Evaluate:
∫π0∫x0√1−x2dydx
I've gotten it done to:
∫π0x√1−x2dx
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
∫10x√1−x2dx=∫π/20sinθcosθcosθdθ=−∫π/20cos2(θ)d(cosθ)=−13cos3(θ)|π/20=13.
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