# Integrating vector functions

1. Jul 31, 2010

### jegues

1. The problem statement, all variables and given/known data
See figure

2. Relevant equations

3. The attempt at a solution

For an integral like this am I allowed to evaluate the cross product between the to vectors,

$$\vec{u}$$ and $$\vec{v}$$?

If so then,

$$\vec{p} = \vec{u}\times\vec{v}$$

and,

$$\int f\vec{p} = -3\int t\vec{p}$$

Does this work?

#### Attached Files:

• ###### VECTORF.JPG
File size:
11.7 KB
Views:
83
2. Jul 31, 2010

### xaos

under what conditions should you think you cannot?

3. Aug 1, 2010

### HallsofIvy

Yes, it is always true that $\left(a\vec{u}\right)\times\vec{v}= a\left(\vec{u}\times\vec{v}\right)$.

(But $\vec{u}\times\left(a\vec{v}\right)= -a\left(\vec{u}\times\vec{v}\right)$.)

4. Aug 2, 2010

### jegues

So,

$$\int f\vec{p}dt = -3\int t\vec{p}dt = \int \left(-3t^{3} - 3t\right)dt\hat{i} + \int \left(-6t^{2} + 3t\right)dt\hat{j} + \int \left(6t + 3t^{4} \right)dt\hat{k}$$

then,

$$\int f\vec{p}dt = -3\int t\vec{p}dt = \left(\frac{-3}{4}t^{4} - \frac{3}{2}t^{2} \right)\hat{i} + \left( -2t^{3} + \frac{3}{2}t^{2} \right)\hat{j} + \left( 3t^{2} + \frac{3}{5}t^{5} \right)\hat{k} + C$$

Is this correct?

5. Aug 2, 2010

### lanedance

what about when a = 1...

may be missing somethig, but I'd say $\vec{u}\times\left(a\vec{v}\right)= a\left(\vec{u}\times\vec{v}\right)$, if you change the order of u & v thats another story

Last edited: Aug 2, 2010
6. Aug 3, 2010

### jegues

Bump, still looking for a quick check on my answer.

7. Aug 3, 2010

### HallsofIvy

Yes, you are right. For some reason, I was thinking $\vec{u}\times(a\vec{v})= -((a\vec{v})\times\vec{u}= -a(\vec{v}\times\vec{u})$ but now we have to swap u and v back again!

$(a\vec{u})\times\vec{v}= \vec{u}\times(a\vec{v})= a(\vec{u}\times\vec{v})$

8. Aug 3, 2010

### jegues

Bump, still looking to see if the route I took is the right one.

9. Aug 4, 2010

### HallsofIvy

What part of "yes", in the second response, do you have trouble with?

10. Aug 4, 2010

### jegues

I was just curious whether or not I had done the integration correctly, that's all.

11. Aug 4, 2010