I am having difficulties understanding the set-up (of the integrals) in evaluating the area between these two curves given below.
Since the question asks for the area with respect to $x$ and $y$, I began by putting the equations in terms of $y = $ (since I will need that later.) So, we now have: $y = \sqrt{x-4}$ and $y = x + 2$.
Then, I proceeded to find the intersection point for the two curves.
$$4 - y^{2} = y - 2$$ $$6 - y^{2} - y = 0$$ $$y^{2} +y - 6 = 0$$ $$ (y+3) (y-2) = 0$$ $$ y = -3, 2$$
I plugged both values of y into the originals to find the x-coordinate. So, the intersection points are: $$ (-5,-3) (0,2)$$
For the first part of the question, why is the area between the curves not simply $$\int_{-5}^{0} x+2-(\sqrt{4-x}) dx $$
For the second part of the question (area with respect to $y$), why is the area given by $$\int_{-3}^{2} 4-y^{2} - (y-2)dy$$
I was taught that, for horizontal slices, the Area = $\int_a^{b} \operatorname{right function} - \operatorname{left function} dy.$ However, I am having a hard time visualizing why $4-y^{2}$ is the "right function" and $y-2$ is the "left function."