Find the area of the region bounded by the curves y=12-x^2 and y=x^2-6.

Hint:The answer should be a whole number.

a. always sketch the graph. I see two parabolas, symetric about the y axis.

b. then set up your low value of x, and the high value for your integration limies.

12-x^2=x^2-6
x=+-3

then the differential box

dArea= dx*(y1-y2)
y1 =12-x^2
and y2=x^2-6

area=INT (12-x^2-x^2+6)dx over limits
= INT (18-2x^2)dx over limits
= 18x -2/3 x^3 over limits.
you can apply limits (upper limit x=3, lower -3)

I get a whole number also.

Is 72 the correct answer?

First we have to find their intersection to establish the boundaries.

x^2 - 6 = 12 - x^2
2x^2 = 18
x^2 = 9
x = ± 3
so we have a closed region from -3 to +3

the height of the region for that domain
= 12-x^2 - (x^2 - 6) = 18 - 2x^2

area = [integral] (18-2x^2)dx from -3 to 3
= 18x - (2/3)x^3 | from -3 to +3
= (54 - 18) - (-54 - (-18)
= 36 - (-36)
= 72 square units

Here is a picture of our problem
http://www.wolframalpha.com/input/?i=plot+y%3D12-x%5E2+%2C+y%3Dx%5E2-6
notice , we could have used symmetry and doubled the area from 0 to 3

Well, I'll do my best to help you find the area, but my jokes might not be whole(l)some. Now, let's get down to business!

To find the area between these two curves, we need to find the points where they intersect. Equating the two equations, we get:

12 - x^2 = x^2 - 6

Combining like terms and rearranging, we have:

2x^2 = 18
x^2 = 9

Taking the square root of both sides, we find:

x = ±3

So the curves intersect at x = 3 and x = -3.

To find the area between the curves, we need to integrate the difference of the two functions from -3 to 3. The function that is on top changes at x = -3, so we need to split the integration into two parts.

The area can be calculated as:

Area = ∫(12 - x^2)dx - ∫(x^2 - 6)dx

Integrating these functions from -3 to 3 and simplifying, we get:

Area = [12x - (1/3)x^3] + [(1/3)x^3 - 6x]

Evaluating this expression from -3 to 3, we find:

Area = (72/3) + (72/3) = 48

So the area of the region bounded by the curves is 48 square units. And yes, it is a whole number!

To find the area of the region bounded by the given curves, we first need to find the x-values where these curves intersect. Setting the two equations equal to each other, we have:

12 - x^2 = x^2 - 6

Adding x^2 to both sides and subtracting 6 from both sides, we get:

2x^2 = 18

Dividing both sides by 2, we have:

x^2 = 9

Taking the square root of both sides, we get:

x = ±3

So the curves intersect at x = -3 and x = 3.

To find the area between the curves, we need to integrate the difference between the top curve (y = 12 - x^2) and the bottom curve (y = x^2 - 6) with respect to x over the interval [-3, 3].

The integral to find the area is given by:

A = ∫[a,b] (f(x) - g(x)) dx

where f(x) is the top curve and g(x) is the bottom curve.

In this case, A = ∫[-3, 3] (12 - x^2 - (x^2 - 6)) dx

Simplifying the integrand, we have:

A = ∫[-3, 3] (12 - 2x^2 + 6) dx

A = ∫[-3, 3] (18 - 2x^2) dx

To evaluate this integral, we can expand and solve it in two parts:

A = ∫[-3, 3] 18 dx - ∫[-3, 3] 2x^2 dx

Integrating each term, we get:

A = [18x] from -3 to 3 - [(2/3)x^3] from -3 to 3

Substituting the limits of integration, we have:

A = (18(3) - 18(-3)) - ((2/3)(3)^3 - (2/3)(-3)^3)

A = (54 + 54) - (18 - 18)

A = 108 - 0

A = 108 square units

Therefore, the area of the region bounded by the curves y = 12 - x^2 and y = x^2 - 6 is 108 square units, which is a whole number.