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9/22/2017 12:11:25 AM
Posted: 11/19/2003 11:22:47 PM EDT
How do you integrate (p^2sin^3(phi))dpd(phi)d(theta)??

When setting up triple integrals, or converting the order of integration from one that is difficult to one that is doable, are there any rules of thumb to which order is best? --dxdydz as opposed to dzdxdy? OR, is there any way to determine which coordinate system is best for integrating? - cylindrical, rectangular or spherical?

TIA
Link Posted: 11/20/2003 3:31:47 AM EDT
btt
Link Posted: 11/21/2003 9:58:47 AM EDT
do you want to differentiate or integrate here? you have a differential as an example but you are asking about integrals...

also, are you talking about definate or indefinat integration/differentiation?

the integration rules i remember:

Where I is integral sign:

I(af(x)+bg(x))dx=aIf(x)dx+bIg(x)dx (this is linearity)

Iudv=uv-Ivdu (this is integration by parts… it can be applied to each part of a difficult equation) this can also be expressed as:

Iu(x)dv=u(x)v(x)-Iv(x)du=u(b)v(b)-u(a)v(a)-Iv(x)du where I is expressed between set limits. I think this is what you are looking for…

If(g(x))g’(x)dx=If(u)du where u=g(x) (this is substitution which can also be applied to complex equations)

Remember also:

Isin(kx)dx=-1/kcos(kx)+c
Link Posted: 11/21/2003 11:18:19 AM EDT


to determine order of integration it usually helps a lot to have a good three dimesnional picture in your head of what you are integrating. Then you can determine where you should start, because hopefully you can simplify your functions considerably. Also, converting from polar to rectangular and vice verse helps sometimes too. I had cal three several years ago so i would have to look at your problem for a bit, i always sucked at polar integration. Gimme a rectangular problem and im all over it.

newar15guy, he is clearly asking how to integrate that function. your cal 1 is too weak for this beast.

-Spaceman

Link Posted: 11/21/2003 11:20:31 AM EDT
Usually I would just type it into my TI-89 and hit enter. Yeah, that's what I'd do.
Link Posted: 11/21/2003 11:24:48 AM EDT

Originally Posted By newar15guy:
Iudv=uv-Ivdu (this is integration by parts… it can be applied to each part of a difficult equation) this can also be expressed as:

Iu(x)dv=u(x)v(x)-Iv(x)du=u(b)v(b)-u(a)v(a)-Iv(x)du where I is expressed between set limits. I think this is what you are looking for…



I've not done an integral in over 10 years, but THAT was the first thing that popped into my mind. I don't recall ever using it on multi-dimensional functions, but that doesn't mean you can't.
Link Posted: 11/21/2003 11:41:21 AM EDT
What math class is that?
Link Posted: 11/21/2003 11:42:53 AM EDT
[Last Edit: 11/21/2003 11:45:56 AM EDT by GaryM]
I'd hit it!... umm....
Buy both!... still not right?
umm....
Its the jews fault!...
No, blame it on the the JBTs!!



Still not the right answer? Damn, I thought those replies were guaranteed to work here..


YES! I get it now...
The way to integrate is with bussing! Yup, still integrating the city with the suburbs since the 1960s....
Link Posted: 11/21/2003 11:48:55 AM EDT
IMHO, having a good picture/idea of the volume you are integrating is more helpful for writing the integral and simplifying it to a form that can be solved easily. You've already got that integral in a reasonably clean form, and it won't make much difference what order you do it in. If it helps you, rewrite them as you do them. For example, you might decide to integrate by phi first. Then, you can rewrite as

k * I(sin^3(phi) * dphi)

where k is p^2

solve it as

k*(-3*cos(phi)/4+cos(3*phi)/12+c1)

then, going back to the original, you have

p^2*(-3*cos(phi)/4+cos(3*phi)/12+c1)*dp*dtheta

do the same thing again, and set k equal to that big mess that results from integrating sin^3. It's effectively a constant for the other integrations because none of it is a function of p or theta.

