# Calculate the line integral of the spin field S=<-y, x> along one twist of the arithmetic spiral x=tcos(t), y=tsin(t), and 0<= t <=2pi

You need to evaluate the following line integral, such that:

int_C f(x,y) ds = int_0^(2pi) f(h(t),g(t))sqrt(((dx)/(dt))^2 + ((dx)/(dt))^2)dt

The problem provides the parametric equations, such that:

x = t*cos t => (dx)/(dt) = cos t - t*sin t => ((dx)/(dt))^2 = (cos t - t*sin t)^2 => ((dx)/(dt))^2 = cos^2 t - 2t*sin t*cos t + t^2*sin^2 t

y = t*sin t => (dy)/(dt) = sin t + t*cos t => ((dy)/(dt))^2 = (sin t + t*cos t)^2 => ((dy)/(dt))^2 = sin^2 t + 2 t*sin t*cos t + t^2*cos^2 t

((dx)/(dt))^2 + ((dx)/(dt))^2 = cos^2 t - 2t*sin t*cos t + t^2*sin^2 t + sin^2 t + 2 t*sin t*cos t + t^2*cos^2 t

Using the Pythagorean identity, yields:

sin^2 t + cos^2 t = 1

((dx)/(dt))^2 + ((dx)/(dt))^2 = 1 + t^2(sin^2 t + cos^2 t)

((dx)/(dt))^2 + ((dx)/(dt))^2 = 1 + t^2

You need to evaluate the following line integral such that:

int_C x ds = int_0^(2pi) t*sin t*(1 + t^2)dt

int_C x ds = int_0^(2pi) (t*sin t + t^3 sin t) dt

You need to use integration by parts to evaluate int_0^(2pi) (t*sin t)dt , such that:

int udv = uv - int vdu

u = t => du = dt

dv = sin t => v = -cos t

int_0^(2pi) (t*sin t) = -t*cos t|_0^(2pi) + int_0^(2pi) cos t dt

int_0^(2pi) (t*sin t) = -t*cos t|_0^(2pi) + sint |_0^(2pi)

int_0^(2pi) (t*sin t) = (-2pi*cos 2pi + sin 2pi + 0 + sin 0)

int_0^(2pi) (t*sin t) = -2pi

You need to use integration by parts to evaluate int_0^(2pi) (t^3*sin t)dt , such that:

u = t^3 => du = 3t^2dt

dv = sin t => v = -cos t

int_0^(2pi) (t^3*sin t) = -t^3*cos t|_0^(2pi) + int_0^(2pi) 3t^2cos t dt

u = t^2 => du = 2tdt

dv = cos t => v = sin t

int_0^(2pi) 3t^2cos t dt = 3t^2sin t|_0^(2pi) - 6int_0^(2pi)tsin tdt

Since int_0^(2pi)tsin tdt = -2pi yields:

int_0^(2pi) 3t^2cos t dt = 3t^2sin t|_0^(2pi) + 12 pi

int_0^(2pi) 3t^2cos t dt = 3(4pi^2 sin 2pi - 0) + 12pi

int_0^(2pi) 3t^2cos t dt = 12pi

int_0^(2pi) (t^3*sin t) = -t^3*cos t|_0^(2pi) + 12pi

int_0^(2pi) (t^3*sin t) = -8pi^3 + 12pi

int_0^(2pi) (t^3*sin t) = 4pi(3 - 2pi^2)

int_C x ds = -2pi + 4pi(3 - 2pi^2)

int_C x ds = 4pi(1 - pi^2)

Hence, evaluating the given line integral, under the given conditions, yields int_C x ds = 4pi(1 - pi^2).

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