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Homework Help: Calculating A Double Integral Using Polar Coords

  1. May 2, 2015 #1
    I'm on a tablet and having trouble with the math symbols so, for clarity, ∫[a,b] xdx is the integral from a to b of x with respect to x, and f(x) |[a,b] is a function of x evaluated from a to b.


    ∫[-1,1]∫[-√(1 - y2),√(1 - y2)] ln(x2 + y2 + 1)
    Relevent Equations:

    x2 + y2 = r2

    ∫udv = uv - ∫vdu

    My Attempt:

    Immediately I recognized that transforming this to a polar integral would make this much easier. The natural log is being integrated over a circular region radius 1 in the x-y plane, so the polar integral would be:

    ∫[0,2π]∫[0,1] ln(r2 + 1)rdrdθ
    u = ln(r2 + 1) dv = rdr
    du = (2r)/(r2 + 1) dr v = 1/2 • r2

    ∫[0,2π] 1/2 • r2 • ln(r2 + 1) |[0,1] -∫[0,2π]∫[0,2π] (r3)/(r2 + 1) • rdrdθ
    Note, I simplified vdu.

    Evaluating the first integral yields:

    ∫[0,2π] 1/2 • ln2 dθ
    The second integral is not in an inegrable form. To get it in integrable form, I performed long division. I can't really display that on here in a neat manor, but I just did r4/(r2 + 1), which equals r2 - 1 + 1/(r2 + 1). So, the new integral is:

    - ∫[0,2π]∫[0,1] [r2 - 1 + 1/(r2 + 1)] drdθ
    The negative came from the earlier integration by parts step. I noticed that the third term in this integral fit the formula for tan-1. Integrating the second integral term by term and distributing the negative yields:

    ∫[0,2π] [-1/3 • r3 + r - tan-1(r)] |[0,1] dθ

    ∫[0,2π] [(-1/3 + 1 - π/4) - (0 + 0 - 0)] dθ
    You can now either combine both integrals or evaluate them term by term. Once you evaluate, you get:

    θ(1/2 • ln2 + 2/3 - π/4) |[0,2π]

    2π(1/2 ln2 + 2/3 - π/4)
    That's my final answer. The correct answer is π(ln4 - 1). These two are not equivalent.
    Last edited: May 2, 2015
  2. jcsd
  3. May 2, 2015 #2

    Simon Bridge

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    When on a tablet, it is usually easier to just type out the LaTeX - which is worth learning anyway.

    You needed to evaluate:

    ... which I read is:
    $$\int_{-1}^1\int_{-\sqrt{y^2-1}}^{\sqrt{y^2-1}} \ln (x^2+y^2+1)\;\mathrm{d}x\mathrm{d}y$$

    Is that correct?
  4. May 2, 2015 #3
    100% correct. Sorry for the messyness. My tablet doesn't like to do the limits, and I'm not that experienced with how the code works.
  5. May 2, 2015 #4

    Simon Bridge

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    OK then - did you sketch those limits?
    note: ##x=\pm\sqrt{y^2-1}:\; y\in [-1,1]## is that a circle?
  6. May 2, 2015 #5
    My fault. I wrote the initial problem backwards. Should be x = +/- √(1 - y2). I will edit it so no one gets confused. I did draw the circle though.
  7. May 2, 2015 #6

    Simon Bridge

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    OK so ... the polar form would be: $$\int_0^{2\pi}\int_0^1 \ln(r^2+1)r\;\mathrm{d}r\mathrm{d}\theta = 2\pi\int_0^1 \ln(r^2+1)r\;\mathrm{d}r$$ ... since there is no ##\theta## dependence in the integrand (always do the easy part first.)

    You combined some crucial steps in your working so I'm having a hard time following what you did ... what did you get for the indefinite integral wrt r?

