Calculating A Double Integral Using Polar Coords

[iismitch55]

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.

Problem:

∫[-1,1]∫[-√(1 - y²),√(1 - y²)] ln(x² + y² + 1)
​

relevant 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.

 

[Simon Bridge]

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

You needed to evaluate:

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

... 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?

 

[iismitch55]

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?

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.

 

[Simon Bridge]

OK then - did you sketch those limits?
note: $x=\pm\sqrt{y^2-1}:\; y\in [-1,1]$ is that a circle?

 

[iismitch55]

OK then - did you sketch those limits?
note: $x=\pm\sqrt{y^2-1}:\; y\in [-1,1]$ is not a circle.

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.

 

[Simon Bridge]

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.

 

[iismitch55]

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

 

[iismitch55]

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.

I'm going to also add in a picture of mywork. It may help since we are having trouble with latex.

 

[Simon Bridge]

Excuse me, I had to step out for a bit.

... we are having trouble with latex...
[vis] \frac{1}{2} ln2 + \frac{2}{3} - \frac{π}{4}

... 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: $$\int r\ln(r^2+1)\;\mathrm{d}r$$

 

[iismitch55]

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: $$\int r\ln(r^2+1)\;\mathrm{d}r$$

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:

$$\frac{1}{2} r^2 \ln(r^2 + 1) - \frac{1}{3} r^3 + r - arctan(r)$$​

 

[Simon Bridge]

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.

 

[iismitch55]

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!

 

[Simon Bridge]

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:
https://www.ctan.org/tex-archive/info/lshort/english/?lang=en
... 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".