Can You Evaluate the Cross Product in Vector Integral Calculations?

[jegues]

Homework Statement

See figure

Homework Equations

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?

 

[xaos]

under what conditions should you think you cannot?

 

[HallsofIvy]

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

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

 

[jegues]

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]

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 that's another story

 

[jegues]

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

 

[HallsofIvy]

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

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

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 that's 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!

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

 

[jegues]

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

 

[HallsofIvy]

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

 

[jegues]

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.

 

[Raskolnikov]

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.