    Note: the latex is not hard.
    int[a,b] becomes \int_{a}^{b} for example, while the natural log symbol is \ln ... it's a bit of a pain having to shift keyboards for the backslashes and curly-brackets but you learn to anticipate and type a row of them while the correct keyboard is active.
  8. May 2, 2015 #7
    Okay, I will give latex a shot. Are you looking for the final indefinite integral with respect to r?

    That would be:

    \frac{1}{2} ln2 + \frac{2}{3} - \frac{π}{4}
    If you would like, I can try retyping it with latex, although I cannot seem to get it to work properly
  9. May 2, 2015 #8
    I'm going to also add in a picture of mywork. It may help since we are having trouble with latex.

  10. May 3, 2015 #9

    Simon Bridge

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    Excuse me, I had to step out for a bit.
    ... that's good - but you have to tell PF to render that as maths instead of code.
    To do that, you put double-dollar signs on either side of it. Then it comes out as: $$\frac{1}{2} ln2 + \frac{2}{3} - \frac{π}{4}$$... see: not a problem :) The "ln" should be "\ln" and the pi symbol is \pi (all Greek letters not already in the Latin charset, are a backslash followed by the letter name i.e. \pi is ##\pi## and \Pi is ##\Pi## ... neat huh?).

    However, that is the result of the definite integral.
    The indefinite integral would be the result of: $$\int r\ln(r^2+1)\;\mathrm{d}r$$
  11. May 3, 2015 #10
    Sorry, fell asleep last night.

    So, the result of the indefinite integral using integration by parts:

    $$\int r\ln(r^2 + 1)dr$$
    $$u = \ln(r^2 + 1); dv = rdr$$
    $$du = \frac{2r}{r^2 + 1} dr; v = \frac{1}{2} r^2$$
    $$\frac{1}{2} r^2 \ln(r^2 + 1) - \int \frac{2r • r^2}{2(r^2 + 1)} rdr$$
    $$\frac{1}{2} r^2 \ln(r^2 + 1) - \int \frac{r^4}{(r^2 + 1)} dr$$
    Using long division on the remaining integral (still not sure how to display that step):

    $$\frac{1}{2} r^2 \ln(r^2 + 1) - \int [r^2 - 1 + \frac{1}{r^2 + 1}] dr$$
    Term by term integration, where the third term is tan-1:

    $$\frac{1}{2} r^2 \ln(r^2 + 1) - \frac{1}{3} r^3 + r - arctan(r)$$
  12. May 4, 2015 #11

    Simon Bridge

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    OK - there's something wrong in that set of calculations... that 1/3 shouldn't appear ferinstance and there is no need for the trig function since ##r\neq \tan\theta##.
    I think your substitution is off but I'm not about to pour through it.

    Why didn't you just set u=r^2+1 so you need only evaluate:
    $$\int \ln(r^2+1)r\;\mathrm{d}r= \frac{1}{2}\int \ln u \; \mathrm{d}u$$ ... then just look u the integral of ln(u).

    LaTeX notes:
    trig functions are their name after a backslash - in general that is how latex tries to do things. so \arctan x gets you ##\arctan x##.
    Sowing long division involves alignments - though you can show the stages in one line each.
  13. May 4, 2015 #12
    I didn't remember the antiderivative of lnu du off hand. I reworked the problem today. The mistake I made was in this step:

    $$\frac{1}{2} \ln(r^2 + 1) - \int \frac{2r • r^2}{2(r^2 + 1)} r dr$$

    I stuck an extra r in next to dr. I assumed that if I integrated again in polar (drdθ) then I needed r. If you don't stick that r in, the problem works out the same as of you did a u sub.

    Thank you for your help, and Latex advice!
  14. May 5, 2015 #13

    Simon Bridge

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    No worries - usually if you know where to look you can find.

    Note: I don't memorize integrals either - I just look them up.
    $$\int \ln(x)\;\mathrm{d}x = x\big(\ln(x)-1\big) + c$$

    For more on the LaTeX typesetting system:
    ... you only need the math section though.
    There is also online documentation and for specific effects you can google for the effect name with "latex".
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