Use Green's Theorem to evaluate the line integral. Orient the curve counterclockwise.
, where C is the triangle with vertices (0, 0), (1, 0), (1, 1).
Recall how to apply Green's Theorem. Let D be a domain whose boundary ∂D (diffD) is a simple closed curve, oriented counterclockwise. Then we have the following equation.
In the figure above, we see that ∂D = C is the triangle bounding the shaded region. In addition, the triangle is oriented counterclockwise as indicated in the problem statement.
Because C = ∂D, we note that is already in the correct form with
F1 =
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and
F2 =
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To apply Green's Theorem to solve , we will need to calculate the appropriate partial derivatives and find the appropriate bounds of integration for the double integral regarding domain of the triangle.
Calculate the partial derivatives (∂F2 / ∂x) and (∂F1 / ∂y) where F1 = and F2 = .
(∂F2 / ∂x) = 1Wh3cvJ2xF4=
(∂F1 / ∂y) = iSba6t70dtA=
Therefore we have (∂F2 / ∂x) − (∂F1 / ∂y) = TyC2GUIutPg=
Find the appropriate bounds of integration for the double integral.
Notice that the triangle is bounded above by a segment of the line y = x and bounded below by the x-axis between x = 0 and x = 1. We also see that the height of the region, y, is between 0 and x as x increases from 0 to 1.
With this information, use Green's Theorem to set up the appropriate double integral.
tY+RVWbpPMOH9fsDk4W+aA==
Use Green's Theorem to evaluate the line integral. Round your answer to three decimal places.
U2GIbglD1oM=