Answer:
a) 2π
b) 0
Step-by-step explanation:
Recall that for a parametrized differentiable curve C = (x(t), y(t)) with the parameter t varying on some interval [a, b]
[tex]\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[/tex]
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)
[tex]\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[/tex]
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)
[tex]\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[\frac{sin(2\pi t)}{2\pi}\right]_2^4=\frac{2}{\pi}(sin(8 \pi)-sin(4\pi))=0[/tex]
