Double integral calculation over a unit disk
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I'm trying to find out the double integral of the function over a unit disk. I have tried putting it in polar coordinated but that becomes too lengthy. Not sure how to move ahead.
Prove that $displaystyle fracpi3 leq iintlimits_D frac{dxdy}{sqrt{x^2+(y-2)^2}} leq pi$ where $D$ is the unit disc.
integration multiple-integral
edited Nov 26 at 7:07
Rócherz
2,7412721
asked Nov 26 at 6:55
Shashank
31
add a comment |
up vote
0
down vote
favorite
I'm trying to find out the double integral of the function over a unit disk. I have tried putting it in polar coordinated but that becomes too lengthy. Not sure how to move ahead.
Prove that $displaystyle fracpi3 leq iintlimits_D frac{dxdy}{sqrt{x^2+(y-2)^2}} leq pi$ where $D$ is the unit disc.
integration multiple-integral
edited Nov 26 at 7:07
Rócherz
2,7412721
asked Nov 26 at 6:55
Shashank
31
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm trying to find out the double integral of the function over a unit disk. I have tried putting it in polar coordinated but that becomes too lengthy. Not sure how to move ahead.
Prove that $displaystyle fracpi3 leq iintlimits_D frac{dxdy}{sqrt{x^2+(y-2)^2}} leq pi$ where $D$ is the unit disc.
integration multiple-integral
edited Nov 26 at 7:07
Rócherz
2,7412721
asked Nov 26 at 6:55
Shashank
31
I'm trying to find out the double integral of the function over a unit disk. I have tried putting it in polar coordinated but that becomes too lengthy. Not sure how to move ahead.
Prove that $displaystyle fracpi3 leq iintlimits_D frac{dxdy}{sqrt{x^2+(y-2)^2}} leq pi$ where $D$ is the unit disc.
integration multiple-integral
integration multiple-integral
edited Nov 26 at 7:07
Rócherz
2,7412721
asked Nov 26 at 6:55
Shashank
31
edited Nov 26 at 7:07
Rócherz
2,7412721
asked Nov 26 at 6:55
Shashank
31
edited Nov 26 at 7:07
Rócherz
2,7412721
edited Nov 26 at 7:07
Rócherz
2,7412721
edited Nov 26 at 7:07
Rócherz
2,7412721
2,7412721
asked Nov 26 at 6:55
Shashank
31
asked Nov 26 at 6:55
Shashank
31
asked Nov 26 at 6:55
Shashank
31
31
add a comment |
add a comment |
1 Answer
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You don't have to calculate the integral exactly because what is required is only lower bound. The statement is equivalent to
$$frac{1}{vol(D)}int_D frac{1}{sqrt{x^2+(y-2)^2}}dxdy geq frac{1}{3}.
$$So try first to show that
$$
frac{1}{sqrt{x^2+(y-2)^2}}geq frac{1}{3}.
$$(This is quite straightforward.)
answered Nov 26 at 7:17
Song
2,380112
Thanks.. i was thinking in a completely different way.
– Shashank
Nov 26 at 7:29
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
You don't have to calculate the integral exactly because what is required is only lower bound. The statement is equivalent to
$$frac{1}{vol(D)}int_D frac{1}{sqrt{x^2+(y-2)^2}}dxdy geq frac{1}{3}.
$$So try first to show that
$$
frac{1}{sqrt{x^2+(y-2)^2}}geq frac{1}{3}.
$$(This is quite straightforward.)
answered Nov 26 at 7:17
Song
2,380112
Thanks.. i was thinking in a completely different way.
– Shashank
Nov 26 at 7:29
add a comment |
up vote
0
down vote
accepted
You don't have to calculate the integral exactly because what is required is only lower bound. The statement is equivalent to
$$frac{1}{vol(D)}int_D frac{1}{sqrt{x^2+(y-2)^2}}dxdy geq frac{1}{3}.
$$So try first to show that
$$
frac{1}{sqrt{x^2+(y-2)^2}}geq frac{1}{3}.
$$(This is quite straightforward.)
answered Nov 26 at 7:17
Song
2,380112
Thanks.. i was thinking in a completely different way.
– Shashank
Nov 26 at 7:29
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
You don't have to calculate the integral exactly because what is required is only lower bound. The statement is equivalent to
$$frac{1}{vol(D)}int_D frac{1}{sqrt{x^2+(y-2)^2}}dxdy geq frac{1}{3}.
$$So try first to show that
$$
frac{1}{sqrt{x^2+(y-2)^2}}geq frac{1}{3}.
$$(This is quite straightforward.)
answered Nov 26 at 7:17
Song
2,380112
You don't have to calculate the integral exactly because what is required is only lower bound. The statement is equivalent to
$$frac{1}{vol(D)}int_D frac{1}{sqrt{x^2+(y-2)^2}}dxdy geq frac{1}{3}.
$$So try first to show that
$$
frac{1}{sqrt{x^2+(y-2)^2}}geq frac{1}{3}.
$$(This is quite straightforward.)
answered Nov 26 at 7:17
Song
2,380112
answered Nov 26 at 7:17
Song
2,380112
answered Nov 26 at 7:17
Song
2,380112
answered Nov 26 at 7:17
Song
2,380112
2,380112
Thanks.. i was thinking in a completely different way.
– Shashank
Nov 26 at 7:29
add a comment |
Thanks.. i was thinking in a completely different way.
– Shashank
Nov 26 at 7:29
Thanks.. i was thinking in a completely different way.
– Shashank
Nov 26 at 7:29
Thanks.. i was thinking in a completely different way.
– Shashank
Nov 26 at 7:29
add a comment |
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