Sum of double integrals

Calculus Level 5

n = 1 R ( y x 2 + y 2 ) n d x d y \large \sum_{n=1}^\infty \iint_{R} \left(\frac y{\sqrt{x^2+y^2}} \right)^n \, dx dy

Given that R = { ( x , y ) R 2 : y x , x 2 + y 2 2 , y 0 } R= \{ (x,y) \in \mathbb R^2 : y \leq x, x^2 + y^2 \leq 2, y \geq 0 \} .


The answer is 0.628.

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1 solution

Otto Bretscher
Feb 21, 2016

Nice problem! I will have to assign it in one of my classes some time...

In polar coordinates, the double integral is 0 π / 4 0 2 sin n ( θ ) r d r d θ = 0 π / 4 sin n ( θ ) d θ \int_{0}^{\pi/4}\int_{0}^{\sqrt{2}}\sin^n(\theta)rdrd\theta=\int_{0}^{\pi/4}\sin^n(\theta)d\theta . Summing the geometric series of sine powers, we find 0 π / 4 sin ( θ ) 1 sin ( θ ) d θ 0.628 \int_{0}^{\pi/4}\frac{\sin(\theta)}{1-\sin(\theta)}d\theta\approx \boxed{0.628}

Thank you Sir. Have a wonderfull day.

Jose Sacramento - 5 years, 3 months ago

Sir I can you give complete solution.I dont know how to use polar co-ordinate in double integral

Kushal Bose - 4 years, 8 months ago

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