Explain how this results in the integral equaling
zero?
a. The unit circle with the specified orientation is described parametrically by r(t) = (cost, sin t), for 0 <t< 27. Therefore, r'(t) = (-sint, cos t) and the circulation of the radial field F = (x,y) is = = = SF.Tds = $2" F:r'(t)dt Evaluation of a line integral с = 82" (cos t, sin t) · (–sin t, cos t)dt Substitute for F and r'. F=(x,y) r'(t) J7 0 dt = 0. Simplify. The tangential component of the radial field is zero everywhere so the
a. The unit circle with the specified orientation is described parametrically by r(t) = (cost, sin t), for 0
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a. The unit circle with the specified orientation is described parametrically by r(t) = (cost, sin t), for 0
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