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puteri [66]
3 years ago
7

Use Stokes' Theorem to evaluate C F · dr where C is oriented counterclockwise as viewed from above. F(x, y, z) = 2yi + xzj + (x

+ y)k, C is the curve of intersection of the plane z = y + 6 and the cylinder x2 + y2 = 1.
Mathematics
1 answer:
Lady_Fox [76]3 years ago
4 0

By Stokes' theorem, the line integral of \vec F over C is given by the surface integral of the curl of \vec F over S, where S is the region of intersection of the plane z=y+6 and the cylinder x^2+y^2=1 with S having positive/upward orientation.

Parameterize S by

\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+(u\sin v+6)\,\vec k

with 0\le u\le1 and 0\le v\le2\pi.

Take the normal vector to S to be

\vec s_u\times\vec s_v=-u\,\vec\jmath+u\,\vec k

The curl of \vec F is

\nabla\times\vec F(x,y,z)=(1-x)\,\vec\imath-\vec\jmath+(z-2)\,\vec k

Then the line integral is equivalent to

\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S

=\displaystyle\int_0^{2\pi}\int_0^1\bigg((1-u\cos v)\,\vec\imath-\vec\jmath+(u\sin v+4)\,\vec k\bigg)\cdot\bigg(-u\,\vec\jmath+u\,\vec k\bigg)\,\mathrm du\,\mathrm dv

=\displaystyle\int_0^{2\pi}\int_0^1(5u+u^2\sin v)\,\mathrm du\,\mathrm dv=\boxed{5\pi}

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