Find all angles θbetween 0° and 2pi satisfying the given equation ? sin(θ)=2√2

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We are to determine <span>all angles between 0° and 2pi satisfying the given equation sin(θ) = 2√2. The answer is that no angles that would fit to the given equation. This is because the values of trigonometric functions such as cosine and sine do not go more than 1. Since </span><span>2√2 is equal to 2.83, then the answer is none.</span>

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Let F=(2x,2y,2x+2z)F=(2x,2y,2x+2z). Use Stokes' theorem to evaluate the integral of FF around the curve consisting of the straig

Brilliant_brown [7]

Stokes' theorem equates the line integral of $\vec F$ along the curve to the surface integral of the curl of $\vec F$ over any surface with the given curve as its boundary. The simplest such surface is the triangle with vertices (1,0,1), (0,1,0), and (0,0,1).

Parameterize this triangle (call it $T$ ) by

$\vec s(u,v)=(1-v)((1-u)(1,0,1)+u(0,1,0))+v(0,0,1)$

$\vec s(u,v)=((1-u)(1-v),u(1-v),1-u+uv)$

with $0\le u\le1$ and $0\le v\le1$ . Take the normal vector to $T$ to be

$\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}=(0,1-v,1-v)$

Divide this vector by its norm to get the unit normal vector. Note that this assumes a "positive" orientation, so that the boundary of $T$ is traversed in the counterclockwise direction when viewed from above.

Compute the curl of $\vec F$ :

$\vec F=(2x,2y,2x+2z)\implies\mathrm{curl}\vec F=(0,-2,0)$

Then by Stokes' theorem,

$\displaystyle\int_{\partial T}\vec F\cdot\mathrm d\vec r=\iint_T\mathrm{curl}\vec F\cdot\mathrm d\vec S$

where

$\mathrm d\vec S=\dfrac{\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}}{\left\|\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}\right\|}\,\mathrm dS$

$\mathrm d\vec S=\dfrac{\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}}{\left\|\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}\right\|}\left\|\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right\|\,\mathrm du\,\mathrm dv$