k*I(p^2*dp)

which gives you

k*(p^3/3+c2)

going back to the original again, you now have

(p^3/3+c2)*(-3*cos(phi)/4+cos(3*phi)/12+c1)*dtheta

which is simple because nothing is a function of theta, so the whole thing can be simplified like before as:

k*I(dtheta)

which is

k*(theta+c3)

or

(thet­a+c3)*(p^3/3+c2)*(-3*cos(phi)/4+cos(3*phi)/12+c1)

That's at least the basic idea. If it's indefinite, then you have to solve for the constants of integration. If it's a definite integral, then you have to plug in the limits and subtract before going on to the next integral instead of generating constants.
Link Posted: 11/21/2003 11:50:16 AM EDT
6.........the answer is 6
Link Posted: 11/21/2003 2:16:28 PM EDT


mace is exactly right, the more importand aspect of three axis integration is understanding exactly what shape you are dealing with. If you know this you can express its volume in a manner that is easiest to integrate for.

practice is the key here, get a feel for how three dimensional shapes are formed by thier corresponding formulae. Setting an axis equal to zero will give you the 2d function of the intercept of the 3d shape on that plane, which helps a lot for me.

-Spaceman

Link Posted: 11/21/2003 2:25:41 PM EDT
I just took a exam in differential calculus today, I think I fucked it up.
Link Posted: 11/21/2003 6:41:47 PM EDT
[Last Edit: 11/21/2003 6:49:28 PM EDT by AeroE]
Try to integrate the function in the order of the differentials. I will try an example; I will us "]" to represent an integral sign -

]]](p^2sin^3(phi))dpd(phi)d(theta) :

1) ] p^2sin^3 = f(p)

hence,
]]](p^2sin^3(phi))dpd(phi)d(theta)

= ]]f(p)sin^3(phi)d(phi)d(theta)

= ]f(p)]sin^3(phi)d(phi)d(theta)

= f(p) ]]sin^3(phi)d(phi)d(theta)

2) sin^3(phi)d(phi) = g(phi)

hence,
]]](p^2sin^3(phi))dpd(phi)d(theta)
= f(p)g(phi) ]d(theta)

(after some substitution)

3) the final integration is

f(p)g(phi) ]d(theta) = f(p)g(phi)h(theta)

[ +C if the integral is indefinite. f, g, and h are new functions yielded by each integration]

I hope this helps. An equation editor would make this clearer; this notation pretty much sucks. Note that after each integration step, the result from that step acts as a fixed coefficient on the following coefficient. Integrate each part from the innermost differential to the outermost. If p is a function of phi or theta (or any of the other variables are not independent), then the answer gets more complicated, but the basic process is the same, and you may need to incorporate separation of variables or other techniques.

As far as rules of thumb, there aren't any - you have to eyeball the function and try to figure out the simplest form, but more importantly, you need to integrate the function over the variable ranges from which you have the physical data for the problem. Write your solutions neatly with notes describing the step you are taking, similar to the method used for presenting example problems, and develop a book keeping method to help you keep track of the solution steps.
Link Posted: 11/21/2003 6:56:05 PM EDT
This thread is useless without pics!!!


ByteTheBullet (-:
Link Posted: 11/21/2003 7:23:54 PM EDT
[Last Edit: 11/21/2003 7:24:12 PM EDT by ColonelKlink]
THe answer to every calculus question:

Link Posted: 11/22/2003 8:48:29 PM EDT
This was for Calc III. Some of you guys are a hoot...!


I understand the necessity of having a drawing of what the function is integrated over in order to find the limits, but I stumbled and didn't realize that this problem required setting the complex function sin^3(phi)d(phi)=
to a f(phi). You can just do that?

AeroE-so you're saying ]]](p^2sin^3(phi))dpd(phi)d(theta)
= f(p)g(phi)(theta)? --Yeah, but at some point you have to actually integrate p^2, which would be p^3/3 and you have to integrate sin^3 (phi) which would be...I don't know...don't you?

