Double integral with polar coordinates

[akleron]

Hello !
I've been trying to figure out this integral all day long but didn't manage to..
This is pretty much what I could do.
in Wolfarm the integral of what I got in the last row integral looks nasty, any way to make it simpler ?

We also received a hint

Question:


My attempt

 

[Cubist]

I think I found a way to use their hint. Please double check my work because I'm not sure. Instead of your "u" substitution you could try...

$\displaystyle \sqrt{\frac{1-r^2}{1+r^2}}=\tan\left(t\right)$

with a bit of trig, this also means...

$\displaystyle r^2=\cos\left(2t\right)$

Differentiate

$\displaystyle 2r\,\mathrm{d}r = -2\sin(2t)\,\mathrm{d}t$

After the change of variable you have...

$\displaystyle -2\int \left(\tan\left(t\right)\cdot \sin\left(2t\right)\right)\,dt$

$\displaystyle = -4\int\sin^{2}\left(t\right)\,dt$

$\displaystyle = -2t+\sin\left(2t\right)+c$

 

[tkhunny]

1) Check du
2) What are those new limits on r?
3) It won't help make it cleaner.

 

[Cubist]

$\displaystyle -\color{red} 2 \color{black}\int \left(\tan\left(t\right)\cdot \sin\left(2t\right)\right)\,dt$

Oops! I just plotted a numeric integral alongside my result and I was out by a factor of 2 I went wrong at the change of variable stage, the red number above is not required. Just divide all the subsequent lines by 2...

$\displaystyle -\int \left(\tan\left(t\right)\cdot \sin\left(2t\right)\right)\,dt$

$\displaystyle = -2\int\sin^{2}\left(t\right)\,dt$

$\displaystyle = -t+\frac{\sin\left(2t\right)}{2}+c$

Do I have to go to the corner?

 

[Deleted member 4993]

Oops! I just plotted a numeric integral alongside my result and I was out by a factor of 2 I went wrong at the change of variable stage, the red number above is not required. Just divide all the subsequent lines by 2...

$\displaystyle -\int \left(\tan\left(t\right)\cdot \sin\left(2t\right)\right)\,dt$

$\displaystyle = -2\int\sin^{2}\left(t\right)\,dt$

$\displaystyle = -t+\frac{\sin\left(2t\right)}{2}+c$

Do I have to go to the corner?

I somehow fell in the same trap and thought we were right.....

2 many 2's in those 2 equations........

 

[Steven G]

I somehow fell in the same trap and thought we were right.....

2 many 2's in those 2 equations........

Corner Corner, Khan is going to the corner (222 minutes)

 

[Steven G]

Hello !
I've been trying to figure out this integral all day long but didn't manage to..
This is pretty much what I could do.
in Wolfarm the integral of what I got in the last row integral looks nasty, any way to make it simpler ?

We also received a hint
View attachment 16236
Question:
View attachment 16234

My attempt
View attachment 16235

Please look at your graph for D. (0,1) is Not on the x-axis and (1.0) is Not on the y-axis. Please be more careful.

 

[akleron]

Please look at your graph for D. (0,1) is Not on the x-axis and (1.0) is Not on the y-axis. Please be more careful.

Yea, silly mistake.. I've been up all night preparing for the upcoming exam..