Where does the $pi$ comes from?
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I have calculated with Mathematica the integral:
$$int_{-h/2}^{h/2}dzint_{-R}^{R}dxint_{-sqrt{R^2-x^2}}^{sqrt{R^2-x^2}}dy(x^2+z^2)$$
the result is: $$dfrac{pi}{12}h^3R^2+dfrac{pi}{4}hR^4$$
I am surprised about the $pi$: where does it comes from?
definite-integrals
asked Jan 24 at 10:25
mattiav27mattiav27
378
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add a comment |
$begingroup$
I have calculated with Mathematica the integral:
$$int_{-h/2}^{h/2}dzint_{-R}^{R}dxint_{-sqrt{R^2-x^2}}^{sqrt{R^2-x^2}}dy(x^2+z^2)$$
the result is: $$dfrac{pi}{12}h^3R^2+dfrac{pi}{4}hR^4$$
I am surprised about the $pi$: where does it comes from?
definite-integrals
asked Jan 24 at 10:25
mattiav27mattiav27
378
$endgroup$
$begingroup$
@ClaudeLeibovici thanks it was a typo. See the edit.
$endgroup$
– mattiav27
Jan 24 at 10:55
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$pi$ somes from the second integration because there is an $tan^{-1}(.)$ in the antiderivative.
$endgroup$
– Claude Leibovici
Jan 24 at 10:58
add a comment |
$begingroup$
I have calculated with Mathematica the integral:
$$int_{-h/2}^{h/2}dzint_{-R}^{R}dxint_{-sqrt{R^2-x^2}}^{sqrt{R^2-x^2}}dy(x^2+z^2)$$
the result is: $$dfrac{pi}{12}h^3R^2+dfrac{pi}{4}hR^4$$
I am surprised about the $pi$: where does it comes from?
definite-integrals
asked Jan 24 at 10:25
mattiav27mattiav27
378
$endgroup$
I have calculated with Mathematica the integral:
$$int_{-h/2}^{h/2}dzint_{-R}^{R}dxint_{-sqrt{R^2-x^2}}^{sqrt{R^2-x^2}}dy(x^2+z^2)$$
the result is: $$dfrac{pi}{12}h^3R^2+dfrac{pi}{4}hR^4$$
I am surprised about the $pi$: where does it comes from?
definite-integrals
definite-integrals
asked Jan 24 at 10:25
mattiav27mattiav27
378
asked Jan 24 at 10:25
mattiav27mattiav27
378
edited Jan 24 at 10:54
mattiav27
asked Jan 24 at 10:25
mattiav27mattiav27
378
asked Jan 24 at 10:25
mattiav27mattiav27
378
asked Jan 24 at 10:25
mattiav27mattiav27
378
378
$begingroup$
@ClaudeLeibovici thanks it was a typo. See the edit.
$endgroup$
– mattiav27
Jan 24 at 10:55
$begingroup$
$pi$ somes from the second integration because there is an $tan^{-1}(.)$ in the antiderivative.
$endgroup$
– Claude Leibovici
Jan 24 at 10:58
add a comment |
$begingroup$
@ClaudeLeibovici thanks it was a typo. See the edit.
$endgroup$
– mattiav27
Jan 24 at 10:55
$begingroup$
$pi$ somes from the second integration because there is an $tan^{-1}(.)$ in the antiderivative.
$endgroup$
– Claude Leibovici
Jan 24 at 10:58
$begingroup$
@ClaudeLeibovici thanks it was a typo. See the edit.
$endgroup$
– mattiav27
Jan 24 at 10:55
$begingroup$
@ClaudeLeibovici thanks it was a typo. See the edit.
$endgroup$
– mattiav27
Jan 24 at 10:55
$begingroup$
$pi$ somes from the second integration because there is an $tan^{-1}(.)$ in the antiderivative.
$endgroup$
– Claude Leibovici
Jan 24 at 10:58
$begingroup$
$pi$ somes from the second integration because there is an $tan^{-1}(.)$ in the antiderivative.
$endgroup$
– Claude Leibovici
Jan 24 at 10:58
add a comment |
1 Answer
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$begingroup$
At some point you are integrating something like:
$$
int_{-1}^{1}sqrt{1-x^2}dx
$$
The graph of the function $f(x)=sqrt{1-x^2}$ looks like this:
So you are trying to integrate (find the area) of half the circle. Of course, you would expect $pi$.
answered Jan 24 at 16:54
Vasily MitchVasily Mitch
2,3691311
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1 Answer
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1 Answer
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active
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active
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active
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votes
$begingroup$
At some point you are integrating something like:
$$
int_{-1}^{1}sqrt{1-x^2}dx
$$
The graph of the function $f(x)=sqrt{1-x^2}$ looks like this:
So you are trying to integrate (find the area) of half the circle. Of course, you would expect $pi$.
answered Jan 24 at 16:54
Vasily MitchVasily Mitch
2,3691311
$endgroup$
add a comment |
$begingroup$
At some point you are integrating something like:
$$
int_{-1}^{1}sqrt{1-x^2}dx
$$
The graph of the function $f(x)=sqrt{1-x^2}$ looks like this:
So you are trying to integrate (find the area) of half the circle. Of course, you would expect $pi$.
answered Jan 24 at 16:54
Vasily MitchVasily Mitch
2,3691311
$endgroup$
add a comment |
$begingroup$
At some point you are integrating something like:
$$
int_{-1}^{1}sqrt{1-x^2}dx
$$
The graph of the function $f(x)=sqrt{1-x^2}$ looks like this:
So you are trying to integrate (find the area) of half the circle. Of course, you would expect $pi$.
answered Jan 24 at 16:54
Vasily MitchVasily Mitch
2,3691311
$endgroup$
At some point you are integrating something like:
$$
int_{-1}^{1}sqrt{1-x^2}dx
$$
The graph of the function $f(x)=sqrt{1-x^2}$ looks like this:
So you are trying to integrate (find the area) of half the circle. Of course, you would expect $pi$.
answered Jan 24 at 16:54
Vasily MitchVasily Mitch
2,3691311
answered Jan 24 at 16:54
Vasily MitchVasily Mitch
2,3691311
answered Jan 24 at 16:54
Vasily MitchVasily Mitch
2,3691311
answered Jan 24 at 16:54
Vasily MitchVasily Mitch
2,3691311
2,3691311
add a comment |
add a comment |
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$begingroup$
@ClaudeLeibovici thanks it was a typo. See the edit.
$endgroup$
– mattiav27
Jan 24 at 10:55
$begingroup$
$pi$ somes from the second integration because there is an $tan^{-1}(.)$ in the antiderivative.
$endgroup$
– Claude Leibovici
Jan 24 at 10:58