Answer
$a) f(x, y) = \frac{1}{3}x^3y^3 + K$
$b) -9$
Work Step by Step
$a)$
$F(x, y) = (x^2y^3)i + (x^3y^2)j$
$\frac{dP}{dy} = 3x^2y^2$
$\frac{dQ}{dx} = 3x^2y^2$
$\frac{dP}{dy} = \frac{dQ}{dx}$, therefore $F(x,y)$ is conservative
$f_x(x,y) = \int (x^2y^3)dx$
$ = \frac{1}{3}x^3y^3 + g(y)$
$f_y(x, y) = \frac{d}{dy}(\frac{1}{3}x^3y^3 + g(y))$
$ = x^3y^2 + g^1(y)$
$f(x,y) = \frac{1}{3}x^3y^3 + K$
$b)$
$0 \leq t \leq 1$
$r(t) = (t^3 - 2t, t^3 + 2t)$
$r(0) = ((0)^3 - 2(0), (0)^3 + 2(0))$
$ = (0, 0)$
$r(1) = ((1)^3 - 2(1), (1)^3 + 2(1))$
$ = (-1, 3)$`
$\int_c Fdr = \int_c \nabla f$
$ = f(r(b)) - f(r(a))$
$ = f(-1, 3) - f(0, 0)$
$ = [(\frac{1}{3}(-1)^3(3)^3) - (\frac{0}{0}(0)^3(0)^3]$
$ = -9$
