Question

Evaluate ∫SF⃗ ⋅dA⃗ , where F⃗ =(bx/a)i⃗ +(ay/b)j⃗ and S is the elliptic cylinder oriented away from the z-axis, and given by x2/a2+y2/b2=1, |z|≤c, where a, b, c are positive constants.

Answer 1

Answer:

Therefore surface integral is $\pi(a^2+b^2)c-0-0=\pi(a^2+b^2)c$.

Step-by-step explanation:

Given function is,

$\vec{F}=\frac{bx}{a}\uvec{i}+\frac{ay}{b}\uvec{j}$

To find,

$\int\int_{S}\vec{F}dS$  

where S=A=surfece of elliptic cylinder we have to apply Divergence theorem so that,

$\int\int_{S}\vec{F}dS$

$=\int\int\int_V\nabla.\vec{F}dV$

$=\int\int\int_V(\frac{b}{a}+\frac{a}{b})dV$  

$=\frac{a^2+b^2}{ab}\int\int\int_VdV$

$=\frac{a^2+b^2}{ab}\times \textit{Volume of the elliptic cylinder}$

$=\frac{a^2+b^2}{ab}\times \pi ab\times 2c=\pi (a^2+b^2)c$

• If unit vector $\cap{n}$ directed in positive (outward) direction then z=c and,

$\int\int_{S_1}\vex{F}.dS_1=\int\int_{S_1} . dA$      

$=\int\int_{S_1}.dA=0$

• If unit vector $\cap{n}$ directed in negative (inward) direction then z=-c and,

$\int\int_{S_2}\vex{F}.dS_2=\int\int_{S_2}. -dA$      

$=\int\int_{S_2}. -dA=0$

Therefore surface integral without unit vector of the surface is,

$\pi(a^2+b^2)c-0-0=\pi(a^2+b^2)c$