Mace-I guess I'm not sure how this: I(sin^3(phi) * dphi) = this: (-3*cos(phi)/4+cos(3*phi)/12+c1)

--I should have clarified; that sin (phi) function is cubed, not (3phi).

Here is a similar problem out of the book:

[[[5p^3 * (sin(phi))^3 dp dphi dtheta

limits of integration, from the outside in:
0 to 3pi/2 ---theta
0 to pi ---phi
0 to 1 ---p

I guess what I'm really wondering is how do you integrate sin^3 (phi)?
Link Posted: 11/23/2003 4:30:52 AM EDT
[Last Edit: 11/23/2003 4:38:06 AM EDT by entropy]
As one who minored in math in college 20 years ago, during which I spent countless hours embroiled in this stuff, real variables, complex variables, graph theory, integer programming.... I am proud to say.....

<­BR>



I don't remember a single fucking thing about it.
Link Posted: 11/23/2003 4:53:50 AM EDT

Entropy,

I actually have a BS in Mathematics, 1986.
Yet I’ve never even used those skills once since graduation.
Although I’m very good with simple math skills, geometry & algebra without the use of paper, I wouldn’t have a clue as to how to do anything more complicated without a textbook.



Lokt, sorry I can't help you.
Link Posted: 11/23/2003 8:41:07 AM EDT
[Last Edit: 11/23/2003 8:44:33 AM EDT by AeroE]
Hell, I don't know the integral of sin^3(phi) off the top of my head, but I can look it up, just like you need to - get a CRC Handbook or some other reference of mathematical functions for a reference if you're going to be doing this for a while.

As long as each variable is independent in this problem, it is mostly a book keeping problem after each step of integration.

Since I'm a nice guy and want to help, the answer from the CRC handbook is:

]sin^3(phi)d(phi) = -(1/3)cos(phi){sin^2(phi) + 2}


This reference www.efunda.com/math/integrals/show_integrals.cfm

gives -

]sin^3(phi)d(phi) = -cos(phi) + cos^3(phi)/3

I'm not going to work through the problem to check that it is the same answer. However, I trust the first answer.

There are several sites on the net that have integration tools, but after looking them over this morning, I would be extremely cautious about their use. Simple integrations all look correct, but due to the complexity of entering the problem statement and any substitutions built into the solver, I get different answers from nearly every solver. Some of them may just be wrong, and I can't figure out how to get some of them to recognize trig functions.

Here's one -
mss.math.vanderbilt.edu/cgi-bin/MSSAgent/~pscrooke/MSS/antiderivative.def

an example problem -

www.calc101.com/trig_power_1.html

If you enter the engineering profession, there is a good chance that you will need calculus at some point - integrating a complex load distribution over an unuusual shape, for example.

Besides, it's fun! Remember, it's just adding - at least that is what my Thermo professor told us.

Edited -
Now I need a shower to wash the geekiness off.
Link Posted: 11/23/2003 7:58:48 PM EDT

Originally Posted By lokt:
Mace-I guess I'm not sure how this: I(sin^3(phi) * dphi) = this: (-3*cos(phi)/4+cos(3*phi)/12+c1)

--I should have clarified; that sin (phi) function is cubed, not (3phi).

[snip]

I guess what I'm really wondering is how do you integrate sin^3 (phi)?



I don't really know off the top of my head, I used my HP48 to get that. If you're looking for a way to solve that integral, then perhaps integration by parts with du=sin(phi) and v=sin^2(phi) would help. Or maybe a trig substitution.

Moral: Manual integration sucks. Computer integration is cool.
Link Posted: 11/23/2003 8:02:10 PM EDT

Originally Posted By ColonelKlink:
THe answer to every calculus question:

home.comcast.net/~ColonelKlink/integral.jpg


a buddy of mine in HS actually wrote that on his AP calc. test
Link Posted: 11/23/2003 8:41:10 PM EDT
Thanks for your time, guys! ...I'm an M.E. major. makes me want to gouge my eyes out some days, but usually things aren't so bad.
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