calculus

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Evaluate the following line integral: ∮C[(4x^2+3x+5y)dx+(6x^2+5x+3y)dy] where C is the path around the square with vertices (0,0),(2,0),(2,2) and (0,2).

  • calculus -

    from point 1 to point 2 dy = 0, y = 0 and x goes from 0 to 2
    so integral one = ∮[(4x^2+3x)dx =(4/3)x^3 + (3/2)x^2 = 32/3+6 = 50/3

    from point 2 to point 3, dx = 0 , x = 2 and y from 0 to 2
    so integral 2 = ∮(34+3y)dy] from 0 to 2

    similar for the last two lines, then add the results. Remember direction for example for integal 3 y does not change but x is from 2 to 0, not 0 to 2

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