Evaluating Polar Int., Find Error: \frac{\pi}{8}-0.25ln(2)

[Juggler123]

Evaluate the integral by changing to polar coordinates

\int\int arctan(y/x)

Given that

0 \leq x \leq 1 and 0 \leq y \leq x

Now I've changed the integral to

\int\int \theta r dr d\theta

Such that 0 \leq \theta \leq \frac{\pi}{2} and 0 \leq r \leq \sqrt{2}

And evaluating this I get \frac{\pi^{2}}{8}

I don't think this is correct though, I have found out that the answer is \frac{\pi}{8} - 0.25ln(2)

Can someone show me where I'm going wrong please, thanks!

Sorry for the poor Latex-ing I've never used it before.

 

[tiny-tim]

Hi Juggler123!

(have an integral: ∫ and a pi: π and a theta: θ and a ≤ )

Sorry, but your limits are toooootally wrong.

0 ≤ x ≤ 1 and 0 ≤ y ≤ x is a 45º triangle with base 1.

So your limits are 0 ≤ θ ≤ 45º and 0 ≤ r ≤ … (depends on θ ) ?

 

[Juggler123]

Would it be correct to say that the upper limit for r is 1/cos\theta ?

 

[tiny-tim]

Hi Juggler123!

(what happened to that θ i gave you? )

Would it be correct to say that the upper limit for r is 1/cos\theta ?

Noooo.

 

[Juggler123]

How can I go about finding the upper limit then? I used the fact that the maximum x=1=max.y then said x=rcos(theta) and rearranged, where have I gone wrong??

 

[Mark44]

Hint: In your triangular region, every x value on the right boundary has the same value.

 

[Juggler123]

Sorry I'm still confused, why does knowing that on the right boundary x is always equal to 1?

 

[tiny-tim]

oops!

How can I go about finding the upper limit then? I used the fact that the maximum x=1=max.y then said x=rcos(theta) and rearranged, where have I gone wrong??

oh no, I had my diagram the wrong way up …

you were right, it is 0 ≤ r ≤ 1/cosθ.

Sorry.

ok, now integrate ∫∫ θr drdθ between those limits. ​

 

[Mark44]

Sorry I'm still confused, why does knowing that on the right boundary x is always equal to 1?

Because, on that boundary x = 1 ==> rcos(θ) = 1, so r = 1/cos(θ).

It looks like you already understand this, based on another reply you gave.

 

[Juggler123]

Now I've stumbled across another problem!

How would I integrate x/(2cos(x)^2) w.r.t x

 

[Mark44]

Now I've stumbled across another problem!

How would I integrate x/(2cos(x)^2) w.r.t x

This is actually pretty simple.
\int \frac{x}{2cos^2x} dx~=~1/2\int x sec^2 x dx
Edit: added missing x factor in right integral.
Hint: There is a trig function whose derivative wrt x is sec^2(x).

 

[Juggler123]

Isn't x/2cos(x)^2 equal to 0.5xsec(x)^2?? Not 0.5sec(x)^2

 

[Mark44]

Right you are. The integral should be
\int \frac{x}{2cos^2x} dx~=~1/2\int x sec^2 x dx
This looks like a natural for integration by parts, with u = x and dv = sec^2(x)dx.

 

[Juggler123]

Yes! That works brilliant, thanks for all the help!