Originally Posted by

**Marconis** We just finished up area between two curves and volume of a solid. I've been having some difficulties with it. This was a homework problem two nights ago, and I thought I understood it (even after doing it in class), but I don't.

The question tells you to find the area between:

$\displaystyle 4x+y^2=12$ and $\displaystyle x=y$

So I solve for x.

$\displaystyle 4x=-y^2+12$

$\displaystyle x=\frac{-y^2+12}{4}$

Now...in class, the interval of the area was shown as -6 to 2 on the y axis. How is that determined? I can get -6 and 2 for values of intersection of x, but not y. What am I missing here?

The answer was eventually shown as (before computation):

$\displaystyle A= \int^{2} [(y) - (\frac{-y^2+12}{4})]dy$

I forgot how to put the lower part of the integral in there, sorry.