+0  
 
0
31
1
avatar+1791 

Help I'm really stuck on this

 

Find the area of the region enclosed by the graph of the equation $x^2-14x+3y+70=15+9y-y^2$ that lies below the line $y=x-3$.

 Aug 27, 2023
 #1
avatar+128732 
+1

x^2-14x+3y+70=15+9y-y^2      simplify as

 

x^2 -14x + y^2 -6y  =  -55         complete the square on x, y

 

x^2 - 14x + 49  + y^2  - 6y + 9  =  -55 + 49 + 9

 

(x - 7)^2  + ( y - 3)^2  =  3

 

This is a  circle with a center of ( 7, 3)  and a  radius of sqrt 3

 

 

The total area is  formed by a  triangle  and  an arc in  the circle  of 360° - 131.81°  = 228.19°

 

The area of the  triangle  = (1/2) (sqrt 3)^2 * sin (131.81°) ≈  1.118

 

The area of the  arc is  pi (sqrt 3)^2  (228.19 / 360)  ≈ 5.974

 

The total area ≈  1.118 + 5.974 ≈  7.092

 

 

cool cool cool

 Aug 27, 2023

1 Online Users