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The potential function φ(x, Y) for the vector field F(x, Y) = (3 + 4xY^2)i + 4x^2Yj is given by φ(x, Y) = 3x + 2x^2Y^2 + C1(Y) + C2(x). To calculate the integral ∫ C F · dr along curve C, we substitute the coordinates of the starting and ending points of the curve into φ(x, Y)

To determine the potential function, we need to find a scalar function φ(x, Y) such that the vector field F(x, Y) can be expressed as the gradient of φ, i.e., F(x, Y) = ∇φ.

Given F(x, Y) = (3 + 4xY^2)i + 4x^2Yj, we can find the potential function φ(x, Y) by integrating the components of F with respect to their respective variables:

φ(x, Y) = ∫ (3 + 4xY^2) dx + ∫ 4x^2Y dy

Integrating the first component with respect to x gives:

∫ (3 + 4xY^2) dx = 3x + 2x^2Y^2 + C1(Y),

where C1(Y) is the constant of integration with respect to x.

Integrating the second component with respect to Y gives:

∫ 4x^2Y dy = 2x^2Y^2 + C2(x),

where C2(x) is the constant of integration with respect to Y.

Combining the results, we have:

φ(x, Y) = 3x + 2x^2Y^2 + C1(Y) + C2(x).

To find the potential function along the curve C, we substitute the given values for x and Y, i.e., (1, 1) and (2, 1/2) into φ(x, Y) and subtract the values at the starting point from the values at the ending point:

∫ C F · dr = φ(2, 1/2) - φ(1, 1)

= (3(2) + 2(2)^2(1/2)^2 + C1(1/2) + C2(2)) - (3(1) + 2(1)^2(1)^2 + C1(1) + C2(1))

Simplifying further will yield the numerical value of the integral along the curve C.

To learn more about the potential function

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