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: $\int_S|y||z|^2d\sigma_3$.

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: $\int_Mf dS = \int\int\int f\circ \phi \sqrt{det(D_\phi*D_\phi^T)}drd\alpha d\beta$

Putting it all together I get that $\int_S|y||z|^2d\sigma_3 = \int_0^{2\pi}\int_0^{2\pi}\int_0^1|rsin(\alpha)|r^2sin(\beta)^2\sqrt{2}r^2drd\alpha d\beta = \frac{\sqrt{2}}{6}\pi\int_0^{2\pi}|sin(\alpha)|d\alpha = \frac{\sqrt{2}}{6}\pi(\int_0^\pi sin(\alpha) d\alpha - \int_\pi^{2\pi} sin(\alpha) d\alpha) = \frac{4\sqrt{2}}{6}\pi = \frac{2\sqrt{2}}{3}\pi$

Is my calculation correct?

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

  • $\begingroup$ 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

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