Is this correct. Evaluate the line integral

[afcwestwarrior]

Evaluate the line integral \int F \circ dr for

(a) F(x,y) = (x - y) * i + xy * j and C is the top half of a circle of radius 2.


Here's green's theorem. Double integral ( dQ/ dx - dP/dy) dA

dQ/dx = y
dP /dy = -1


It becomes ∫∫ (y +1) dA

y = r sin Θ




so then it becomes ∫∫ (r sin Θ + 1) r dr dΘ
I changed it into cylindrical coordinates


then 0 < r < 2
and 0 < Θ < pi


then we integrate respect to r

and we get ∫ r^3/3 sin Θ + r^2/ 2 d Θ r is from 0 to 2

we plug in r

∫ 2^3/3 sin Θ + 2^2/ 2 - 0 d Θ = ∫ 4 sin Θ + 2 d Θ

then we integrate for theta

and we get
- 8/3 cos Θ + 2Θ from 0 to pi = - 8/3 cos pi + 2 pi - -8/3 cos 0 + 2(0)


= -8/3(-1) + 2pi - [ -8/3 -0]



= 8/3 + 2pi + 8/3 = 16/3 + 2pi

[HallsofIvy]

Evaluate the line integral \int F \circ dr for

(a) F(x,y) = (x - y) * i + xy * j and C is the top half of a circle of radius 2.


Here's green's theorem. Double integral ( dQ/ dx - dP/dy) dA

dQ/dx = y
dP /dy = -1


It becomes ∫∫ (y +1) dA

y = r sin Θ




so then it becomes ∫∫ (r sin Θ + 1) r dr dΘ
I changed it into cylindrical coordinates


then 0 < r < 2
and 0 < Θ < pi


then we integrate respect to r

and we get ∫ r^3/3 sin Θ + r^2/ 2 d Θ r is from 0 to 2

we plug in r

∫ 2^3/3 sin Θ + 2^2/ 2 - 0 d Θ = ∫ 4 sin Θ + 2 d Θ
2^3/3= 8/3, not 4!

then we integrate for theta

and we get
- 8/3 cos Θ + 2Θ from 0 to pi = - 8/3 cos pi + 2 pi - -8/3 cos 0 + 2(0)


= -8/3(-1) + 2pi - [ -8/3 -0]



= 8/3 + 2pi + 8/3 = 16/3 + 2pi
Greens theorem equates \int\int ( dQ/ dx - dP/dy) dA to the integral around the closed path forming the boundary of the region. Your original problem " C is the top half of a circle of radius 2" does not have a closed path. You could, after correcting this calculation, find the integral from (-2, 0) back to (2, 0), along the x-axis, and subtract it off.

Are you required to use Green's theorem? Just a straight path integral does not look difficult.