»

Mathematics

2 answers:

Ede4ka [16]3 years ago

6 0

Answer:

It is rhombuses

Step-by-step explanation:

natita [175]3 years ago

4 0

Answer:Rhombuses :)

Step-by-step explanation:

You might be interested in

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

Lady_Fox [76]

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

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

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