# Find the area of the region enclosed between the curves x=y^2 - 12 and y=x.

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justaguide | Certified Educator

The area enclosed by the curves x = y^2 -12 and y=x has to be determined.

x = y^2 - 12

substitute x = y

=> y^2 - y - 12 = 0

=> y^2 - 4y + 3y - 12 = 0

=> y(y - 4) + 3(y - 4) = 0

=> y = -3 and y = 4

The points of intersection are (-3, -3) and (4, 4)

The graph of x = y^2 - 12 above and below the x-axis is symmetrical.

The area enclosed by the two curves is:

`int_(-12)^4 sqrt(x + 12) dx - (1/2)*4*4 + int_(-12)^-3 sqrt(x +12) dx + (1/2)*3*3`

=> `(sqrt(x+12)*(2*x+24))/3 |_(-12)^4 - 8 + (sqrt(x+12)*(2*x+24))/3|_(-12)^-3 + 4.5`

=> `(4*32)/3 - 8 + (3*18)/3 + 4.5`

`~~57.167`

**The area enclosed by the curves x=y^2 - 12 and y=x is approximately 57.167**

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