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Multiple integrals

  1. Jan 23, 2010 #1
    I have some conceptual problems regarding multiple integrals,out of which some often make me do sums wrong. Please help me out!

    1. If we triple integrate a function f(x,y,z) within appropiate limits (there are some sums of this kind in my book) are we integrating in 4-dimensions?---(since a function of x,y and z can only be described in 4 dimensions)......if so,integrations of higher order must involve higher dimensions!!

    2. I found a sum (involving cylindrical coordinates),in which z is integrated within the limits r and r^2...I suppose this is just a numerical point of view,as z does not take the values of r....am I right?

    3. Suppose we integrate a function xy ( limits are x: 0 to y and y:0 to 1), it says in my book,that we can separately integrate x and y (within appropiate limits), and then multiply the separate answers...but I don't understand how this should give us the right answer.....afterall, each of the integrals will have separate (constant) answers,and multiplying them will give the area of a rectangle (with those dimensions),and not the rqd area under the curves.

    4. In my book, it says that when we're double integrating a function,we sum up all the insinitesimal units (dx or dy) in one direction,and then in the other direction.....now,we usually find the integral by integrating between variable limits first and then for the 2nd integration step,we take constant limits........if we did it in the opposite order,we get an answer with a variable in it...but there's nothing conceptually wrong in that,is there?
  2. jcsd
  3. Jan 23, 2010 #2
    1. Yes, but I don't see what's the problem with that (and I would actually say that in this case it is a three-dimensional integration)

    2. Why not? When Integrating you're essentialy evaluating some function at z=r and z=r^2. Actually for this integral to be a correct representation of the volume of the cylinder, it must at least intersect with the planes z=r and z=r^2 (I guess r is the radius of the cylinder)

    3. Either your book was wrong, or you got it wrong. In your example it's not correct to integrate seperatly, because the integral of x depends on y. In this case you first integrate wrt x, getting an expression of y, and then you integrate wrt to y.
    If the integrals are independent (the boundaries are constant) and they're seperable by multiplication (you're integrating f(x,y)=X(x)Y(y)) then you can seperate the doule-integration to three seperate integrals that multiplies.

    4. Yes there is, integration of f(x,y) wrt x and y, must result only with a constant number. The integration with the variable in the limits, and the specific order of integration, can be called a parametrization of the domain of integration. So by switching order of integration you've chosen a wrong parametrization of the domain, and therefore everything is conceptually wrong about it. (Also your answer must be independent of the way you chose to get to it)
  4. Jan 23, 2010 #3
    Thanks,elibj123 for your help...I think I've got it now,though I'll have to understand what parametrization of domain of integration is exactly and find out why I wasn't aware of it before.

    By the way,one extra point that I came accross while solving some sums today...in polar coordinates,there is nothing such as a -ve or +axis,so if we have a curve like a 'leminiscate',we should get the correct answer if we integrate theta between 0 to (2pi)....but it doesn't work out!
    Also,in regard to this issue,suppose we integrate sine of theta between 0 and pi directly,we get 0....however if we convert this integral into (2 times the integration of sine of theta betn. 0 and pi/2),we get a particular value(2 in this case). Should the latter method be used only for finding area below the curve?
  5. Jan 23, 2010 #4
    What are you integrating on your curve (the leminiscate) and how did you do it?

    About the sine, I'm not sure how you get zero when integrating from 0 to pi:

  6. Jan 24, 2010 #5
    Sorry, I made a little mistake....I meant to say that if we integrate cos theta within the limits 0 to pi,we get 0...but the area under the curve isn't zero.That's why I thought in such cases,there must be a different procedure to calculate the area under the curve and to find value of the integral(not considering the area)...is that right?

    I integrated r^2cos(2theta) [the leminiscate] between 0 to 2pi...and I got zero!
  7. Jan 24, 2010 #6

    Char. Limit

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    Absolutely, the net area of cos(theta) from 0 to pi is 0... half of the area is above the axis and half of the area is below it.

    The total area from 0 to pi is 2 of course.
  8. Jan 24, 2010 #7
    Thanks Char. Limit ,so I suppose you agree that we do follow a different procedure for finding area under a curve in such situations?

    Also,please help me about the leminiscate problem I mentioned in my last post.
  9. Jan 24, 2010 #8


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    The area isn't automatically the integral! It is true that the integral is introduced in term of "area under the curve" but that is assuming that the graph is above the x-axis. For other figures you have to be more careful.
  10. Jan 24, 2010 #9

    Char. Limit

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    I didn't help you with your lemniscate problem because I can't. Polar equations beyond the
    simple ones are beyond me.
  11. Jan 25, 2010 #10
    Thanks HallsofIvy and Char. Limit .

    My question on the leminiscate was rather about why we don't get the correct answer by integrating from 0 to 2pi for certain curves,even though there is no chance of the 'areas' cancelling out in polar coordinates (because there is no negative and positive for polar coordinates).....it would be nice if someone clarified this point for me.
  12. Feb 5, 2010 #11
    Hi everyone,I just wanted to put in a quick question that occurred to me all of a sudden....in calculating area in terms of polar coordinates,using double integrals,we always use rdrd(theta)...that is,in polar coordinates,the area is not drd(theta)...but
    drd(theta) with r multiplied with it...now,when we simply integrate a function
    f(r,theta),within a region,that is expressed in polar coordinates area (e.g. integration of f(r,theta) within the cardiod) within certain limits of r and theta,we are really calculating the area within the cardioid...but here,we don't use rdrd(theta).

    This is especially evident if f(r,theta)= 1 (constant function).....then,we are really calculating the area of the cardiod...but as I said,her we don't use
    rdrd(theta)....please explain why.
  13. Feb 5, 2010 #12
    How did you calculate the area? (And what's the parametrization of your cardiod)
  14. Feb 6, 2010 #13
    I calculated the area of the cardioid by integrating rdrd(theta) within the limits (r: 0 to a(1+cos(theta)) and (theta: 0 to 2pi).

    But in the the general description of double integrals in polar coordinates,it says we simply integrate drd(theta) within the required limits.Even by this method,we get the area (conceptually....I mean by doing this,we're doing nothing but integrating the area within the limits).

    Now,if both the methods basically calculate the area under a curve,then what's the difference?Why is it that as soon as we see "calculate the area" we automatically put in rdrd(theta) whereas if the question does not specify that er're calculating the area,we simply integrate drd(theta) ?
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