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Jason Quinn (talk | contribs) →Elementary example: trying to improve example |
Jason Quinn (talk | contribs) →Weight 0 example: +clarity fixes |
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:<math> \int_0^2 \! \int_{0}^{\pi/2} \! \int_0^2 \! \bar{f}(r,t,h) \, dh \, dt \, dr = 12 + 10 \pi</math>.<ref>[http://www.wolframalpha.com/input/?i=int_0^2+int_0^%28pi%2F2%29+int_0^2+%282+r+cos%28t%29+%2B+r+cos%28t%29+%2B+5%29+dh+dt+dr]</ref>
They are not equal. The integral of temperature is not independent of the coordinate
system used. It is non-physical in that sense, hence "strange". Note that if the integral of <math>\bar{f}</math> included a factor of the Jacobian (which is just <math>r</math>),
we get
:<math> \int_0^2 \! \int_{0}^{\pi/2} \! \int_0^2 \! \bar{f}(r,t,h) r \, dh \, dt \, dr = 16 + 10 \pi</math>,<ref>[http://www.wolframalpha.com/input/?i=int_0^2+int_0^%28pi%2F2%29+int_0^2+%282+r+cos%28t%29+%2B+r+cos%28t%29+%2B+5%29+r^2+dh+dt+dr]</ref>
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