# Integrating vector functions

See figure

## 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?

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under what conditions should you think you cannot?

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

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

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

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

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?

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

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

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

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

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

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
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!

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

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

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

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

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

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?

Yep, that's right. Just make sure your C is a constant vector, not a real number constant.