Question

Let C be the boundary of the square with vertices (0,0),(1,0),(1,1) and (0,1) oriented the in the counter clockwise sense. Then the value of the line integral
`\oint_ { \rm C} \rm {x}^(2) {y}^(2) dx + ( {x}^(2) - {y}^(2) )dy \\` is ________. (rounded of two decimal places)​

Answer 1

By Green's theorem,

`\displaystyle \oint_C x^2y^2 \, dx + (x^2-y^2) \, dy = \iint_([0,1]^2) \left((\partial(x^2-y^2))/(\partial x) - (\partial(x^2y^2))/(\partial y)\right) \, dx \, dy \\\\ ~~~~~~~~ = \int_0^1 \int_0^1 (2x - 2x^2y) \, dx \, dy \\\\ ~~~~~~~~ = \int_0^1 \left(1-\frac23 y\right) \, dy \\\\ ~~~~~~~~ = 1 - \frac13 = \frac23 \approx \boxed{0.67}`