$int_{[0,1]} int_{[0,1]} frac {x^2-y^2}{(x^2+y^2)^2}dxdy$ and $int_{[0,1]} int_{[0,1]} frac...
$begingroup$
I am facing problem in calculating $$I_1=int_{[0,1]} int_{[0,1]} frac {x^2-y^2}{(x^2+y^2)^2} dxdy$$ and $$I_2=int_{[0,1]} int_{[0,1]} frac {x^2-y^2}{(x^2+y^2)^2} dydx$$ after substituting $x=rcos theta$ and $y=rsin theta$ we have $I_1=int_{[-pi,pi]} int_{[0,1]}frac{cos(2theta)}{r^2}rdr dtheta$ so I am getting $infty times 0=0$ So I am not getting $pi/4$ and $-pi/4$ resp.
Maybe there is a silly point I am missing. Please help.
calculus integration multivariable-calculus definite-integrals
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|
show 4 more comments
$begingroup$
I am facing problem in calculating $$I_1=int_{[0,1]} int_{[0,1]} frac {x^2-y^2}{(x^2+y^2)^2} dxdy$$ and $$I_2=int_{[0,1]} int_{[0,1]} frac {x^2-y^2}{(x^2+y^2)^2} dydx$$ after substituting $x=rcos theta$ and $y=rsin theta$ we have $I_1=int_{[-pi,pi]} int_{[0,1]}frac{cos(2theta)}{r^2}rdr dtheta$ so I am getting $infty times 0=0$ So I am not getting $pi/4$ and $-pi/4$ resp.
Maybe there is a silly point I am missing. Please help.
calculus integration multivariable-calculus definite-integrals
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1
$begingroup$
How about just doing the integrals, as asked, and not "substituting"?
$endgroup$
– Lord Shark the Unknown
Dec 16 '18 at 7:45
$begingroup$
Yeah I am confused in not getting. May be a result of doing a lot of pure math :( .. sorry
$endgroup$
– Gimgim
Dec 16 '18 at 7:54
$begingroup$
Yeah I might get downvote but as I asked this question what I didn't get. So I am not going to erase the question. Thanks all of you for helping
$endgroup$
– Gimgim
Dec 16 '18 at 8:00
1
$begingroup$
Its a good question and you showed some effort. Clearly a "trick" was needed.
$endgroup$
– RRL
Dec 16 '18 at 8:01
1
$begingroup$
@Gimgim: The anti-derivatives don't exactly jump out at you so we can all move on.
$endgroup$
– RRL
Dec 16 '18 at 8:16
|
show 4 more comments
$begingroup$
I am facing problem in calculating $$I_1=int_{[0,1]} int_{[0,1]} frac {x^2-y^2}{(x^2+y^2)^2} dxdy$$ and $$I_2=int_{[0,1]} int_{[0,1]} frac {x^2-y^2}{(x^2+y^2)^2} dydx$$ after substituting $x=rcos theta$ and $y=rsin theta$ we have $I_1=int_{[-pi,pi]} int_{[0,1]}frac{cos(2theta)}{r^2}rdr dtheta$ so I am getting $infty times 0=0$ So I am not getting $pi/4$ and $-pi/4$ resp.
Maybe there is a silly point I am missing. Please help.
calculus integration multivariable-calculus definite-integrals
$endgroup$
I am facing problem in calculating $$I_1=int_{[0,1]} int_{[0,1]} frac {x^2-y^2}{(x^2+y^2)^2} dxdy$$ and $$I_2=int_{[0,1]} int_{[0,1]} frac {x^2-y^2}{(x^2+y^2)^2} dydx$$ after substituting $x=rcos theta$ and $y=rsin theta$ we have $I_1=int_{[-pi,pi]} int_{[0,1]}frac{cos(2theta)}{r^2}rdr dtheta$ so I am getting $infty times 0=0$ So I am not getting $pi/4$ and $-pi/4$ resp.
Maybe there is a silly point I am missing. Please help.
calculus integration multivariable-calculus definite-integrals
calculus integration multivariable-calculus definite-integrals
asked Dec 16 '18 at 7:44
GimgimGimgim
25713
25713
1
$begingroup$
How about just doing the integrals, as asked, and not "substituting"?
