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

1 Answer | Add Yours

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

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).`

1 reply Hide Replies

amy93's profile pic

amy93 | Student, College Freshman | eNotes Newbie

Posted on

I didn't get this: ∫ xds= ∫ tsint(1+t^2) dt

Could you explain it?

We’ve answered 318,928 questions. We can answer yours, too.

Ask a question