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.
Yay, I (almost) I figured it out.
Green's thm says that if we have a simply connected region, R with a boundary C (smooth), and which is positively oriented, then we have M and N (that is, if they have continuous partial derivatives) where they are in an open region containing R...then,
int(Mdx + Ndy), with C as the limit of integration = int(int(N_x - M_y))dydx ..
M = 2xy - e^(cos(x^2)+3x) ; N = sin(3y^2 + e^y) + xy^2
N_x = y^2 ; M_y = 2x
So, int(y^3 - 2xy dx), but I need limits of integration now .. so it should be zero is what you're saying?
No, it should be: iint(y^2 - 2x) dA
The limits of integration is over the region which is a circle at the origin with radius 2.
If you express what I have in polar coordinates you should get,
INT(theta=0 , theta = 2*pi) INT(r=0, r=2) [r^2*sin^2(theta) - 2r*cos(theta)]rdr dtheta
Note, for #2 we can use the fabolous fundamental theorem of line integrals.
We start at theta =0 so r=0. But that means we start at the origin.
We end at theta = 8*pi. Now, x=r*cos theta and y=r*sin theta. So x=8*pi *cos (8pi) = 8*pi and y=8*pi sin(8*pi)=0
So you end at (8pi,0).
1)Evaluate scalar potential at (8pi,0)
2)Evaluate scalar potential at (0,0)
3)Substract #2 from #1.
4)#3 is the answer.