+0  
 
0
1423
5
avatar+218 

Let R be the circle centered at (0,0) with radius  10. The lines x=6 and y=5 divide R into four regions R1, R2, R3 , and R4. Let  R_i denote the area of region R_i If R1>R2>R3>R4,
then find R1-R2-R3+R4

 

Thank you so much!!

 Oct 11, 2020
 #1
avatar
+1

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] = 92.

 Oct 11, 2020
 #2
avatar+218 
0

I don't think that that is correct. Thanks though! Can someone else help?

Noori  Oct 11, 2020
 #3
avatar
+1

https://web2.0calc.com/questions/help_98205#r3

 

urw laugh

 Oct 11, 2020
 #4
avatar+218 
0

thx!

Noori  Oct 11, 2020
 #5
avatar+118687 
+1

Use this to help

 

\(R_1=Pink,\quad R_2=Blue,\quad R_3=Green, \quad R_4=Purple\)

 

Area of circle is    100pi

Area of rectangle in the middle is 12*10=120

 

B+B+G+G+Pu-Pu+120 = 100pi

simply to get

B+G=50pi-60

 

Pink=G+B-Purple +120

Pink+ purple = G+B+120

substitute

Pink+ purple = 50pi-60+120 = 50pi+60

 

(Pink+Purple) - (Blue+green) = (50pi+60) - ( 50pi-60) = 120 units squared.

 

 

 

I really did not need to do all this.  

Just looking at the pic I can see that         pink - blue - green +Purple = 120

 

 Oct 12, 2020

2 Online Users

avatar