Answer :
The value of the line integral [tex]f(x,y) = \dfrac{x^2y^2}{2}[/tex] is 2
What is line integral?
Integration is defined as the summation of the small units to find the total value so line integration is the summation of the small parts of the line to find the total length of the line.
h(x,y) = xy²i + yx²j
Note that if you derivate the first part over the variable y and the second part over the variable x, then in both cases you obtain 2xy, therefore there must be a function f whose gradient is h, because the cross derivates are equal.
In order to find such f, you can calculate a primitive of both expressions, the first one over the variable x and the second one over the variable y.
A general primitive of xy² i (over x) is
[tex]f_1 (x,y) = \dfrac{x^2y^2}{2} +a(y)[/tex]
With a(y) a function that depends only on y. A general primitive of yx² j (over y) is
[tex]f_2(x,y)=\dfrac{x^2y^2}{2}+b(x)[/tex]
With b(x) only depending on x
The function f(x,y) whose gradient is h is obtained by equating the expressions of f₁ and f₂. f₁ and f₂ are equal when a(x) = b(x) = 0, therefore
[tex]f(x,y)=\dfrac{x^2y^2}{2}[/tex]
note that
fx(x,y) = xy²
fy(x,y) = yx²
As we wanted. Let's find the endpoints of the C
r(u) = u i + 2 u² j
r(0) = (0,0)
f(1) = (1,2)
Therefore,
[tex]\int_cxy^2dx+yx^2dy=f(1,2)-f(0,0)=\dfrac{1^2\times 2^2}{2}-0=\dfrac{4}{2}-0[/tex]
[tex]\int_cxy^2dx+yx^2dy=f(1,2)-f(0,0)=2[/tex]
Therefore the value of the line integral over C is 2.
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