Find the line integral below.

Question

Find the line integral below.

 

∫_c^ ¦(x−y+2z)ds 

where C is the circle

r(t)=(1, 3 cost, 3 sin t) for 0≤t<2pi.

Answer 1

For a general line integral, we need to arrange the given information in such a way that we are able to integrate over the parameter (t) involved.

 

The function f(x,y,z)=x-y+2z will assume different values depending on where you are located in space. In particular, your x,y,z coordinates will be given by r(t) for the allowable values of t (restricted to [0, 2pi] in this case). Therefore we can say:

f(x,y,z)=f(r(t))=x(t)-y(t)+2z(t)=1-3cos(t)+6sin(t).

and the line integral becomes

∫_c (1-3cos(t)+6sin(t)) ds

 

 

We have a function of t but we are integrating over ds (which represents a piece of C's arc length distance). We can express the differential ds using the chain rule:

ds = (ds/dt)dt

∫_c (1-3cos(t)+6sin(t))(ds/dt) dt

Here, ds/dt represents the speed with which you traverse the curve C. Speed is the magnitude of your velocity vector, and velocity is the rate of change of r(t):

dr/dt= [0, -3 sin(t), 3 cos(t)] = velocity

(ds/dt)= sqrt[(0)^2+(-3 sin(t))^2+(3 cos(t))^2]=sqrt[9]=3=speed

 

Plugging everything in we can evaluate the integral over the range of t:

∫ (1-3cos(t)+6sin(t))(3)dt[0,2pi]

=3t-9sin(t)-18cos(t) [0,2pi]

=(6pi-0-18)-(0-0-18)=6pi.