Saturday, August 20, 2016

calculus - Evaluating intpi0intx0sqrt1x2:dydx



Evaluate:

π0x01x2dydx



I've gotten it done to:



π0x1x2dx



Should I now change to polar coordinates because of the pi, or how should I proceed?


Answer



I think you made a typo, the upper limit of the integral looks more like 1 instead of π (otherwise the square root is not well defined), I will give my answer based on this correction.







Try to make the change of variable x=sinθ, the dx=cosθdθ, and the integral becomes
10x1x2dx=π/20sinθcosθcosθdθ=π/20cos2(θ)d(cosθ)=13cos3(θ)|π/20=13.



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