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Let $$\mathcal{R}$$ be the circle centered at $$(0,0)$$ with radius $$10.$$ The lines $$x=6$$ and $$y=5$$ divide $$\mathcal{R}$$ into four regions $$\mathcal{R}_1$$$$\mathcal{R}_2$$$$\mathcal{R}_3$$, and $$\mathcal{R}_4$$. Let $$[\mathcal{R}_i]$$ denote the area of region $$\mathcal{R}_i$$. If

$$[\mathcal{R}_1] > [\mathcal{R}_2] > [\mathcal{R}_3] > [\mathcal{R}_4],$$

then find $$[\mathcal{R}_1] - [\mathcal{R}_2] - [\mathcal{R}_3] + [\mathcal{R}_4]$$.

I know there was already an answer to this question, but it was wrong. I've been trying to do this problem for ages, but I can't get anywhere. Any help is appreciated!

Oct 13, 2020

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By calculus,

$$[\mathcal{R}_1] = 30 + \frac{1}{4} \cdot \pi \cdot 10^2 + \int_0^5 \sqrt{100 - x^2} \ dx + \int_0^6 \sqrt{100 - x^2} \ dx.$$

We can write out the areas similarly, to get [R_1] - [R_2] - [R_3] + [R_4] = 80.

Oct 13, 2020