Respuesta :
Answer:
Step-by-step explanation:
given a field of the form F = (P(x,y),Q(x,y) and a simple closed curve positively oriented, then
[tex]\int_{C} F \cdot dr = \int_A \frac{dQ}{dx} - \frac{dP}{dy} dA[/tex] where A is the area of the region enclosed by C.
In this case, by the description we can assume that C starts at (0,0). Then it goes the point (pi,0) on the path giben by y = sin(x) and then return to (0,0) along the straigth line that connects both points. Note that in this way, the interior the region enclosed by C is always on the right side of the point. This means that the curve is negatively oriented. Consider the path C' given by going from (0,0) to (pi,0) in a straight line and the going from (pi,0) to (0,0) over the curve y = sin(x). This path is positively oriented and we have that
[tex] \int_{C} F\cdot dr = - \int_{C'} F\cdot dr[/tex]
We use the green theorem applied to the path C'. Taking [tex] P = x+4y^3, Q = 4x^2+y[/tex] we get
[tex] \int_{C'} F\cdot dr = \int_{A} 8x-12y^2dA[/tex]
A is the region enclosed by the curves y =sin(x) and the x axis between the points (0,0) and (pi,0). So, we can describe this region as follows
[tex]0\leq x \leq \pi, 0\leq y \leq \sin(x)[/tex]
This gives use the integral
[tex] \int_{A} 8x-12y^2dA = \int_{0}^{\pi}\int_{0}^{\sin(x)} 8x-12y^2 dydx[/tex]
Integrating accordingly, we get that [tex]\int_{C'} F\cdot dr = 8\pi - \frac{16}{3}[/tex]
So
[tex] \int_{C} F cdot dr = - (8\pi - \frac{16}{3}) = \frac{16}{3} - 8\pi [/tex]
