Find the area of the region bounded by $y = 1$, $y = -1$, $x = y^2-2$, $x = e^y$ This is problem #3 in section 6.1. of the current edition of Stewart's calculus text.
So it's been a while since I've done Calc. II material. I am tempted to try double integrals with this question, but of course, double integrals are not covered in Calc. II. 
Here is my proposed solution:
$$\int\limits_{-1}^{1}\int\limits_{y^2-2}^{\exp(y)}\text{ d}x\text{ d}y = \int\limits_{-1}^{1}\left(e^{y}-y^{2}+2\right)\text{ d}y = \left.e^{y}-\dfrac{y^3}{3}+2y\right|^{y=1}_{y=-1} = e-\dfrac{1}{3}+2-e^{-1}-\dfrac{1}{3}+2=e-e^{-1}+4-\dfrac{2}{3} \approx 5.684\text{.}$$ 
According to Stewart, this answer is correct. I however, do not have the detailed solutions manual. Is there a way to do this problem WITHOUT double integration, in case I were to teach this to someone in Calc. II?
 A: Look to see if the domain is x-y convenient or y-x convenient (meaning that it is easier to integrate in respect to y or x depending on the amount of solutions for the relevant equations). We have $-1≤y≤1$ , and since the relevant equations are $ x = y^2 - 2$ and $ x = e^y$ we have less solutions for x than for y. This is y-x convenient, so now let us find which function is larger over this domain. We know that y^2 - 2 is a parabola open upwards with vertex at (0, -2) and e^y is always positive,  so e^y is larger over this domain. Now we integrate  $$\int\limits_{-1}^{1}(e^y - y^2 + 2)\text{ d}y$$ this results in the same numerical value you have arrived at. You do not need double integration for this problem, if you are not familiar with x-y or y-x convenient domains then I suggest you research a little bit -- Stewart's calculus book will not include this. X-y or y-x convenience is also very helpful in volume problems as well. 
A: Sure there is. I think the possible confusion might come from thinking that the functions are defined in terms of what we usually use as the dependant variable. 
Since we are calculating the area it doesn't matter if we do any rotation to them. How about considering the area defined by:
$x \in [-1,1]$ and bounded by the curves $f_1(x) = x^2 -2$ and $f_2(x) = e^x$ Then, since $f_2 > f_1$ in the given interval, the area would be:
$$\int_{-1}^{1} e^x dx - \int_{-1}^{1}( x^2 -2 )dx$$
EDIT: Corrected a couple of typos.
