Calculating Line Integral in xy-Plane

[KUphysstudent]

Homework Statement

Calculate the line integral ° v ⋅ dr along the curve y = x³ in the xy-plane when -1 ≤ x ≤ 2 and v = xy i + x² j.

Note: Sorry the integral sign doesn't seem to work it just makes a weird dot, looks like a degree sign, ∫.2. The attempt at a solution
I have to write something along with the calculations for exams so I do that here as well, it is probably not correct.

parameterize x and y in terms of t because we need to take the integral of the dot product of the position vector dr with the vector field v. In this case we can say that x = t which in return gives us y = x³ = t³. Now taking the derivative with of x and y with respect to t, dx/dt = 1 dy/dt = 3t² which we can put into the equation for dr, which is dr = dx i+ dy j+ dz k → dr = 1 dt i+ 3t² dt j + 0 dt k
insert in (integralsign)∫ v ⋅ dr → (integralsign)∫ xy i+x² j ⋅ 1 i+ 3t² j , with the lower bound of integration being t = -1 and the upper bound t = 2.
The dot product of [a, b, c] ⋅ [d, e, f] = [ad, be, cf]. therefore (integralsign)∫ v ⋅ dr = xy + x² ⋅ 1 dt i+ 3t² dt j = xy dt + 3t² x² dt

3. Relevant equations

But now I'm completely lost and don't understand what I'm supposed to do with x's and y's, shouldn't they have been in terms of t's as well? My book has one example which is much easier than this one since the vector field is just v = yi in that one.

Anyone who can teach me what to do please let me know.

→
Do I just substitute xy and x² with my parameter x = t and y = t³, so I get xy = t³ and x² = t² → (integralsign)∫ t³ dt + 3t² t² dt ?

 

[tnich]

Homework Statement

Calculate the line integral ° v ⋅ dr along the curve y = x³ in the xy-plane when -1 ≤ x ≤ 2 and v = xy i + x² j.

Note: Sorry the integral sign doesn't seem to work it just makes a weird dot, looks like a degree sign, ∫.2. The attempt at a solution
I have to write something along with the calculations for exams so I do that here as well, it is probably not correct.

parameterize x and y in terms of t because we need to take the integral of the dot product of the position vector dr with the vector field v. In this case we can say that x = t which in return gives us y = x³ = t³. Now taking the derivative with of x and y with respect to t, dx/dt = 1 dy/dt = 3t² which we can put into the equation for dr, which is dr = dx i+ dy j+ dz k → dr = 1 dt i+ 3t² dt j + 0 dt k
insert in (integralsign)∫ v ⋅ dr → (integralsign)∫ xy i+x² j ⋅ 1 i+ 3t² j , with the lower bound of integration being t = -1 and the upper bound t = 2.
The dot product of [a, b, c] ⋅ [d, e, f] = [ad, be, cf]. therefore (integralsign)∫ v ⋅ dr = xy + x² ⋅ 1 dt i+ 3t² dt j = xy dt + 3t² x² dt

3. Relevant equations

But now I'm completely lost and don't understand what I'm supposed to do with x's and y's, shouldn't they have been in terms of t's as well? My book has one example which is much easier than this one since the vector field is just v = yi in that one.

Anyone who can teach me what to do please let me know.

→
Do I just substitute xy and x² with my parameter x = t and y = t³, so I get xy = t³ and x² = t² → (integralsign)∫ t³ dt + 3t² t² dt ?

Yes. Everything you have done looks correct so far except for your final substitution. You have the right idea, but check for errors.

 

[KUphysstudent]

if you calculate the integral with this step, (integralsign)∫ t³ dt + 3t² t² dt from -1 to 2 you get 25.25

I forgot to add that if you graph y=x^3 and look at the area between x = 1 and x = 2 it looks like it adds to about 4, which is quite a bit away from 25.25