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Use stokes' theorem to evaluate c f · dr where c is oriented counterclockwise as viewed from above. f(x, y, z) = yzi + 3xzj + exyk, c is the circle x2 + y2 = 16, z = 7.

Respuesta :

LammettHash
Stokes' theorem states that the line integral of [tex]\mathbf f(x,y,z)[/tex] along [tex]C[/tex] is equivalent to the surface integral of the curl of [tex]\mathbf f(x,y,z)[/tex] over the disk [tex]D[/tex] bounded by [tex]C[/tex]:

[tex]\displaystyle\int_C\mathbf f(x,y,z)\cdot\mathrm d\mathbf r=\iint_D\nabla\times\mathbf f(x,y,z)\cdot\mathrm dS[/tex]

The disk can be parameterized by

[tex]\mathbf s(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+7\,\mathbf k[/tex]

where [tex]0\le u\le4[/tex] and [tex]0\le v\le2\pi[/tex]. We have

[tex]\mathbf s_u=\cos v\,\mathbf i+\sin v\,\mathbf j[/tex]
[tex]\mathbf s_v=-u\sin v\,\mathbf i+u\cos v\,\mathbf j[/tex]
[tex]\implies\mathbf s_u\times\mathbf s_v=u\,\mathbf k[/tex]
[tex]\implies\mathrm dS=(u\,\mathbf k)\,\mathrm du\,\mathrm dv[/tex]

The curl of the given vector field is

[tex]\nabla\times\mathbf f(x,y,z)=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\\\\frac\partial{\partial x}&\frac\partial{\partial y}&\frac\partial{\partial z}\\\\yz&3xz&e^{xy}\end{vmatrix}=x(e^{xy}-3)\,\mathbf i+(1-e^{xy})y\,\mathbf j+2z\,\mathbf k[/tex]
[tex]=u\cos v(e^{u^2\cos v\sin v}-3)\,\mathbf i+(1-e^{u^2\cos v\sin v})u\sin v\,\mathbf j+14\,\mathbf k[/tex]

So the surface integral is equivalent to

[tex]\displaystyle\iint_D\nabla\times\mathbf f(x(u,v),y(u,v),z(u,v))\cdot(u\,\mathbf k)\,\mathrm du\,\mathrm dv[/tex]
[tex]=\displaystyle\int_{u=0}^{u=4}\int_{v=0}^{v=2\pi}14u\,\mathrm dv\,\mathrm du[/tex]
[tex]=28\pi\displaystyle\int_{u=0}^{u=4}u\,\mathrm du[/tex]
[tex]=224\pi[/tex]
shristiparmar1221

This question is based on the stokes' theorem.Therefore,  the [tex]\int_c f. dr = 224\pi[/tex]  by using stokes theorem.

Given:

f(x, y, z) = yzi + 3xzj + [tex]e^x[/tex]yk, c is the circle [tex]x^2 + y^2 = 16,[/tex] z = 7.

According to the question,

Stokes' theorem states that, the line integral of f(x, y, z) along C is equivalent to the surface integral of the curl f(x, y, z) of  over the disk D bounded by :

[tex]\int_c f(x, y, z) \; dr = \int\int_D \bigtriangledown \times f(x, y, z). \; dS[/tex]

Now, the disk can be parameterized by,

s(u,v) = u cos v i + u sin v j + 7 k

Where, u lies between 0 to 4 and v lies between 0 to 2[tex]\bold{\pi }[/tex].

[tex]s_u = cosv \; i + sin v \; j\\s_v = -u \; sinv \; i + u\; cos v \; j + 7 \; k\\\\\Rightarrow s_u \time s_v = u \; k\\\\\Rightarrow dS = (u \; k) du \; dv[/tex]

The curl of the given vector field is,

[tex]\bigtriangledown \times f(x, y, z) = \begin{vmatrix}i & j & k \\ \frac{\partial}{\partial x} &\frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ yz & 3xz & e^{xy}\end{vmatrix}[/tex]

[tex]= u \; cosv \; (e^{u^2 cosv \; sinv } -3 ) i + (1 - e^{u^2 cosv \; sinv }) u \; sinv \; j + 14 k[/tex]

Hence, the surface integral is,

[tex]\int \int _D \bigtriangledown \times f(x(u,v), y (u,v) , z( u, v) . ( u,k) du \; dv \\\\= \int\limits^4_0\int\limits^{v = 2\pi}_{v = o} 14u \, dv \, du\\\\= 28\pi \int\limits^{u = 4} _{u =0} udu\\\\= 224\pi[/tex]

Therefore,  the [tex]\int_c f. dr = 224\pi[/tex] by using stokes theorem.

For more details, prefer this link:

https://brainly.com/question/8130922

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