Calculate the following line integrals by hand
1.) Note: this integral is a closed curve (so it has the o in the middle of the integral):
int_(C)(2xy – e^(cos(x^2) + 3x)dx + (sin(3y^2 + e^y) + xy^2)dy, where C is the circle of radius 2 centered at the origin oriented counterclockwise.
2.) These below are NOT a closed curve (so DOESN’T have a o in the middle)
int_(C)(F*dr) (note F and r are bold since they are vectors.. also note, * means dot product),
where F(x,y) = <2xy – 3y^2 + 15x^2, x^2 – 6xy + 2>, where C is the polar spiral r = theta for 0 <= theta <= 8Pi
3.) int_(C)(F*dr), where F(x,y) = <xe^(y) + 2xy, cos(y) – x> and C is the portion of the parabola y = x^2 for -1 <= x <= 2 moving from left to right
WORK: Okay, so I tried #2, but I doubt it's this easy. Especially since I think it has to be done using polar coordinates. I'm stuck. This was my thinking:
F(x,y) = <F_(1)(x,y), F_(2)(x,y)>
x = x(t) ... y = y(t)
dr = <dx, dy>
Line integral int_(C)(F*dr), then, is int_(C)[F_(1)(x,y)dx] + int_(C)[F_(2)(x,y)dy]
So, int_(C)[2xy - 3y^2 + 15x^2dx] + int_(C)[x^2 - 6x + 2dy]
= x^2y - 3xy^2 + 5x^3 + x^2y - 6x + 2y, evaluated from whatever the limits of integration are (C).
= 5x^3 + 2x^2y - 3xy^2 - 6xy + 2y, again evaluated from whatever limits of integration are.