- #1

KUphysstudent

- 40

- 1

## Homework Statement

Calculate the line integral °

**v ⋅**d

**r**along the curve y = x

^{3}in the xy-plane when -1 ≤ x ≤ 2 and

**v**= xy

**i**+ x

^{2}

**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 d

**r**with the vector field

**v**. In this case we can say that x = t which in return gives us y = x

^{3}= t

^{3}. Now taking the derivative with of x and y with respect to t, dx/dt = 1 dy/dt = 3t

^{2}which we can put into the equation for d

**r**, which is d

**r**= dx

**i**+ dy

**j**+ dz

**k**→ d

**r**= 1 dt

**i**+ 3t

^{2}dt

**j**+ 0 dt

**k**

insert in (integralsign)∫

**v ⋅**d

**r →**(integralsign)∫ xy

**i**+x

^{2}

**j ⋅**1

**i**+ 3t

^{2}

**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 ⋅**d

**r**= xy + x

^{2}

**⋅**1 dt

**i**+ 3t

^{2}dt

**j**= xy dt + 3t

^{2}x

^{2}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**= y

**i**in that one.

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

→

Do I just substitute xy and x

^{2}with my parameter x = t and y = t

^{3}, so I get xy = t

^{3}and x

^{2}= t

^{2}→ (integralsign)∫ t

^{3}dt + 3t

^{2}t

^{2}dt ?