex] = 1/3, [tex]\alpha z / \alpha y[/tex] = - 2/3
[tex]\int\limits^d_d \sqrt {[-1/3]^{2} . [-2/3]^{2} +1da[/tex]
= [tex]\int\limits^d_d \sqrt{14} / 9 da[/tex]
= [tex]\sqrt{14}/3 \int\limits^0_ {2^{\pi }[/tex] [tex]\int\limits^\sqrt{5[/tex]r dr dθ
Evaluating the integral we have
[tex]\sqrt{14}/3 \int\limits^0_ {2^{\pi }[/tex][tex]\int\limits^\sqrt{5[/tex]rdrdθ = [tex]\sqrt{14} /3[/tex] 2 [tex]\pi[/tex] [ 5/2]
= 5/3 [tex]\sqrt{14}[/tex][tex]\pi[/tex]
To learn more about the area of surface
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