# Suppose vector F(x, y) equals 2y, minus sin(y) and C is the circle of radius 4 centered at the...

## Question:

Suppose {eq}\vec{F}(x, y) = (2y, - \sin(y)) {/eq} and {eq}C {/eq} is the circle of radius 4 centered at the origin oriented counterclockwise.

(a) Find a vector parametric equation {eq}\vec{ r}(t) {/eq} for the circle {eq}C {/eq} that starts at the point (4, 0) and travels around the circle once counterclockwise for {eq}0 \le t \le 2 \pi {/eq}.

{eq}\vec{r}(t) = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ {/eq}

(b) Using your parametrization in part (a), set up an integral for calculating the circulation of {eq}\vec{F} {/eq} around {eq}C {/eq}.

{eq}{\displaystyle\int_C} \vec{F} \cdot \vec{dr} = {\displaystyle\int_a^b} \vec{F} (\vec{r}(t)) \cdot \vec{r}'(t) dt = {\displaystyle\int_a^b} \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ dt {/eq} with limits of integration {eq}a = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ {/eq}

and {eq}b = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ {/eq}

(c) Find the circulation of {eq}\vec{F} {/eq} around {eq}C {/eq}.

Circulation = {eq}\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ {/eq}

## Green's Theorem:

We will solve the line integral by using Green's Theorem where we will find the partial derivatives and then find the area of the circle where the radius of the circle is 4.

## Answer and Explanation:

We will solve the problem we will find x and y in terms of parameter t:

{eq}x=2\cos t\\ y=2\sin t {/eq}

Now let us set up the integral:

{eq}\int \vec{F}(r(t))r'(t)\\ =\int_{0}^{2\pi}4\sin t(-2\sin t)dt-\sin (2\sin t)2\cos tdt {/eq}

Now let us use Green's Theorem to solve the problem:

{eq}\oint F_{1}dx+F_{2}dy=\int \int \left ( \frac{\partial F_{2}}{\partial x}-\frac{\partial F_{1}}{\partial y} \right )dxdy {/eq}

Now let us find the partial derivatives:

{eq}\frac{\partial F_{2}}{\partial x}=0\\ \frac{\partial F_{1}}{\partial y}=2 {/eq}

The integral will be:

{eq}=-2\pi (4)^{2}\\ =-32\pi {/eq}

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