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G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Jan 30, 2010 #2 along parabola y=x2\displaystyle y=x^{2}y=x2 F(x,y)=2xy2i+2(x2y+y)j\displaystyle F(x,y)=2xy^{2}i+2(x^{2}y+y)jF(x,y)=2xy2i+2(x2y+y)j Let x=t be the parameter. Then, the path C: x=t, y=t2, (0≤t≤2)\displaystyle x=t, \;\ y=t^{2}, \;\ (0\leq t \leq 2)x=t, y=t2, (0≤t≤2) ∫CA⋅dr=∫C(2xy2⋅i+2(x2y+y)j)⋅(dx⋅i+dy⋅j)\displaystyle \int_{C}A \cdot dr=\int_{C}\left(2xy^{2}\cdot i+2(x^{2}y+y)j\right)\cdot (dx\cdot i+dy\cdot j)∫CA⋅dr=∫C(2xy2⋅i+2(x2y+y)j)⋅(dx⋅i+dy⋅j) =∫02(2t5+2t2(t2+1)(2t))dt\displaystyle =\int_{0}^{2}\left(2t^{5}+2t^{2}(t^{2}+1)(2t)\right)dt=∫02(2t5+2t2(t2+1)(2t))dt
along parabola y=x2\displaystyle y=x^{2}y=x2 F(x,y)=2xy2i+2(x2y+y)j\displaystyle F(x,y)=2xy^{2}i+2(x^{2}y+y)jF(x,y)=2xy2i+2(x2y+y)j Let x=t be the parameter. Then, the path C: x=t, y=t2, (0≤t≤2)\displaystyle x=t, \;\ y=t^{2}, \;\ (0\leq t \leq 2)x=t, y=t2, (0≤t≤2) ∫CA⋅dr=∫C(2xy2⋅i+2(x2y+y)j)⋅(dx⋅i+dy⋅j)\displaystyle \int_{C}A \cdot dr=\int_{C}\left(2xy^{2}\cdot i+2(x^{2}y+y)j\right)\cdot (dx\cdot i+dy\cdot j)∫CA⋅dr=∫C(2xy2⋅i+2(x2y+y)j)⋅(dx⋅i+dy⋅j) =∫02(2t5+2t2(t2+1)(2t))dt\displaystyle =\int_{0}^{2}\left(2t^{5}+2t^{2}(t^{2}+1)(2t)\right)dt=∫02(2t5+2t2(t2+1)(2t))dt
