Calculate the integral of the following function over the sphere in $mathbb{R}^4$












1












$begingroup$



Calculate the integral of $f(x,y,w,z) = |y||z|^2$ over $S={(x,y,w,z)|x^2+y^2+z^2+w^2=1}$, i.e. calculate: $int_S|y||z|^2dsigma_3$.




My attempt -



I'll define a parameterization as follows - $phi(r,alpha,beta)=(rcos(alpha),rsin(alpha),rcos(beta),rsin(beta))$.



Using the following formula: $int_Mf dS = intintint fcirc phi sqrt{det(D_phi*D_phi^T)}drdalpha dbeta$



Putting it all together I get that $int_S|y||z|^2dsigma_3 = int_0^{2pi}int_0^{2pi}int_0^1|rsin(alpha)|r^2sin(beta)^2sqrt{2}r^2drdalpha dbeta = frac{sqrt{2}}{6}piint_0^{2pi}|sin(alpha)|dalpha = frac{sqrt{2}}{6}pi(int_0^pi sin(alpha) dalpha - int_pi^{2pi} sin(alpha) dalpha) = frac{4sqrt{2}}{6}pi = frac{2sqrt{2}}{3}pi$



Is my calculation correct?



And if not, what is the right way to calculate it?






calculus integration multivariable-calculus manifolds spheres







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you asking how to do the single variable integral that you've arrived at?
    $endgroup$
    – jgon
    Dec 15 '18 at 16:11










  • $begingroup$
    @jgon Edited... Is my final result correct?
    $endgroup$
    – ChikChak
    Dec 15 '18 at 16:15










  • $begingroup$
    Well the final result certainly can't be zero, so no, and you haven't done the single variable integral correctly, since that integral also must be positive.
    $endgroup$
    – jgon
    Dec 15 '18 at 16:16










  • $begingroup$
    @jgon Though, is the way up to the single variable integral correct?
    $endgroup$
    – ChikChak
    Dec 15 '18 at 16:19










  • $begingroup$
    Not sure, I didn't want to compute the determinant of the matrix.
    $endgroup$
    – jgon
    Dec 15 '18 at 16:20
















1












$begingroup$



Calculate the integral of $f(x,y,w,z) = |y||z|^2$ over $S={(x,y,w,z)|x^2+y^2+z^2+w^2=1}$, i.e. calculate: $int_S|y||z|^2dsigma_3$.




My attempt -



I'll define a parameterization as follows - $phi(r,alpha,beta)=(rcos(alpha),rsin(alpha),rcos(beta),rsin(beta))$.



Using the following formula: $int_Mf dS = intintint fcirc phi sqrt{det(D_phi*D_phi^T)}drdalpha dbeta$



Putting it all together I get that $int_S|y||z|^2dsigma_3 = int_0^{2pi}int_0^{2pi}int_0^1|rsin(alpha)|r^2sin(beta)^2sqrt{2}r^2drdalpha dbeta = frac{sqrt{2}}{6}piint_0^{2pi}|sin(alpha)|dalpha = frac{sqrt{2}}{6}pi(int_0^pi sin(alpha) dalpha - int_pi^{2pi} sin(alpha) dalpha) = frac{4sqrt{2}}{6}pi = frac{2sqrt{2}}{3}pi$



Is my calculation correct?



And if not, what is the right way to calculate it?






calculus integration multivariable-calculus manifolds spheres







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you asking how to do the single variable integral that you've arrived at?
    $endgroup$
    – jgon
    Dec 15 '18 at 16:11










  • $begingroup$
    @jgon Edited... Is my final result correct?
    $endgroup$
    – ChikChak
    Dec 15 '18 at 16:15










  • $begingroup$
    Well the final result certainly can't be zero, so no, and you haven't done the single variable integral correctly, since that integral also must be positive.
    $endgroup$
    – jgon
    Dec 15 '18 at 16:16










  • $begingroup$
    @jgon Though, is the way up to the single variable integral correct?
    $endgroup$
    – ChikChak
    Dec 15 '18 at 16:19










  • $begingroup$
    Not sure, I didn't want to compute the determinant of the matrix.
    $endgroup$
    – jgon
    Dec 15 '18 at 16:20














1












1








1





$begingroup$



Calculate the integral of $f(x,y,w,z) = |y||z|^2$ over $S={(x,y,w,z)|x^2+y^2+z^2+w^2=1}$, i.e. calculate: $int_S|y||z|^2dsigma_3$.




My attempt -



I'll define a parameterization as follows - $phi(r,alpha,beta)=(rcos(alpha),rsin(alpha),rcos(beta),rsin(beta))$.



Using the following formula: $int_Mf dS = intintint fcirc phi sqrt{det(D_phi*D_phi^T)}drdalpha dbeta$



Putting it all together I get that $int_S|y||z|^2dsigma_3 = int_0^{2pi}int_0^{2pi}int_0^1|rsin(alpha)|r^2sin(beta)^2sqrt{2}r^2drdalpha dbeta = frac{sqrt{2}}{6}piint_0^{2pi}|sin(alpha)|dalpha = frac{sqrt{2}}{6}pi(int_0^pi sin(alpha) dalpha - int_pi^{2pi} sin(alpha) dalpha) = frac{4sqrt{2}}{6}pi = frac{2sqrt{2}}{3}pi$



Is my calculation correct?