$endgroup$
– Lord Shark the Unknown
Dec 16 '18 at 7:45
$begingroup$
Yeah I am confused in not getting. May be a result of doing a lot of pure math :( .. sorry
$endgroup$
– Gimgim
Dec 16 '18 at 7:54
$begingroup$
Yeah I might get downvote but as I asked this question what I didn't get. So I am not going to erase the question. Thanks all of you for helping
$endgroup$
– Gimgim
Dec 16 '18 at 8:00
1
$begingroup$
Its a good question and you showed some effort. Clearly a "trick" was needed.
$endgroup$
– RRL
Dec 16 '18 at 8:01
1
$begingroup$
@Gimgim: The anti-derivatives don't exactly jump out at you so we can all move on.
$endgroup$
– RRL
Dec 16 '18 at 8:16
|
show 4 more comments
1
$begingroup$
How about just doing the integrals, as asked, and not "substituting"?
$endgroup$
– Lord Shark the Unknown
Dec 16 '18 at 7:45
$begingroup$
Yeah I am confused in not getting. May be a result of doing a lot of pure math :( .. sorry
$endgroup$
– Gimgim
Dec 16 '18 at 7:54
$begingroup$
Yeah I might get downvote but as I asked this question what I didn't get. So I am not going to erase the question. Thanks all of you for helping
$endgroup$
– Gimgim
Dec 16 '18 at 8:00
1
$begingroup$
Its a good question and you showed some effort. Clearly a "trick" was needed.
$endgroup$
– RRL
Dec 16 '18 at 8:01
1
$begingroup$
@Gimgim: The anti-derivatives don't exactly jump out at you so we can all move on.
$endgroup$
– RRL
Dec 16 '18 at 8:16
1
1
$begingroup$
How about just doing the integrals, as asked, and not "substituting"?
$endgroup$
– Lord Shark the Unknown
Dec 16 '18 at 7:45
$begingroup$
How about just doing the integrals, as asked, and not "substituting"?
$endgroup$
– Lord Shark the Unknown
Dec 16 '18 at 7:45
$begingroup$
Yeah I am confused in not getting. May be a result of doing a lot of pure math :( .. sorry
$endgroup$
– Gimgim
Dec 16 '18 at 7:54
$begingroup$
Yeah I am confused in not getting. May be a result of doing a lot of pure math :( .. sorry
$endgroup$
– Gimgim
Dec 16 '18 at 7:54
$begingroup$
Yeah I might get downvote but as I asked this question what I didn't get. So I am not going to erase the question. Thanks all of you for helping
$endgroup$
– Gimgim
Dec 16 '18 at 8:00
$begingroup$
Yeah I might get downvote but as I asked this question what I didn't get. So I am not going to erase the question. Thanks all of you for helping
$endgroup$
– Gimgim
Dec 16 '18 at 8:00
1
1
$begingroup$
Its a good question and you showed some effort. Clearly a "trick" was needed.
$endgroup$
– RRL
Dec 16 '18 at 8:01
$begingroup$
Its a good question and you showed some effort. Clearly a "trick" was needed.
$endgroup$
– RRL
Dec 16 '18 at 8:01
1
1
$begingroup$
@Gimgim: The anti-derivatives don't exactly jump out at you so we can all move on.
$endgroup$
– RRL
Dec 16 '18 at 8:16
$begingroup$
@Gimgim: The anti-derivatives don't exactly jump out at you so we can all move on.
$endgroup$
– RRL
Dec 16 '18 at 8:16
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Hint:
$$frac {x^2-y^2}{(x^2+y^2)^2} = frac{partial}{partial y} left(frac{y}{x^2 + y^2}right) = -frac{partial}{partial x} left(frac{x}{x^2 + y^2}right) $$
$endgroup$
$begingroup$
Thanks a lot....
$endgroup$
– Gimgim
Dec 16 '18 at 7:58
1
$begingroup$
You're welcome. Perhaps you didn't need the second clue.
$endgroup$
– RRL
Dec 16 '18 at 7:59
$begingroup$
Yeah I might get downvote but as I asked this question what I didn't get. So I am not going to erase the question. Thanks all of you for helping
$endgroup$
– Gimgim
Dec 16 '18 at 8:00
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
Hint:
$$frac {x^2-y^2}{(x^2+y^2)^2} = frac{partial}{partial y} left(frac{y}{x^2 + y^2}right) = -frac{partial}{partial x} left(frac{x}{x^2 + y^2}right) $$
$endgroup$
$begingroup$
Thanks a lot....
$endgroup$
– Gimgim
Dec 16 '18 at 7:58
1
$begingroup$
You're welcome. Perhaps you didn't need the second clue.
