Find the area of the region R, enclosed by:
x+2y=0 and x+y^2=3
First let's see where they intersect...
Subtracting the second equation from the first gives
$\displaystyle \displaystyle \begin{align*} x + y^2 - (x + 2y) &= 3 - 0 \\ y^2 - 2y &= 3 \\ y^2 - 2y - 3 &= 0 \\ (y + 1)(y - 3) &= 0 \\ y &= -1 \textrm{ or }y = 3 \end{align*} $
and the corresponding $\displaystyle \displaystyle \begin{align*} x \end{align*} $ values are $\displaystyle \displaystyle \begin{align*} 2\end{align*} $ and $\displaystyle \displaystyle \begin{align*} -6 \end{align*} $ respectively.
Now, if we decide to use horizontal strips, they are bounded on the left by the line $\displaystyle \displaystyle \begin{align*} x = -2y \end{align*} $ and they are bounded on the right by the curve $\displaystyle \displaystyle \begin{align*} x = 3 - y^2 \end{align*} $. When we sum all these strips, they are bounded below by $\displaystyle \displaystyle \begin{align*} y = -1 \end{align*} $ and bounded above by $\displaystyle \displaystyle \begin{align*} y = 3 \end{align*} $.
So the double integral is
$\displaystyle \displaystyle \begin{align*} \int_{-1}^3{\int_{-2y}^{3 - y^2}{\,dx}\,dy} &= \int_{-1}^3{\left[x\right]_{-2y}^{3-y^2}\,dy} \\ &= \int_{-1}^3{\left(3 - y^2\right) - (-2y)\,dy} \\ &= \int_{-1}^3{3 + 2y - y^2\,dy} \\ &= \left[3y + y^2 - \frac{y^3}{3}\right]_{-1}^3 \\ &= \left[3(3) + 3^2 - \frac{3^3}{3}\right] - \left[3(-1) + (-1)^2 - \frac{(-1)^3}{3}\right] \\ &= \left(9 + 9 - 9\right) - \left(-3 + 1 + \frac{1}{3}\right) \\ &= 9 - \left(-\frac{5}{3}\right) \\ &= \frac{32}{3} \end{align*} $
Now why not try using vertical strips to see if you get the same answer?