
Yefim S. answered 12/10/23
Math Tutor with Experience
x = cosθ; y = sinθ; r = 1; and θ from π/2 to 0.
lC xy^(3)dx-4xdy, = ∫π/20[cosθsin3θ(-sinθ) - 4cos2θ]dθ = ∫π/20(- sin4θcosθ - 2 - 2cos2θ)dθ =
= (- sin5/5 - 2θ - sin2θ)π/20 = 1/5 +π
Abdul A.
asked 12/10/23Evaluate IntegralC xy^(3)dx-4xdy, where C is the portion of the circle centered at the origin of radius 1 in the first quadrant with clockwise rotation.
Thanks in advance!
Yefim S. answered 12/10/23
Math Tutor with Experience
x = cosθ; y = sinθ; r = 1; and θ from π/2 to 0.
lC xy^(3)dx-4xdy, = ∫π/20[cosθsin3θ(-sinθ) - 4cos2θ]dθ = ∫π/20(- sin4θcosθ - 2 - 2cos2θ)dθ =
= (- sin5/5 - 2θ - sin2θ)π/20 = 1/5 +π
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Abdul A.
why is the bound pi/2 to 0 and not 0 to pi/212/10/23