Area of a region between curves and a point. The question is to find the area between the curve  $y=x^3$, the straight lines $y=8$, $y=-8$ and the $y$-axis.
I either get $8$ or $24$ not sure which one is correct.
Any help is much appreciated.
 A: You have undoubtedly drawn a picture. Note that the region we want to find the area of consists of two equal parts, in the first and third quadrant. We find the first quadrant area and double.
We want the area of the region below $y=8$ and above $y=x^3$, and to the right of the $y$-axis.  
The line $y=8$ meets the curve at $(2,8)$. So half our area is
$$\int_0^2(8-x^3)\,dx.$$
Calculate and double. We get $24$.
Remark: Let us find the area of the region below $y=x^3$, above the $x$-axis, from $x=0$ to $x=2$.  This turns out to be $4$, so if we double we get the $8$ you had obtained along with $24$. The $8$ is the correct answer to an area problem, but not to the area problem that was asked.
We can however use the $4$, by noting that our first quadrant region is a $2\times 8$ rectangle, with the region below $y=x^3$ removed. 
A: The curves intersect at $x=-2$ and $x=2$ but the area is symetric about the y-axis so we can multiply the integral of the difference in functions from $0\leq x\leq2$ by $2$.
$A=2*\int \limits_{0}^{2} (top function)-(lower function)dx$
$= 2*\int \limits_{0}^{2} 8-x^3dx$
$=2*(8x-x^4/x)|_0^2$
$=2*(16-4)=24$
