28. Suppose F(x,y) = (2y, -sin(y) ) and C is the circle of radius 7 centered at the origin...


28. Suppose {eq}\vec{F}(x,y) = \langle 2y, -\sin(y) \rangle {/eq} and C is the circle of radius 7 centered at the origin oriented counterclockwise.

(a) Find a vector parametric equation {eq}\vec{r}(t) {/eq} for the circle C that starts at the point (7,0) and travels around the circle once counterclockwise for 0 {eq}\leq t \leq 2\pi. \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 C.{eq}\displaystyle \int_C \vec{F} \cdot d \vec{r} = \int_a^b \vec{F}(\vec{r}(t)) \cdot \vec{r}\,'(t) \, dt = \int_a^b dt {/eq} with limits of integration a = and b =

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

Line Integrals of Vector Field

To evaluate the line integral of a vector field along a given curve (or circulation of the vector field along the curve)

we will convert the line integral in a definite integral by obtaining a parametrization of the curve, first.

If the curve is a circle, then we will use sine and cosine function for x and y variables from the standard form of a circle centered at the origin.

Answer and Explanation:

To evaluate the line integral {eq}\displaystyle \int_{\mathcal{C}} \mathbf{F}\cdot d\mathbf{r}, {/eq} where {eq}\displaystyle \mathcal{C} {/eq} is the circle centered at the origin and radius 7, in counterclockwise direction

and {eq}\displaystyle \mathbf{F}=\langle 2y,-\sin y\rangle , {/eq}

we need a parametrization of the curve, which is given as

(a) A parametrization of the circle is given in terms of sine and cosine function, traveled counterclockwise, we will choose cosine for x and sine for y.

So the parametrization is given as {eq}\displaystyle \mathcal{C}: \boxed{\mathbf{r}(t)=\langle 7\cos t, 7\sin t\rangle, 0 \leq t\leq 2\pi}, {/eq} and verifies that when {eq}\displaystyle t=0, x=7\text{ and } y=0. {/eq}

(b) To set up the line integral, we need the tangent vector to the curve {eq}\displaystyle \mathbf{r}'(t)=\langle -7\sin t, 7\cos t\rangle, {/eq} and the vector field in the curve parametrization is

{eq}\displaystyle \mathbf{F}=\langle 14\sin t, -\sin (7\sin t )\rangle {/eq}

Therefore, the line integral is given as

{eq}\displaystyle \begin{align}\int_{\mathcal{C}} \mathbf{F}\cdot d\mathbf{r}&=\int_0^{2\pi} \langle 14\sin t, -\sin (7\sin t )\rangle \cdot \langle -7\sin t, 7\cos t\rangle \ dt=\boxed{\int_0^{2\pi} \left[-98\sin^2 t- 7\cos t\sin(7\sin t) \right]\ dt}. \end{align} {/eq}

(c) To evaluate the integral obtained in (b), we will do the following integrations,

{eq}\displaystyle \begin{align} \int_{\mathcal{C}} \text{Circulation}=\mathbf{F}\cdot d\mathbf{r}&=\int_0^{2\pi} \left[-98\sin^2 t- 7\cos t\sin(7\sin t) \right]\ dt\\ &=\int_0^{2\pi} \left( -49\left[1-\cos(2 t)\right]- 7\cos t\sin(7\sin t) \right) \ dt, &\left[\text{ using the double angle formula } \sin^2 t=\frac{1}{2} \left(1-\cos(2 t)\right)\right]\\ &= \left( -49\left[t-\frac{1}{2}\sin(2 t)\right]+\cos(7\sin t) \right) \bigg\vert_0^{2\pi} &\left[\text{using the substitution } 7\sin t=u, 7\cos t\ dt=du \text{ and } \int \sin u\ du=-\cos u+C\right]\\ &=\boxed{-98\pi}. \end{align} {/eq}

Learn more about this topic:

Graphs of Parametric Equations

from Precalculus: High School

Chapter 24 / Lesson 5

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