Friday

January 30, 2015

January 30, 2015

Posted by **Max** on Sunday, July 7, 2013 at 3:46pm.

- Calculus -
**Steve**, Sunday, July 7, 2013 at 4:04pm∫[0,6] ∫[0,√(36-x^2)] xy dy dx

∫[0,6] 1/2 xy^2 [0,√(36-x^2)] dx

∫[0,6] 1/2 x (36-x^2) dx

9x^2 - 1/8 x^4 [0,6]

162

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