Answer:
Explanation:
From the information:
[tex]F(x,y) = 2xy^3i+3x^2y^2j;P(-9,4), B(10,5)[/tex]
[tex]W = \int ^{10,5}_{-9,4} f .dn \\ \\ W = \int ^{10,5}_{-9,4} (2xy^3i) + 3x^2y^2j) *(dxi+dyj) \\ \\ f = 2xy^3\ \ ,\ \ g = 3x^2y^2 \\ \\ \dfrac{\partial f}{\partial y} = \dfrac{\partial g}{\partial x} \\ \\ \text{We wil realize that f is conservative; as a result, there is a potential function } \phi ;\\\\ \dfrac{\partial \phi}{dx}= 2xy^3 \\ \\ \phi= \dfrac{2x^2}{2}y^3+f(y) \\ \\ \phi = x^2y^3 + f(y) \\ \\ \dfrac{\partial \phi}{\partial y } = 3x^2y^2 + f'(y) \\ \\[/tex]
[tex]\dfrac{\partial \phi}{\partial y } = 3x^2y^2 + f'(y) = 3x^2y^2 \\ \\ f'(y) = 0 \\ \ f(y) = k \\ \\ \phi = x^2y^3 + k \\ \\ Recall: \int^{10,5}_{-9,4 } \ F* dn = W = \phi(10,5) - \phi (-9,4) \\ \\ = (10)^2(5)^3 + k - (-9)^2(4)^3 - k \\ \\ = (100*125) - (81*64) \\ \\ = 12500 - 5184 \\ \\ =7316[/tex]
