1. ## Area between Curves

I want to find the area between the curves $\displaystyle y=12-x^2$ and $\displaystyle y=x^2-6$. If you graph these functions, you can see they intersect at $\displaystyle (-3,3)$ and $\displaystyle (3,3)$. The graph is also symmetric about the y-axis, so I will find the area by finding the area of the left side, and mulitplying it by two. I will chop the left hand side, into two areas $\displaystyle A_1$ and $\displaystyle A_2$ where $\displaystyle A_1=\int_{3}^{12}\sqrt{12-y}dy$ and $\displaystyle A_2=\int_{-6}^{3}\sqrt{y+6}dy$. The total area will be:

$\displaystyle A=2(A_1+A_2)$

$\displaystyle =2\left(\int_{3}^{12}\sqrt{12-y}dy+\int_{-6}^{3}\sqrt{y+6}dy\right)$

Am I setting this up right?

I want to find the area between the curves $\displaystyle y=12-x^2$ and $\displaystyle y=x^2-6$. If you graph these functions, you can see they intersect at $\displaystyle (-3,3)$ and $\displaystyle (3,3)$. The graph is also symmetric about the y-axis, so I will find the area by finding the area of the left side, and mulitplying it by two. I will chop the left hand side, into two areas $\displaystyle A_1$ and $\displaystyle A_2$ where $\displaystyle A_1=\int_{3}^{12}\sqrt{12-y}dy$ and $\displaystyle A_2=\int_{-6}^{3}\sqrt{y+6}dy$. The total area will be:
$\displaystyle A=2(A_1+A_2)$
$\displaystyle =2\left(\int_{3}^{12}\sqrt{12-y}dy+\int_{-6}^{3}\sqrt{y+6}dy\right)$
$\displaystyle A = 2 \int_0^3 (12-x^2) - (x^2-6) \, dx$