Integrate Fun Math Challenges for New Year's Celebration

[samalkhaiat]

Happy new year for you all,
This is very nice place, So girls and boys, let me give you something to play with:
Integrate (r^.a)(r^.b)(r^.c)(r^.d) d(cos8)
from cos8=1 to cos8=-1.
a,b,c and d are constant vectors, r^ is the unit vector
(iX+jY+kZ)/|r|, 8 is the spherical angle between r^ and Z-axis.
The dot represents the scalar product.
When you have done that, do this one:

(a.r^)/(1+K.r^) d(cos8)
the limits of integration same as befor, a and K are constant vectors.

cheers

sam

 

[abdo375]

pleasezzzzzzzzzzzzzzzzzzzz take sometime and learn latex

 

[quantumdude]

pleasezzzzzzzzzzzzzzzzzzzz take sometime and learn latex

I have to agree, I can't really tell what's supposed to be going on there with the carets. And integrating with respect to the number 8? Is that really what you meant?

 

[Lisa!]

I've not noticed we could use $$8$$ instead of $$\theta$$ !

 

[Tide]

$$\frac {2}{5} a b c d$$

I was tempted to say 0 because 8 is usually a constant so that d cos(8) = 0. ;)

 

[samalkhaiat]

[

QUOTE=abdo375]pleasezzzzzzzzzzzzzzzzzzzz take sometime and learn latex

[/QUOTE]

1) I don't have that "sometime".
2) It is to complicated for my little brain.

sam

 

[fargoth]

[

1) I don't have that "sometime".
2) It is to complicated for my little brain.
sam[/QUOTE]

just click on the equations and see how they wrote it... I am sure youd figure it out...

 

[Pengwuino]

1) I don't have that "sometime".
2) It is to complicated for my little brain.
sam

Find time :)

And if you can integrate, i don't think its too complicated for you... especially with a program such as texaide

 

[samalkhaiat]

[

QUOTE=Tom Mattson]
I have to agree, I can't really tell what's supposed to be going on there with the carets.

The carets used to distinguish between a vector and its "unit vector"

And integrating with respect to the number 8? Is that really what you meant?

[/QUOTE]
No. The 8 , as I said, is the spherical angle(do you remember theta), it is just a symbol

sam

 

[samalkhaiat]

I've not noticed we could use $$8$$ instead of $$\theta$$ !

Yes darling, you could if you think of 8 as symbol.:zzz:


sam

 

[samalkhaiat]

$$\frac {2}{5} a b c d$$

Is this meant to be the answer? well it is wrong

I was tempted to say 0 because 8 is usually a constant so that d cos(8) = 0. ;)

Yes, now this answer is write, if 8 was a number, but it is not. So,even your temptation is wrong.

good luck

sam

 

[Tide]

samal,

Is this meant to be the answer? well it is wrong

Prove it! :)

 

[samalkhaiat]

samal,
Prove it!

Not just yet. I want to let others try to do it.
But the answer is:

2/15 [(a.b)(c.d)+(a.c)(b.d)+(a.d)(b.c)]
To avoid misunderstanding, the dot(.) represents the scalar product between vectors.

 

[fargoth]

samal,
Prove it! :)

the final answer has mixed parts of a b c and d in it, not only $$a_xb_x$$ etc. so it can't be abcd

 

[fargoth]

Not just yet. I want to let others try to do it.
But the answer is:
2/15 [(a.b)(c.d)+(a.c)(b.d)+(a.d)(b.c)]
To avoid misunderstanding, the dot(.) represents the scalar product between vectors.

are you sure? represting the r usint vector in spherical coordinates, youd get for the first part
$$a_xb_xc_xd_xcos^4(\Theta)cos^2(\phi)+a_yb_yc_yd_ycos^4(\Theta)sin^2(\phi)+a_zb_zc_zd_zsin^4(\Theta)$$
you should atleast get a solution that depends on $$\phi$$

 

[samalkhaiat]

just click on the equations and see how they wrote it... I am sure youd figure it out

let me try it,
$$\int_{-1}^{+1} \ (a.r)(b.r)(c.r)(d.r) d(cos{\theta})$$
r is a unit vector.
let us see.

this nice, thank you very much.

sam

 

[samalkhaiat]

Integrate (r^.a)(r^.b)(r^.c)(r^.d) d(cos8)
from cos8=1 to cos8=-1.
a,b,c and d are constant vectors, r^ is the unit vector
(iX+jY+kZ)/|r|, 8 is the spherical angle between r^ and Z-axis.
The dot represents the scalar product.

this is fun, so I asked you to integrate this:
$$J(a,b,c,d)=\int_{-1}^{+1} (a.r)(b.r)(c.r)(d.r) d(cos{\theta})$$

When you have done that, do this one:
(a.r^)/(1+K.r^) d(cos8)
the limits of integration same as befor, a and K are constant vectors.

and this:

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


cheers


sam

 

[samalkhaiat]

are you sure?

Yes, I am.

represting the r usint vector in spherical coordinates, youd get for the first part
$$a_xb_xc_xd_xcos^4(\Theta)cos^2(\phi)+a_yb_yc_yd_ycos^4(\Theta)sin^2(\phi)+a_zb_zc_zd_zsin^4(\Theta)$$
you should atleast get a solution that depends on $$\phi$$

No, you should not.
take the case when:
a=b=c=d=r
you will have

$$J(r,r,r,r)=\int_{-1}^{+1}(1)(1)(1)(1) d(cos{\theta})=2$$

regards

sam

 

[Lisa!]

