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3 months ago

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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.

1 answer:

Inessa [12.5K]3 months ago

5 0

Respuesta:

Por lo tanto, la integral de superficie es $\pi(a^2+b^2)c-0-0=\pi(a^2+b^2)c$ .

Explicación paso a paso:

La función dada es,

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

Para encontrar,

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

donde S=A=superficie del cilindro elíptico debemos aplicar el teorema de divergencia, así que,

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

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

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

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

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

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

Si el vector unitario $\cap{n}$ está dirigido en dirección positiva (hacia afuera), entonces z=c y,

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

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

Si el vector unitario $\cap{n}$ está dirigido en dirección negativa (hacia adentro), entonces z=-c y,

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

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

Por lo tanto, la integral de superficie sin el vector unitario de la superficie es,

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

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