Transforming double integrals into Polar coordinates

[lmstaples]

Homework Statement

Show that:

$I = \int\int_{T}\frac{1}{(1 + x^{2})(1 + y^{2})}dxdy = \int^{1}_{0}\frac{arctan(x)}{(1 + x^{2})}dx = \frac{\pi^{2}}{32}$

where T is the triangle with successive vertices $(0,0), (1,0), (1,1)$.

*By transforming to polar coordinates $(r,θ)$ show that:*

$I = \int^{\frac{\pi}{4}}_{0}\frac{log(\sqrt{2}cos(θ))}{cos(2θ)}dθ$

Homework Equations

The Attempt at a Solution

Part one is fine:

$I = \int\int_{T}\frac{1}{(1 + x^{2})(1 + y^{2})}dxdy = \int^{1}_{0}\underbrace{\int^{x}_{0}\frac{1}{(1 + x^{2})(1 + y^{2})}dy}_{A}dx$

$A: \frac{1}{(1 + x^{2})}\int^{x}_{0}\frac{1}{(1 + y^{2})}dy = \frac{1}{(1 + x^{2})}\left[arctan(y)\right]^{x}_{0} = \frac{arctan(x)}{(1 + x^{2}}$

$\Rightarrow I = \int^{1}_{0}\frac{arctan(x)}{(1 + x^{2})}dx$

$Let:$ $u = arctan(x)$

$\frac{du}{dx} = \frac{1}{(1 + x^{2})} \Rightarrow du = \frac{1}{(1 + x^{2})}dx$

$x = 0 \Rightarrow u = 0$
$x = 1 \Rightarrow u = \frac{\pi}{4}$

$\Rightarrow I = \int^{\frac{\pi}{4}}_{0}udu = \left[\frac{1}{2}u^{2}\right]^{\frac{\pi}{4}}_{0} = \frac{\pi^{2}}{32}.$


PART 2

$x = rcos(\vartheta)$
$y = rsin(\vartheta)$

$Boundaries:$
$(i) y = 0 \Rightarrow rsin(θ) = 0$
$(ii) x = 1 \Rightarrow rcos(θ) = 1$
$(iii) y = x \Rightarrow rsin(θ) = rcos(θ) \Rightarrow tan(θ) = 1 \Rightarrow θ = \frac{\pi}{4}$

From then on everything seems to get very complex and I'm not really sure what the limits are either.

I was wondering if it was possible to go simply from:

$I = \int^{1}_{0}\frac{arctan(x)}{(1 + x^{2})}dx$

to:

$I = \int^{\frac{\pi}{4}}_{0}\frac{log(\sqrt{2}cos(θ))}{cos(2θ}dθ$

Please help :(

 

[Simon Bridge]

OK - you want to show that transforming to polar coordinates does this:$$I = \int\int_{T}\frac{1}{(1 + x^{2})(1 + y^{2})}dxdy \rightarrow \int^{\frac{\pi}{4}}_{0}\frac{\log(\sqrt{2}\cos \theta)}{\cos 2 \theta }d\theta$$Is that supposed to be a natural logarithm on the RHS?

You have noticed that:
$x = r\cos\theta$
$y = r\sin\theta$

Boundaries:
$y = 0 \Rightarrow r\sin\theta = 0$
$x = 1 \Rightarrow r\cos\theta = 1$
$y = x \Rightarrow r\sin\theta = r\cos(\theta) \Rightarrow \tan\theta = 1 \Rightarrow \theta = \frac{\pi}{4}$

You also need to realize that:
$dx.dy = dA = r.dr.d\theta$ (check)

Transforming the integrand would involve using the right trig identities - equip yourself with a table of them, and see that you are looking for something with $\cos 2 \theta$ ... the logarithm probably comes from integrating over $r$.

Please show your working.

 

[lmstaples]

$dx.dy=dA=r.dr.dθ$ that's a given

not sure about the limits on r...

$(expanding gives:$

$\int^{\frac{\pi}{4}}_{0}\int^{?}_{0}\frac{r}{(1 + r^{2}sin^{2}(θ) + r^{2}cos^{2}(θ) + r^{4}sin^{2}(θ)cos^{2}(θ))}drdθ$

$= \int^{\frac{\pi}{4}}_{0}\int^{?}_{0}\frac{r}{(1 + r^{2} + r^{4}sin^{2}(θ)cos^{2}(θ))}drdθ$

$= \int^{\frac{\pi}{4}}_{0}\int^{?}_{0}\frac{1}{\frac{1}{r} + r + r^{3}sin^{2}(θ)cos^{2}(θ))}drdθ$

then it gets confusing unless maybe you make a nice substitution?

 

[voko]

Try u = r^2

Regarding the limits, r is the distance from (0, 0) to (x, y) on the the x = 1 line.

 

[lmstaples]

When me and my friend first tried we got the upper limit as sqrt(2)?

Unless its still a function in θ

 

[lmstaples]

My friend and I*

 

[voko]

sqrt(2) can be the distance to ONE point; you have a segment from (1, 0) to (1, 1), so the distance must be a function of the polar angle.

 

[lmstaples]

The boundary x = 1 gives:
rcos(θ) = 1 which is:
r = 1/cos(θ)

Hmmm

 

[voko]

That seems OK.

 

[lmstaples]

integral = -1/2 sec(2 theta) (log(r^2 (-cos(2 theta))+r^2+2)-log(r^2 cos(2 theta)+r^2+2))+constant

 

[voko]

You have demonstrated mastery of LATEX, so please use it for complex expressions.

Please show how you obtained that result.

 

[lmstaples]

LATEX is very difficult on a mobile sorry :(

Thanks for the mastery compliment :)

As for how I obtained the result... WolframAlpha

But it times out when you try to put limits from 0 to 1/cos(θ)

 

[voko]

I am sure you can plug in the integration limits manually.

 

[lmstaples]

It came out perfectly :D