Question

find the area of region enclosed between the two circles X square + Y square equals to 9 and x minus 3 whole square + Y square equals to 9​

Answer 1

Answer:

6∏ -9√3/2 Square units

Step-by-step explanation:

The two circle equations are given below.

x^2+y^2=9

(x-3)^2+y^2=9

The centers of above circles are (0,0) and (3,0). The are over lapped as shown in figure.  

Let us find the point on X axis, where the segments of each circle were attached to each other.  

Subtracting both equations we get

x^2+y^(2 )- (x-3)^2+y^2=0

6x  - 9 = 0

X = 3/2.  

The point on X axis is (3/2, 0)

Now the over lapped area is shown in shaded color.  

The over lapped area can be calculated by  

Integrating each circles equation on x – axis coordinates.  

Hence the area of shaded area =  

2(integration from 0 to 3/2 of circle2 equaion + integration of 3/2 to 3 of circle 1 equation)  (***following integration expression might seam in correct.*** but text is correct)

=2($\int\limits^0_3/2 {y} \, dx$  + $\int\limits^3/2_3 {y} \, dx$)

Substitute y = √(9-(x-3)²  for first y and y = √(9-x²) for second y.

Simplifying integration, we get

= 2 ((x -3)√(9 – (x-3)²)/2 + 9sin^(-1) ((x-3)/3)/2) (Boundaries 0 to 3/2) + 2(x√(9 – x²)/2 + 9sin^(-1)⁡(x/3)/2) boundaries  3/2 to 3)

= 2(-3√(9 – 9/4)/4 + 9sin^(-1) ((-1)/2)/2 –(1/2) - 9sin^(-1) (⁡-1)/2) + 2(- 9sin^(-1) (⁡1)/2 -3√(9 – 9/4)/4 - - 9sin^(-1) (-1/2)/2)

= 2(-3 * 3√3/8 - 9∏/12 + 9∏/4) + 2(9∏/4 -3 * 3√3/8 - 9∏/12)

Simplifying above equation, we get

=6∏ -9√3/2 Square units.