$endgroup$
– RRL
Dec 16 '18 at 7:59
$begingroup$
Yeah I might get downvote but as I asked this question what I didn't get. So I am not going to erase the question. Thanks all of you for helping
$endgroup$
– Gimgim
Dec 16 '18 at 8:00
add a comment |
$begingroup$
Hint:
$$frac {x^2-y^2}{(x^2+y^2)^2} = frac{partial}{partial y} left(frac{y}{x^2 + y^2}right) = -frac{partial}{partial x} left(frac{x}{x^2 + y^2}right) $$
$endgroup$
$begingroup$
Thanks a lot....
$endgroup$
– Gimgim
Dec 16 '18 at 7:58
1
$begingroup$
You're welcome. Perhaps you didn't need the second clue.
$endgroup$
– RRL
Dec 16 '18 at 7:59
$begingroup$
Yeah I might get downvote but as I asked this question what I didn't get. So I am not going to erase the question. Thanks all of you for helping
$endgroup$
– Gimgim
Dec 16 '18 at 8:00
add a comment |
$begingroup$
Hint:
$$frac {x^2-y^2}{(x^2+y^2)^2} = frac{partial}{partial y} left(frac{y}{x^2 + y^2}right) = -frac{partial}{partial x} left(frac{x}{x^2 + y^2}right) $$
$endgroup$
Hint:
$$frac {x^2-y^2}{(x^2+y^2)^2} = frac{partial}{partial y} left(frac{y}{x^2 + y^2}right) = -frac{partial}{partial x} left(frac{x}{x^2 + y^2}right) $$
edited Dec 16 '18 at 7:58
answered Dec 16 '18 at 7:55
RRLRRL
51k42573
51k42573
$begingroup$
Thanks a lot....
$endgroup$
– Gimgim
Dec 16 '18 at 7:58
1
$begingroup$
You're welcome. Perhaps you didn't need the second clue.
$endgroup$
– RRL
Dec 16 '18 at 7:59
$begingroup$
Yeah I might get downvote but as I asked this question what I didn't get. So I am not going to erase the question. Thanks all of you for helping
$endgroup$
– Gimgim
Dec 16 '18 at 8:00
add a comment |
$begingroup$
Thanks a lot....
$endgroup$
– Gimgim
Dec 16 '18 at 7:58
1
$begingroup$
You're welcome. Perhaps you didn't need the second clue.
$endgroup$
– RRL
Dec 16 '18 at 7:59
$begingroup$
Yeah I might get downvote but as I asked this question what I didn't get. So I am not going to erase the question. Thanks all of you for helping
$endgroup$
– Gimgim
Dec 16 '18 at 8:00
$begingroup$
Thanks a lot....
$endgroup$
– Gimgim
Dec 16 '18 at 7:58
$begingroup$
Thanks a lot....
$endgroup$
– Gimgim
Dec 16 '18 at 7:58
1
1
$begingroup$
You're welcome. Perhaps you didn't need the second clue.
$endgroup$
– RRL
Dec 16 '18 at 7:59
$begingroup$
You're welcome. Perhaps you didn't need the second clue.
$endgroup$
– RRL
Dec 16 '18 at 7:59
$begingroup$
Yeah I might get downvote but as I asked this question what I didn't get. So I am not going to erase the question. Thanks all of you for helping
$endgroup$
– Gimgim
Dec 16 '18 at 8:00
$begingroup$
Yeah I might get downvote but as I asked this question what I didn't get. So I am not going to erase the question. Thanks all of you for helping
$endgroup$
– Gimgim
Dec 16 '18 at 8:00
add a comment |
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1
$begingroup$
How about just doing the integrals, as asked, and not "substituting"?
$endgroup$
– Lord Shark the Unknown
Dec 16 '18 at 7:45
$begingroup$
Yeah I am confused in not getting. May be a result of doing a lot of pure math :( .. sorry
$endgroup$
– Gimgim
Dec 16 '18 at 7:54
$begingroup$
Yeah I might get downvote but as I asked this question what I didn't get. So I am not going to erase the question. Thanks all of you for helping
$endgroup$
– Gimgim
Dec 16 '18 at 8:00
1
$begingroup$
Its a good question and you showed some effort. Clearly a "trick" was needed.
$endgroup$
– RRL
Dec 16 '18 at 8:01
1
$begingroup$
@Gimgim: The anti-derivatives don't exactly jump out at you so we can all move on.
$endgroup$
– RRL
Dec 16 '18 at 8:16