- #26

OlderDan

Science Advisor

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They are indeed the same.Hey I just did both integrals (mine and yours) and I get different answers. I have checked them and the same. Mine gives I = (8-3sqrt2)mc^2/(12sqrt2)

Yours is giving me a negative number which seems illogical. What is happening?

Your way:

[tex] I = 2\sigma \int_0^{c/\sqrt 2 } {x^2 \int_x^{c\sqrt 2 - x} {dydx} } [/tex]

[tex] I = 2\sigma \int_0^{c/\sqrt 2 } {x^2 dx} \left( {c\sqrt 2 - 2x} \right) [/tex]

[tex] I = 2\sigma \int_0^{c/\sqrt 2 } {\left( {c\sqrt 2 x^2 - 2x^3 } \right)dx} [/tex]

[tex] I = 2\sigma \left( {c\sqrt 2 \frac{{\left( {c/\sqrt 2 } \right)^3 }}{3} - \frac{{\left( {c/\sqrt 2 } \right)^4 }}{2}} \right) = 2\sigma \left( {\frac{2}{3} - \frac{1}{2}} \right)\left( {c/\sqrt 2 } \right)^4 [/tex]

[tex] I = 2\sigma \left( { - \frac{{\left( {c/\sqrt 2 } \right)^4 }}{2} + c\sqrt 2 \frac{{\left( {c/\sqrt 2 } \right)^3 }}{3}} \right) = \sigma \left( {\frac{1}{{12}}} \right)c^4 = \frac{M}{{12}}c^2 [/tex]

My way:

[tex] I = 4\sigma \int_0^{c/\sqrt 2 } {x^2 \int_x^{c/\sqrt 2 } {dydx} } [/tex]

[tex] I = 4\sigma \int_0^{c/\sqrt 2 } {\left( {c/\sqrt 2 x^2 - x^3 } \right)dx} [/tex]

[tex] I = 4\sigma \left( {\frac{{\left( {c/\sqrt 2 } \right)^4 }}{3} - \frac{{\left( {c/\sqrt 2 } \right)^4 }}{4}} \right) = 4\sigma \left( {\frac{1}{3} - \frac{1}{4}} \right)\left( {c/\sqrt 2 } \right)^4 = \sigma \left( {\frac{1}{{12}}} \right)c^4 = \left( {\frac{M}{{12}}} \right)c^2 [/tex]