Calculating Work of Force Field F on Curve C

[user3]

Find the work done by the Force Field F to make a displacement on the curve C.

F= <-y^2 , x>

C: semicircle x^2 + y^2 = 1 , y<=0 , from (-1,0) to (1,0)since y<=0 , then it's the semi circle under the x-axis. and according to the solution I have:

Work=integral[sin t - sin t cos^2 t +(1+cos 2t)/2] dt form t= -pi to t=0

Now I understand everything going on except the limits. why from -pi to 0? why not from pi to 2pi?

And it may not make a difference here, but if the integral had a constant inside it would.

 

[tiny-tim]

hi user3! welcome to pf!

Now I understand everything going on except the limits. why from -pi to 0? why not from pi to 2pi?

either will do

And it may not make a difference here, but if the integral had a constant inside it would.

no, if the field is a function of x and y, the integrand would always be the same whenever you add 2π to the angle

 

[user3]

Ok...let's say you have the following double integral: integral[integral(x^2 + y)] in region D, where D is the semicircle below the x axis.

If one decides to use polar coordinates: x=rcos(theta) and y=rsin(theta)

the integral becomes integral[ integral(r^3 cos^2 (theta) + r^2 sin(theta)) dr) d(theta)] , what are the proper limits of the integral?

please note that if you rewrite the cos^2(theta) term using the half angle rule, you would have a free (theta) term in the outer integral, in which case the choice of limits would make a difference.

Thank you :)

 

[tiny-tim]

hi user3!

please note that if you rewrite the cos^2(theta) term using the half angle rule, you would have a free (theta) term in the outer integral, in which case the choice of limits would make a difference.

you mean, 1/2 + 1/2cos2θ ?

yes, the 1/2 makes a difference then, but so does the 1/2cos2θ (because it's 2θ not θ) …

and they should cancel out! ​

 

[user3]

Thank you! I was pretty confused.