Question

Sketch the region enclosed by the given curves and find its area.

Sketch the region enclosed by the given curves.

\( 11. y = 12 - {x^2},\quad y = {x^2} - 6\)

Short answer

The resulting area is \(a = 72\)

Step-by-step solution

Let us define the following functions:

\(f(x) = {x^2} - 6\)

\(g(x) = 12 - {x^2}\)

Let us sketch the graph presenting those functions:

To find the area enclosed between those two functions, we need to calculate:

\(f(x) = g(x)\)

\({x^2} - 6 = 12 - {x^2}\)

\(2{x^2} = 18\)

\({x^2} = 9\)

Hence, either \(x = 3\)or\(x = - 3\)

Final Proof

The symmetry provided by the graphs will allow us to calculate one side of the area from\((0,3)\) and multiply it by \(2\)

\(a = 2\int_0^3 g (x) - f(x)dx\)

\(a = 2\int_0^3 1 2 - {x^2} - {x^2} + 6\)

\(a = 2\int_0^3 - 2{x^2} + 18dx\)

\(a = 2\left( {\frac{{ - 2}}{3}{x^3} + 18x} \right)_0^3\)

\(a = 2\left( {\frac{{ - 2}}{3} \times {3^3} + 18 \times 3 - 0} \right)\)

\(a = 2( - 18 + 54)\)

\(a = 72\)

The resulting area is \(a = 72\)