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.
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.
3 answers
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
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
4 answers