Gradient theorem
Example 1

Suppose γ ⊂ R is the circular arc oriented counterclockwise from (5, 0) to (−4, 3). Using the definition of a line integral,

∫ γ y d x + x d y = ∫ 0 π − tan − 1 (3 4) ((5 sin t) (− 5 sin t) + (5 cos t) (5 cos t)) d t = ∫ 0 π − tan − 1 (3 4) 25 (− sin 2 t + cos 2 t) d t = ∫ 0 π − tan − 1 (3 4) 25 cos (2 t) d t = 25 2 sin (2 t) | 0 π − tan − 1 (3 4) = 25 2 sin (2 π − 2 tan − 1 (3 4)) = − 25 2 sin (2 tan − 1 (3 4)) = − 25 (3 / 4) (3 / 4) 2 + 1 = − 12. {\displaystyle {\begin{aligned}\int _{\gamma }y\,\mathrm {d} x+x\,\mathrm {d} y&=\int _{0}^{\pi -\tan ^{-1}\!\left({\frac {3}{4}}\right)}((5\sin t)(-5\sin t)+(5\cos t)(5\cos t))\,\mathrm {d} t\\&=\int _{0}^{\pi -\tan ^{-1}\!\left({\frac {3}{4}}\right)}25\left(-\sin ^{2}t+\cos ^{2}t\right)\mathrm {d} t\\&=\int _{0}^{\pi -\tan ^{-1}\!\left({\frac {3}{4}}\right)}25\cos(2t)\mathrm {d} t\ =\ \left.{\tfrac {25}{2}}\sin(2t)\right|_{0}^{\pi -\tan ^{-1}\!\left({\tfrac {3}{4}}\right)}\\&={\tfrac {25}{2}}\sin \left(2\pi -2\tan ^{-1}\!\!\left({\tfrac {3}{4}}\right)\right)\\&=-{\tfrac {25}{2}}\sin \left(2\tan ^{-1}\!\!\left({\tfrac {3}{4}}\right)\right)\ =\ -{\frac {25(3/4)}{(3/4)^{2}+1}}=-12.\end{aligned}}}

This result can be obtained much more simply by noticing that the function f (x, y) = x y {\displaystyle f(x,y)=xy} has gradient ∇ f (x, y) = (y, x) {\displaystyle \nabla f(x,y)=(y,x)}, so by the Gradient Theorem:

∫ γ y d x + x d y = ∫ γ ∇ (x y) ⋅ (d x, d y) = x y | (5, 0) (− 4, 3) = − 4 ⋅ 3 − 5 ⋅ 0 = − 12. {\displaystyle \int _{\gamma }y\,\mathrm {d} x+x\,\mathrm {d} y=\int _{\gamma }\nabla (xy)\cdot (\mathrm {d} x,\mathrm {d} y)\ =\ xy\,|_{(5,0)}^{(-4,3)}=-4\cdot 3-5\cdot 0=-12.}
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