Integrate Fun Math Challenges for New Year's Celebration

[samalkhaiat]

show that
$$\int\int\int \left( \vec{a}\cdot\vec{r}\right) \left( \vec{b}\cdot\vec{r}\right) \left( \vec{c}\cdot\vec{r}\right) dV =\frac{ \left( \alpha\beta\gamma\right) ^2}{8\left| \vec{a}\cdot\left( \vec{b}\times\vec{c}\right) \right|}$$
(this one comes out of Stewart, Calculus 4th ed., ch. 15 Problems Plus #4 on pg. 1038).
Enjoy

The integral is scalar. a.(bxc) is the only scalar formed from three vectors. But this scalar is antisymmetric, while the integral is symmetric. So, we must have |a.(bxc)|.
Thus, we can write,
$$\int (\vec{a}.\vec{r})(\vec{b}.\vec{r})(\vec{c}.\vec{r}) dV=A[\left| \vec{a}\cdot\left( \vec{b}\times\vec{c} \right) \right|]^n$$
now take,
$$\vec{a}=\hat{i}|\vec{a}|, \vec{b}=\hat{j}|\vec{b}|, and, \vec{c}=\hat{k}|\vec{c}|$$
and find;
n=-1 and $$A=\frac{(\alpha\beta\gamma)^2}{8}$$


regards

sam

 

[samalkhaiat]

integrate this:
$$I( \vec{a}, \vec{K})= \int_{-1}^{+1} \frac{( \vec{a}. \hat{r})} {(1+ \vec{K}. \hat{r})} d(cos{\theta})$$

I will do this around this time tomorow, till then,(hint); think about this
$$J(\vec{K})=\int_{-1}^{+1} \frac{\hat{r} }{(1+ \vec{K}. \hat{r})}d(cos{\theta})$$
If you know the nature of this integral, the rest will be very easy

regards

sam

 

[samalkhaiat]

I will do this around this time tomorow, till then,(hint); think about this
$$J(\vec{K})=\int_{-1}^{+1} \frac{\hat{r} }{(1+ \vec{K}. \hat{r})}d(cos{\theta})$$
If you know the nature of this integral, the rest will be very easy
regards
sam

Hi every one,

Note that, $\vec{J(K)}=\int_{-1}^{+1}\frac{\hat{r} }{(1+ \vec{K}. \hat{r})}d(cos{\theta})$, is a vector. Since K is the only available vector, we must have, for some constant A;
$$\vec{J(K)}=A\vec{K}$$

Thus,

$$A=\frac{\vec{K}.\vec{J(K)}}{K^2}=\frac{1}{K^2} \int_{-1}^{+1}\frac{\vec{K}. \hat{r}}{1+ \vec{K}. \hat{r}} d(cos{\theta})$$

this can be written as,

$$A=\frac{1}{K^2}\int_{-1}^{+1}[1-\frac{1}{1+ \vec{K}.\hat{r}}]d(cos{\theta})$$

note that, $\vec{K}. \hat{r}=|K|cos{\theta}$,thus

$$A=\frac{1}{K^2}[2- \frac{1}{|K|} ln( \frac{1+|K|}{1-|K|})]$$

Now you have A, you get

$$\vec{J(K)}=A\vec{K}$$

and,

$$I(\vec{a},\vec{K})= \vec{J}. \vec{a}=\frac{\vec{a}.\vec{K}}{K^2}[2-\frac{1}{|K|}ln(\frac{1+|K|}{1-|K|})]$$

Job done

regards

sam

 

[samalkhaiat]

This, very challenging one, is for all PF members and mentors, Integrate this

$$\Gamma(\vec{k},\vec{p})=\int_{-1}^{+1} \frac{d(cos{\theta})}{(1+ \vec{k}. \hat{r})(1+ \vec{p}. \hat{r})}$$


regards


sam

 

[mathwonk]

why do i feeel i am talikng to billy criystal?

 

[samalkhaiat]

why do i feeel i am talikng to billy criystal?

talikng?, Do you mean "talking"?
criystal?, Do you mean "cristall", "crystal" or what?
Mathwonk, I hope you spelt this one right , are you a comedian by any chance? I must admit, you made me laugh

You think this thread is turning to a one-man-show. Yes, so what? This does not bother me because many people, I believe, have gained some useful mathematical tricks from this (hot) thread. And this makes me happy
Look, if you can not do the integrals, you could, either ask questions or don't reply to the posts.
I remember, I spent four hours trying to do the last integral. So, it is not a shameful thing, if you can not do it! But again, the problem is you call yourself mathwonk and you are supposed to be a science advisor
Ok, mathwonk-science advisor, I consider myself lucky, for I will never need your advice.

Byeeee

sam

 

[leon1127]

talikng?, Do you mean "talking"?
criystal?, Do you mean "cristall", "crystal" or what?
Mathwonk, I hope you spelt this one right , are you a comedian by any chance? I must admit, you made me laugh
You think this thread is turning to a one-man-show. Yes, so what? This does not bother me because many people, I believe, have gained some useful mathematical tricks from this (hot) thread. And this makes me happy
Look, if you can not do the integrals, you could, either ask questions or don't reply to the posts.
I remember, I spent four hours trying to do the last integral. So, it is not a shameful thing, if you can not do it! But again, the problem is you call yourself mathwonk and you are supposed to be a science advisor
Ok, mathwonk-science advisor, I consider myself lucky, for I will never need your advice.
Byeeee
sam

The job of an advisor is not to answer any challenge problem (or show-off) problem. Their job is to help individual who has difficulty in certain field (since this is a Sciences/Maths forum.) Since all your difficulties of these questions have been solved by yourself, I personally don’t think that they are responsible to answer solved problem of yours. If you really have difficulty to do your homework or make an important decision of your life, I am sure they are more than happy to give you advice.

Sorry for my broken English

 

[enigma]

I think this thread is about done.