Referring to the figure, suppose that
Use Green's Theorem to determine the circulation of F around C1, assuming that (∂F2 / ∂x) − (∂F1 / ∂y) = $c on the shaded region.
We wish to use Green's Theorem to calculate .
We cannot compute the line integral over C1 directly because the region D which it encloses is also bounded by the simple closed curves C2 and C3. However, we can still apply Green's Theorem on the region D provided we take into account all of the simple closed curves that bound the region and their orientations.
Recall the definition of Green's Theorem for more valid general domains of this type.
We will apply this form of Green's Theorem to eventually solve for , but first we must write the boundary of D in terms of the oriented curves C1, C2, and C3. Recall that a curve will have positive orientation if the region D is to the left of the curve as we traverse it. Write ∂D in terms of C1, C2, and C3.
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Use properties of vector line integrals to choose the correct expansion of .
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 the expanded integral from Step 2 into the definition of Green's Theorem from Step 1.
Now substitute the given information , and (∂F2 / ∂x) − (∂F1 / ∂y) = $c.
U2GIbglD1oM= π
Use geometry and the fact that D is composed of circular discs to find the area of D.
Area D = dA = π · 52 − (π · 12 + π · 12) = HOObzUzdvKo=π
Compute the double integral
= SFgqQUkJGdg=23π
= 5dWJhLEUU4Y=π
Solve for the circulation of F around C1.
or6dmYWrEbA=π