+0  
 
0
24
1
avatar

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
 #1
avatar
0

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

35 Online Users

avatar