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G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Jul 4, 2010 #2 Did you use ∮C[(−1y3)dx+(1x3)dy]=3a23∫02π∫0adydx\displaystyle \oint_{C}\left[\left(\frac{-1}{\sqrt[3]{y}}\right)dx+\left(\frac{1}{\sqrt[3]{x}}\right)dy\right]=3a^{\frac{2}{3}}\int_{0}^{2\pi}\int_{0}^{a}dydx∮C[(3y−1)dx+(3x1)dy]=3a32∫02π∫0adydx ?. Where x=a⋅cos3(t), y=a⋅sin3(t)\displaystyle x=a\cdot cos^{3}(t), \;\ y=a\cdot sin^{3}(t)x=a⋅cos3(t), y=a⋅sin3(t) This is using Green's Theorem. One can achieve the same result.
Did you use ∮C[(−1y3)dx+(1x3)dy]=3a23∫02π∫0adydx\displaystyle \oint_{C}\left[\left(\frac{-1}{\sqrt[3]{y}}\right)dx+\left(\frac{1}{\sqrt[3]{x}}\right)dy\right]=3a^{\frac{2}{3}}\int_{0}^{2\pi}\int_{0}^{a}dydx∮C[(3y−1)dx+(3x1)dy]=3a32∫02π∫0adydx ?. Where x=a⋅cos3(t), y=a⋅sin3(t)\displaystyle x=a\cdot cos^{3}(t), \;\ y=a\cdot sin^{3}(t)x=a⋅cos3(t), y=a⋅sin3(t) This is using Green's Theorem. One can achieve the same result.
