Integration over Sphere












0












$begingroup$


I have a question to a pretty basic integration problem. I was pretty sure about my solution but my tutor had a different one such that I am confused now. The integral is the following:



$$int_{S_r(0)} z^2 d S_r(0)$$



Using the standard parameterization, I obtain:



$$d S = r^2 sin(theta) d theta d phi$$



$$int_{S_r(0)} z^2 d S_r(0) = int_{0}^{2 pi} int_{0}^{pi} r^2 cos^2(theta) r^2 sin(theta) d theta d phi = dots = frac{4 r^4pi}{3}$$



I would be very glad if someone could verify or falsify the result!



Thanks much in advance



Andreas






integration proof-verification spheres







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  • $begingroup$
    Can you encase your tex syntax in dollar and double dollar cases?
    $endgroup$
    – Kevin
    Mar 15 '18 at 15:42
















0












$begingroup$


I have a question to a pretty basic integration problem. I was pretty sure about my solution but my tutor had a different one such that I am confused now. The integral is the following:



$$int_{S_r(0)} z^2 d S_r(0)$$



Using the standard parameterization, I obtain:



$$d S = r^2 sin(theta) d theta d phi$$



$$int_{S_r(0)} z^2 d S_r(0) = int_{0}^{2 pi} int_{0}^{pi} r^2 cos^2(theta) r^2 sin(theta) d theta d phi = dots = frac{4 r^4pi}{3}$$



I would be very glad if someone could verify or falsify the result!



Thanks much in advance



Andreas






integration proof-verification spheres







share|cite|improve this question











$endgroup$












  • $begingroup$
    Can you encase your tex syntax in dollar and double dollar cases?
    $endgroup$
    – Kevin
    Mar 15 '18 at 15:42














0












0








0





$begingroup$


I have a question to a pretty basic integration problem. I was pretty sure about my solution but my tutor had a different one such that I am confused now. The integral is the following:



$$int_{S_r(0)} z^2 d S_r(0)$$



Using the standard parameterization, I obtain:



$$d S = r^2 sin(theta) d theta d phi$$



$$int_{S_r(0)} z^2 d S_r(0) = int_{0}^{2 pi} int_{0}^{pi} r^2 cos^2(theta) r^2 sin(theta) d theta d phi = dots = frac{4 r^4pi}{3}$$



I would be very glad if someone could verify or falsify the result!



Thanks much in advance



Andreas






integration proof-verification spheres







share|cite|improve this question











$endgroup$




I have a question to a pretty basic integration problem. I was pretty sure about my solution but my tutor had a different one such that I am confused now. The integral is the following:



$$int_{S_r(0)} z^2 d S_r(0)$$



Using the standard parameterization, I obtain:



$$d S = r^2 sin(theta) d theta d phi$$



$$int_{S_r(0)} z^2 d S_r(0) = int_{0}^{2 pi} int_{0}^{pi} r^2 cos^2(theta) r^2 sin(theta) d theta d phi = dots = frac{4 r^4pi}{3}$$



I would be very glad if someone could verify or falsify the result!



Thanks much in advance



Andreas






integration proof-verification spheres




integration proof-verification spheres






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 11:20









J.G.

24.2k22539




24.2k22539










asked Mar 15 '18 at 15:39









Andreas MuellerAndreas Mueller

11




11












  • $begingroup$
    Can you encase your tex syntax in dollar and double dollar cases?
    $endgroup$
    – Kevin
    Mar 15 '18 at 15:42


















  • $begingroup$
    Can you encase your tex syntax in dollar and double dollar cases?
    $endgroup$
    – Kevin
    Mar 15 '18 at 15:42
















$begingroup$
Can you encase your tex syntax in dollar and double dollar cases?
$endgroup$
– Kevin
Mar 15 '18 at 15:42




$begingroup$
Can you encase your tex syntax in dollar and double dollar cases?
$endgroup$
– Kevin
Mar 15 '18 at 15:42










1 Answer
1






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$begingroup$

$I=r^4displaystyle int_{0}^{2pi} dphi .left(-displaystyle int_{0}^{pi}cos^2theta
d(costheta)right)$



$I=dfrac{1}{3}r^4displaystyle int_{0}^{2pi} dphi .left[cos^3thetaright]_{pi}^{0}$



