The arc length integral can be generalized to the integral of any
function.
b. Line Integrals of Scalars
Line Integral of a Scalar Function
The line integral of a function, f(x,y,z),
along a parametric curve, r(t)=(x(t),y(t),z(t)), between
A=r(a) and B=r(b) is given by
∫ABfds=∫abf(r(t))∣v∣dt
where f(r(t)) is the composition of f and r,
i.e. the function f(x,y,z) with x, y and z replaced by
x(t), y(t) and z(t). The composition f(r(t)) is
called the value of the function along the curve
or the restriction of the function to the curve.
We sometimes write f(t) to mean f(r(t)), the value of f at
time t, even though this is not mathematically correct. Consequently,
this is sometimes called abuse of notation.
Notice that we again switch the limits on the integral from the abstract
points A and B to the values a and b of the parameter t when we
switch from the integral with the general differential ds to the integral
with the parameter differential dt.
Compute ∫(0,0,0)(1,1,2/3)(3xz+y2)ds along the
twisted cubic r(t)=(t,t2,32t3).
The value of the function along the curve is
f(r(t))=3x(t)z(t)+y(t)2=3(t)(32t3)+(t2)2=3t4
The velocity is v=(1,2t,2t2) and its length is
∣v∣=1+4t2+4t4=(1+2t2)2=1+2t2.
Finally, the endpoints are:
(0,0,0)=r(0)(1,1,32)=r(1)
So
∫(0,0,0)(1,1,2/3)(3xz+y2)ds=∫01f(r(t))∣v∣dt=∫013t4(1+2t2)dt=[53t5+76t7]01=53+76=3551
Compute ∫x2ds along the helix
x=4cos3z,y=4sin3z
between z=0 and z=6π. You can rotate this plot with your mouse.
INFO: Time for setup and init of GL element no. 0: 128 ms.
The derivatives are:
dzdx=−34sin3zanddzdy=34cos3z
So the differential of arc length is
ds=(dzdx)2+(dzdy)2+1dz=(−34sin3z)2+(34cos3z)2+1dz=916+1dz=35dz
Thus, using the identity cos2A=21+cos2A:
∫x2ds=∫06π16cos2(3z)35dz=380∫06π21+cos32zdz=340[z+23sin32z]06π=80π
We first identify the integrand and evaluate it on the curve:
ff(r(t))=x2+y2+z2=16cos2t+16sin2t+9t2=16+9t2
The velocity is v(t)=(−4sint,4cost,3) and its length is
∣v∣=5.
Also, the point (4,0,0) occurs at t=0 and the point (4,0,6π)
occurs at t=2π. So the integral becomes:
∫(4,0,0)(4,0,6π)(x2+y2+z2)ds=∫02π(16+9t2)5dt=5[16t+3t3]02π=160π+120π3
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