Yes darling, you could if you think of 8 as symbol.:zzz:
sam

I meant I hadn't noticed te similarity, dear!

 

[fargoth]

Yes, I am.
No, you should not.
take the case when:
a=b=c=d=r
you will have
$$J(r,r,r,r)=\int_{-1}^{+1}(1)(1)(1)(1) d(cos{\theta})=2$$
regards
sam

in the case a=b=c=d=r these vectors arent constant.
the r unit vector changes its direction according to $$\phi and \Theta$$
and another thing: $$sin(\Theta)$$ is orthogonal to $$cos(\Theta)$$ when you integrate over 0-2pi so the parts with any sin in them are thrown out (including all the Z components of these vectors)

 

[samalkhaiat]

in the case a=b=c=d=r these vectors arent constant.

You are right, it was a silly mistake.I tried to avoid letting
a=b=c=d=k(the unit vector along the Z-axis),because this would answer the question! Ok consider this as a hint.

regard

sam

 

[samalkhaiat]

so I asked you to integrate this:
$$J(a,b,c,d)=\int_{-1}^{+1} (a.r)(b.r)(c.r)(d.r) d(cos{\theta})$$

Ok, let me show you how "physicists" do this integral.
1) notice that J(a,b,c,d) is a scalar, thus it must be constructed out of all possible scalar products of the form (a.b)(c.d).
2) J is linear in each vector, thus there is no (a.a),(b.b),etc.factors.
So you can write;
J(a,b,c,d)= A+B(a.b)(c.d)+C(a.c)(b.d)+D(a.d)(b.c)
3) J(0,0,0,0)=0, therefore A=0.
4) J(a,b,c,d) is invariant under any interchange of these vectors, thus
B=C=D, i.e
J(a,b,c,d) = B[(a.b)(c.d)+(a.c)(b.d)+(a.d)(b.c)]
5) now let,
a=b=c=d=k, where k is the unit vector along the Z-axis;
3B = J(k,k,k,k) = 2/5, ==> B = 2/15
AND YOU HAVE IT.
Ok, you try to do this one;

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


cheers,

sam

 

[samalkhaiat]

Find time :)
And if you can integrate, i don't think its too complicated for you... especially with a program such as texaide

OK, now you find time and integrate this:

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


sam

 

[benorin]

Here's a similar integral

If $\vec{a},\vec{b},\mbox{ and }\vec{c}$ are constant vectors, $\vec{r}$ is the position vector $\vec{r}=\left< x,y,z\right> ,$ and E is given by the inequalities

$$0\leq\vec{a}\cdot\vec{r}\leq\alpha,0\leq\vec{b}\cdot\vec{r}\leq\beta,0\leq\vec{c}\cdot\vec{r}\leq\gamma,$$

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 .

 

[Integral]

1) I don't have that "sometime".
2) It is to complicated for my little brain.
sam

So because you are to lazy to learn LaTex the rest of use must expend extra energy to translate what you write. This shows no respect for others, please show some respect, learn to use the tools provided to make this forum a better place for all.

Personally I am reluctant to bother responding to a post such as yours, other then to as learn LaTex.

 

[samalkhaiat]

So because you are to lazy to learn LaTex the rest of use must expend extra energy to translate what you write. This shows no respect for others, please show some respect

Sir, I am sorry to put you under the impression that I was disrespectful. But I was not. At the time , I was mixing truth (I didn't know how to use LaTex) with humour

learn to use the tools provided to make this forum a better place for all.

With all due respect sir, your angry response came one week to late. I think you should read all my posts first. If you do, you will see that I thanked "Fargoth" very much for telling me how to use LaTex.

Personally I am reluctant to bother responding to a post such as yours, other then to as learn LaTex

When I realized that somebody called "Integral" replied to my post, I said to myself; It will be interesting to see how "Integral integrate my integral"! What a disappointment.
Anyway, can Integral integrate this integral;

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


regards

sam

 

[leon1127]

People, who think this is interesting, shall find the solution in a normal calculus book. I personally don't think this is a challenging problem of all. What is the point of this thread?
Now, I shall go back and memorise my analysis book just for the fun of it. I am fascinated to proof 1+1=2. May you please give this little lost student a hint to solve this problem?

 

[GCT]

perhaps this is homework

 

[samalkhaiat]

People, who think this is interesting, shall find the solution in a normal calculus book.

No, These integrals and many similar ones pop up when you study;
The interactions between photons and atoms (Compton effect, photoelectric effect, multipole transition,...), Many electrons systems, Scattering theory and almost all QED processes. So, they are not trivial, normal calculus book exercises

I personally don't think this is a challenging problem of all. What is the point of this thread?

They are challenging because one can, almost always, solve them without going through the messy process of integration! and to be able to do this, one needs to know something more than "I know how to integrate"

Now, I shall go back and memorise my analysis book just for the fun of it. I am fascinated to proof 1+1=2. May you please give this little lost student a hint to solve this problem?

I can not help you on this one, But you could ask Integral


sam

 

[samalkhaiat]

perhaps this is homework

No, it is not. You will have to do some of them when you study Quantum Theory of H2, NH3 and others.

sam

 

[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?