Find the point where the area of a graph is split evenly I searched and found one other answer that was similar to my question, but it's still not enough detail for me to understand.
I need to find $a$ such that the line $x = a$ evenly divides the region bounded by the graphs of
$$y^2 = 64 − x$$ and $$x = 0.$$
I know that I must find the area first, here is what I got for the area:
$$A = 682.66$$
And now I must divide it in half, but I'm not clear on the next step.
I'm looking for guidance, not an answer, please help me out!
Thanks for reading!
 A: Draw a picture. We get a backward opening parabola, with axis the $x$-axis and vertex at $(64,0)$. Note the symmetry about the $x$-axis.. 
To find the area, we can integrate with respect to $x$ or with respect to $y$. With respect to $y$ is somewhat easier, but with respect to $x$ is natural for the second part, so that's what we will do.  
But I will take advantage of symmetry, because I always do. The area of the top half is
$$\int_0^{64}\sqrt{64-x}\,dx.$$
An antiderivative is $-\frac{2}{3}(64-x)^{3/2}$. Plug in the endpoints. We get $\frac{1024}{3}$. It is not useful to give a decimal form.  Remember, this is the area of the top half of our region.
We want to choose $a$ so that the area of the top half, up to $a$, is $\frac{512}{3}$. Draw a vertical line that you think splits the region into two equal parts. Note that $a$ should be closer to $0$ than to $64$.  We want
$$\int_0^a \sqrt{64-x}\,dx=\frac{512}{3}.$$
The integral is 
$$\frac{2}{3}\left(512-(64-a)^{3/2}\right).$$
Set this equal to $\frac{512}{3}$ and solve for $a$.  The equation simplifies to $(64-a)^{3/2}=256$, giving $a=64-32\sqrt[3]{2}$.
Remark: The arithmetic would have been somewhat simpler if we calculated the integral from $x=a$ to $x=64$, and set it equal to $\frac{512}{3}$, but we did it the slightly more awkward way because that's what most students would do.
Instead of calculating the area at the beginning, we could set
$$\int_0^a (64-x)^{1/2}\,dx=\int_a^{64}(64-x)^{1/2}\,dx$$
and solve for $a$. The arithmetic is about the same.
