+0

0
112
3

What is the area of the region between y = 2x^2 and y = 12 - x^2?

A. 0

B. 16

C. 32

D. 48

Apr 25, 2019

#1
+1

Find the intersection points

2x^2  = 12 - x^2

3x^2  = 12

x^2  = 4

x  = - 2  and x  = 2

So we have, using symmetry

2                                              2                                                 2

2 * ∫  12 - x^2 - 2x^2 dx  =   2    ∫  12 - 3x^2  dx   =  2 [ 12x - x^3]      =  2 [ 12(2) - (2)^3 ]  =

0                                              0                                                 0

2 [ 24 - 8]  =

2 * 16  =

32 units^2   Apr 25, 2019
#2
0

I'm sorry, I'm just a bit confused. When I did it, I got 16. Where does the other 2 come from? (Also, thank you very much for your help!)

Guest Apr 25, 2019
#3
+2

We are using the fact that   both functions are symmetric about the y axis

So

2                                                          2

∫ 12 - 3x^2   dx  is  the same  as    2 * ∫  12 - 3x^2  dx

-2                                                         0   CPhill  Apr 25, 2019