$I=dfrac{2}{3}r^4displaystyle int_{0}^{2pi} dphi $



$I=left(dfrac{2}{3}r^4.2piright)=dfrac{4pi}{3}r^4 $






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you!.. As I stated, that’s exactly my result as well. Does it make a difference when I write $int_{S_r(0)} z^2 dvol_{S_r(0)}$ ?
    $endgroup$
    – Andreas Mueller
    Mar 15 '18 at 21:45













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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

$I=r^4displaystyle int_{0}^{2pi} dphi .left(-displaystyle int_{0}^{pi}cos^2theta
d(costheta)right)$



$I=dfrac{1}{3}r^4displaystyle int_{0}^{2pi} dphi .left[cos^3thetaright]_{pi}^{0}$



$I=dfrac{2}{3}r^4displaystyle int_{0}^{2pi} dphi $



$I=left(dfrac{2}{3}r^4.2piright)=dfrac{4pi}{3}r^4 $






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you!.. As I stated, that’s exactly my result as well. Does it make a difference when I write $int_{S_r(0)} z^2 dvol_{S_r(0)}$ ?
    $endgroup$
    – Andreas Mueller
    Mar 15 '18 at 21:45


















0












$begingroup$

$I=r^4displaystyle int_{0}^{2pi} dphi .left(-displaystyle int_{0}^{pi}cos^2theta
d(costheta)right)$



$I=dfrac{1}{3}r^4displaystyle int_{0}^{2pi} dphi .left[cos^3thetaright]_{pi}^{0}$



$I=dfrac{2}{3}r^4displaystyle int_{0}^{2pi} dphi $



$I=left(dfrac{2}{3}r^4.2piright)=dfrac{4pi}{3}r^4 $






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you!.. As I stated, that’s exactly my result as well. Does it make a difference when I write $int_{S_r(0)} z^2 dvol_{S_r(0)}$ ?
    $endgroup$
    – Andreas Mueller
    Mar 15 '18 at 21:45
















0












0








0





$begingroup$

$I=r^4displaystyle int_{0}^{2pi} dphi .left(-displaystyle int_{0}^{pi}cos^2theta
d(costheta)right)$



$I=dfrac{1}{3}r^4displaystyle int_{0}^{2pi} dphi .left[cos^3thetaright]_{pi}^{0}$



$I=dfrac{2}{3}r^4displaystyle int_{0}^{2pi} dphi $



$I=left(dfrac{2}{3}r^4.2piright)=dfrac{4pi}{3}r^4 $






share|cite|improve this answer









$endgroup$



$I=r^4displaystyle int_{0}^{2pi} dphi .left(-displaystyle int_{0}^{pi}cos^2theta
d(costheta)right)$



$I=dfrac{1}{3}r^4displaystyle int_{0}^{2pi} dphi .left[cos^3thetaright]_{pi}^{0}$



$I=dfrac{2}{3}r^4displaystyle int_{0}^{2pi} dphi $



$I=left(dfrac{2}{3}r^4.2piright)=dfrac{4pi}{3}r^4 $







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Mar 15 '18 at 19:35









deleteprofiledeleteprofile

1,153316




1,153316












  • $begingroup$
    Thank you!.. As I stated, that’s exactly my result as well. Does it make a difference when I write $int_{S_r(0)} z^2 dvol_{S_r(0)}$ ?
    $endgroup$
    – Andreas Mueller
    Mar 15 '18 at 21:45




















  • $begingroup$
    Thank you!.. As I stated, that’s exactly my result as well. Does it make a difference when I write $int_{S_r(0)} z^2 dvol_{S_r(0)}$ ?
    $endgroup$
    – Andreas Mueller
    Mar 15 '18 at 21:45


















$begingroup$
Thank you!.. As I stated, that’s exactly my result as well. Does it make a difference when I write $int_{S_r(0)} z^2 dvol_{S_r(0)}$ ?
$endgroup$
– Andreas Mueller
Mar 15 '18 at 21:45






$begingroup$
Thank you!.. As I stated, that’s exactly my result as well. Does it make a difference when I write $int_{S_r(0)} z^2 dvol_{S_r(0)}$ ?
$endgroup$
– Andreas Mueller
Mar 15 '18 at 21:45




















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