Calculate \int_C \vecF \cdot d \vec s where the vector field \vec F (x,y)=(-y,x) and C the...


Calculate {eq}\int_C \vec F \cdot \vec {ds} {/eq} where the vector field {eq}\vec F (x,y)=(-y,x) {/eq} and {eq}C {/eq} the closed curve consisting of the upper half semicircle of radius {eq}1 {/eq} centered at the origin together with the line segment along the {eq}x {/eq} axis forming its diameter, oriented counterclockwise.

Green's Theorem:

Whenever we are dealing with a closed integral over a closed path in the plane, we should be thinking about Green's Theorem. With it, we can change that line integral into an area integral, which usually makes things much easier to do. The theorem is

{eq}\begin{align*} \oint_C \vec F \cdot d\vec s &= \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \ dA \end{align*} {/eq}

Answer and Explanation:

The difference of the partials is

{eq}\begin{align*} \frac{\partial }{\partial x} \left( x \right) - \frac{\partial }{\partial y} \left( -y \right) &= 1 - (-1) \\ &= 2 \end{align*} {/eq}

And so we use Green's Theorem to write (note we are positively oriented)

{eq}\begin{align*} \oint_C \vec F \cdot d\vec s &= \iint_D (2)\ dA \\ &= 2A \end{align*} {/eq}

But the region enclosed by our path is half of the unit disk, so {eq}A = \frac\pi2 {/eq} and we have

{eq}\begin{align*} \oint_C \vec F \cdot d\vec s &= 2 \left( \frac\pi2 \right) \\ &= \pi \end{align*} {/eq}

Learn more about this topic:

Double Integrals: Applications & Examples

from AP Calculus AB & BC: Help and Review

Chapter 12 / Lesson 14

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