contestada

Suppose F(x,y)=(2x−4y)i+2xj and C is the counter-clockwise oriented sector of a circle centered at the origin with radius 2 and central angle π/6. Use Green's theorem to calculate the circulation of F around C

Respuesta :

LammettHash

[tex]\dfrac{\partial(2x)}{\partial x}-\dfrac{\partial(2x-4y)}{\partial y}=6[/tex]

so that by Green's theorem,

[tex]\displaystyle\int_C\vec F\cdot\mathrm d\vec r=6\iint_D\mathrm dx\,\mathrm dy[/tex]

i.e. 6 times the area of the sector [tex]D[/tex] with boundary [tex]C[/tex].

The area of the entire circle occurs with the area of the sector [tex]D[/tex] in the following ratio:

[tex]\dfrac{4\pi}{2\pi\,\rm rad}=\dfrac{\displaystyle\iint_D\mathrm dx\,\mathrm dy}{\frac\pi6\,\rm rad}\implies\displaystyle\iint_D\mathrm dx\,\mathrm dy=\frac\pi3[/tex]

so the circulation of [tex]\vec F[/tex] around [tex]C[/tex] is [tex]\dfrac{6\pi}3=\boxed{2\pi}[/tex].
  • Green´s theorem establishes the relationship between the circulation about the curve C, and the sum of all the circulations inside the region enclosed by C.
  • Green´s theorem is one of the 4 fundamentals theorems of vector calculus

The answer to the question is:

∫C) F ( x , y ) dr =  2×π

Green's Theorem establishes:

∫ P(x,y)i ×dx + Q(x,y)j × dy   = ∫∫ (δQ/δx - δP/δy ) dA

In the question:

F ( x , y ) = ( 2×x - 4×y )×i  + 2×x×j

P ( x , y ) = (2×x - 4× y )    and . Q ( x , y ) = 2x

Then

δQ/δy = 2    and     δP/δy  = - 4

δP/δy -  δP/δy = 2 - ( - 4 ) = 6

Therefore

∫ P(x,y)i ×dx + Q(x,y)j × dy   = ∫∫ 6 dx×dy  = 6 ∫∫R) dxdy

Now by rule of three :

4×π/2π rad  = ∫∫R) / (π/6) rad

[(1/6)π rad × 4×π]/2π rad  = ∫∫R)

∫∫R) = π/3

And the circulation of the field vector over C is:

∫C)  F dr  = 6× π/3 = 2×π.

Related link: https://brainly.com/question/2758275

Otras preguntas