Respuesta :
Answer:
Therefore surface integral is [tex]\pi(a^2+b^2)c-0-0=\pi(a^2+b^2)c[/tex].
Step-by-step explanation:
Given function is,
[tex]\vec{F}=\frac{bx}{a}\uvec{i}+\frac{ay}{b}\uvec{j}[/tex]
To find,
[tex]\int\int_{S}\vec{F}dS[/tex]
where S=A=surfece of elliptic cylinder we have to apply Divergence theorem so that,
[tex]\int\int_{S}\vec{F}dS[/tex]
[tex]=\int\int\int_V\nabla.\vec{F}dV[/tex]
[tex]=\int\int\int_V(\frac{b}{a}+\frac{a}{b})dV[/tex]
[tex]=\frac{a^2+b^2}{ab}\int\int\int_VdV[/tex]
[tex]=\frac{a^2+b^2}{ab}\times \textit{Volume of the elliptic cylinder}[/tex]
[tex]=\frac{a^2+b^2}{ab}\times \pi ab\times 2c=\pi (a^2+b^2)c[/tex]
- If unit vector [tex]\cap{n}[/tex] directed in positive (outward) direction then z=c and,
[tex]\int\int_{S_1}\vex{F}.dS_1=\int\int_{S_1}<\frac{bx}{a}, \frac{ay}{b}, 0> . <-z_x,z_y,1>dA[/tex]
[tex]=\int\int_{S_1}<\frac{bx}{a},\frac{ay}{b}, 0>.<0,0,1>dA=0[/tex]
- If unit vector [tex]\cap{n}[/tex] directed in negative (inward) direction then z=-c and,
[tex]\int\int_{S_2}\vex{F}.dS_2=\int\int_{S_2}<\frac{bx}{a}, \frac{ay}{b}, 0>. -<-z_x,z_y,1>dA[/tex]
[tex]=\int\int_{S_2}<\frac{bx}{a},\frac{ay}{b}, 0>. -<0,0,1>dA=0[/tex]
Therefore surface integral without unit vector of the surface is,
[tex]\pi(a^2+b^2)c-0-0=\pi(a^2+b^2)c[/tex]
The value of ∫SF ⋅dA where F =(bx/a)i +(ay/b)j and S is the elliptic cylinder oriented away from the z-axis is 2πc(a² + b²).
How to solve the elliptic cylinder?
From the information, F = (b/ax) + (a/by)j and S is the elliptic cylinder.
To evaluate ∫F.dA goes thus:
divF = (I'd/dx + jd/dx + kd/dx) × (b/ax)i + (a/by)j
= b/a + a/b
= (a² + b²)/ab
Using Gauss divergence theorem, this will be further solved below:
∫∫∫v(a² + b²/ab)dV
= (a² + b²/ab)∫∫∫vdV
= (a² + b²/ab) × Volume of cylinder
= (a² + b²/ab) × πab(2c)
= 2πc(a² + b²)
Learn more about cylinder on:
https://brainly.com/question/76386
