Question

Evaluate each of the following line integrals.

(a)

integral.gif
C
x dy − y dx, c(t) = (cos(t), sin(t)), 0 ≤ t ≤ 2π

(b)

integral.gif
C
x dy + y dx, c(t) = (2 cos(πt), 2 sin(πt)), 2 ≤ t ≤ 4

Answer 1

a) 2π

b) 0

Explanation:

Recall that for a parametrized differentiable curve C = (x(t), y(t)) with the parameter t varying on some interval [a, b]

`\large \int_(C)[P(x,y)dx+Q(x,y)dy]=\int_(a)^(b)[P(x(t),y(t))x'(t)+Q(x(t),y(t))y'(t)]dt`

Where P, Q are scalar functions

a)

C(t) = (x(t), y(t)) = (cos(t), sin(t)), 0 ≤ t ≤ 2π

P(x,y) = -y ==> P(x(t),y(t)) = -y(t) = -sin(t)

Q(x,y) = x ==> Q(x(t),y(t)) = x(t) = cos(t)

x'(t) = -sin(t)

y'(t) = cos(t)

`\large \int_(C)[-ydx+xdy]=\int_(0)^(2\pi)[(-sin(t))(-sin(t))+cos(t)cos(t)]dt=\\\\\int_(0)^(2\pi)[sin^2(t)+cos^2(t)]dt=\int_(0)^(2\pi)dt=2\pi`

b)

C(t) = (x(t), y(t)) = (2cos(πt), 2sin(πt)), 2 ≤ t ≤ 4

P(x,y) = y ==> P(x(t),y(t)) = y(t) = 2sin(πt)

Q(x,y) = x ==> Q(x(t),y(t)) = x(t) = 2cos(πt)

x'(t) = -2πsin(πt)

y'(t) = 2πcos(πt)

`\large \int_(C)[ydx+xdy]=\int_(2)^(4)[(2sin(\pi t))(-2\pi sin(\pi t))+2cos(\pi t)2\pi cos(\pi t)]dt=\\\\4\int_(2)^(4)[cos^2(\pi t)-sin^2(\pi t)]dt=4\int_(2)^(4)cos(2\pi t)dt=\\\\4\left[(sin(2\pi t))/(2\pi)\right]_2^4=(2)/(\pi)(sin(8 \pi)-sin(4\pi))=0`