And if not, what is the right way to calculate it?






calculus integration multivariable-calculus manifolds spheres







share|cite|improve this question











$endgroup$





Calculate the integral of $f(x,y,w,z) = |y||z|^2$ over $S={(x,y,w,z)|x^2+y^2+z^2+w^2=1}$, i.e. calculate: $int_S|y||z|^2dsigma_3$.




My attempt -



I'll define a parameterization as follows - $phi(r,alpha,beta)=(rcos(alpha),rsin(alpha),rcos(beta),rsin(beta))$.



Using the following formula: $int_Mf dS = intintint fcirc phi sqrt{det(D_phi*D_phi^T)}drdalpha dbeta$



Putting it all together I get that $int_S|y||z|^2dsigma_3 = int_0^{2pi}int_0^{2pi}int_0^1|rsin(alpha)|r^2sin(beta)^2sqrt{2}r^2drdalpha dbeta = frac{sqrt{2}}{6}piint_0^{2pi}|sin(alpha)|dalpha = frac{sqrt{2}}{6}pi(int_0^pi sin(alpha) dalpha - int_pi^{2pi} sin(alpha) dalpha) = frac{4sqrt{2}}{6}pi = frac{2sqrt{2}}{3}pi$



Is my calculation correct?



And if not, what is the right way to calculate it?






calculus integration multivariable-calculus manifolds spheres




calculus integration multivariable-calculus manifolds spheres






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 15 '18 at 17:05







ChikChak

















asked Dec 15 '18 at 16:06









ChikChakChikChak

764418




764418












  • $begingroup$
    Are you asking how to do the single variable integral that you've arrived at?
    $endgroup$
    – jgon
    Dec 15 '18 at 16:11










  • $begingroup$
    @jgon Edited... Is my final result correct?
    $endgroup$
    – ChikChak
    Dec 15 '18 at 16:15










  • $begingroup$
    Well the final result certainly can't be zero, so no, and you haven't done the single variable integral correctly, since that integral also must be positive.
    $endgroup$
    – jgon
    Dec 15 '18 at 16:16










  • $begingroup$
    @jgon Though, is the way up to the single variable integral correct?
    $endgroup$
    – ChikChak
    Dec 15 '18 at 16:19










  • $begingroup$
    Not sure, I didn't want to compute the determinant of the matrix.
    $endgroup$
    – jgon
    Dec 15 '18 at 16:20


















  • $begingroup$
    Are you asking how to do the single variable integral that you've arrived at?
    $endgroup$
    – jgon
    Dec 15 '18 at 16:11










  • $begingroup$
    @jgon Edited... Is my final result correct?
    $endgroup$
    – ChikChak
    Dec 15 '18 at 16:15










  • $begingroup$
    Well the final result certainly can't be zero, so no, and you haven't done the single variable integral correctly, since that integral also must be positive.
    $endgroup$
    – jgon
    Dec 15 '18 at 16:16










  • $begingroup$
    @jgon Though, is the way up to the single variable integral correct?
    $endgroup$
    – ChikChak
    Dec 15 '18 at 16:19










  • $begingroup$
    Not sure, I didn't want to compute the determinant of the matrix.
    $endgroup$
    – jgon
    Dec 15 '18 at 16:20
















$begingroup$
Are you asking how to do the single variable integral that you've arrived at?
$endgroup$
– jgon
Dec 15 '18 at 16:11




$begingroup$
Are you asking how to do the single variable integral that you've arrived at?
$endgroup$
– jgon
Dec 15 '18 at 16:11












$begingroup$
@jgon Edited... Is my final result correct?
$endgroup$
– ChikChak
Dec 15 '18 at 16:15




$begingroup$
@jgon Edited... Is my final result correct?
$endgroup$
– ChikChak
Dec 15 '18 at 16:15












$begingroup$
Well the final result certainly can't be zero, so no, and you haven't done the single variable integral correctly, since that integral also must be positive.
$endgroup$
– jgon
Dec 15 '18 at 16:16




$begingroup$
Well the final result certainly can't be zero, so no, and you haven't done the single variable integral correctly, since that integral also must be positive.
$endgroup$
– jgon
Dec 15 '18 at 16:16












$begingroup$
@jgon Though, is the way up to the single variable integral correct?
$endgroup$
– ChikChak
Dec 15 '18 at 16:19




$begingroup$
@jgon Though, is the way up to the single variable integral correct?
$endgroup$
– ChikChak
Dec 15 '18 at 16:19












$begingroup$
Not sure, I didn't want to compute the determinant of the matrix.
$endgroup$
– jgon
Dec 15 '18 at 16:20




$begingroup$
Not sure, I didn't want to compute the determinant of the matrix.
$endgroup$
– jgon
Dec 15 '18 at 16:20










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