Double integral with polar coordinates

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akleron

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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
1579728767083.png
Question:
1579727133651.png

My attempt
1579727325078.png
 
Last edited:
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...

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

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

\(\displaystyle = -2t+\sin\left(2t\right)+c \)
 
1) Check du
2) What are those new limits on r?
3) It won't help make it cleaner.
 
Cubist said:
\(\displaystyle -\color{red} 2 \color{black}\int \left(\tan\left(t\right)\cdot \sin\left(2t\right)\right)\,dt \)
Click to expand...

Oops! I just plotted a numeric integral alongside my result and I was out by a factor of 2 :oops: 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?
 
Cubist said:
Oops! I just plotted a numeric integral alongside my result and I was out by a factor of 2 :oops: 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?
Click to expand...
I somehow fell in the same trap and thought we were right.....

2 many 2's in those 2 equations........
 
akleron said:
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
Click to expand...
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.
 
Jomo said:
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.
Click to expand...

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