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."
Best Answer
(1) When you integrate with respect to $x$, you’re chopping the region into vertical slices. Look at the shaded region in your picture. First of all, $x$ ranges from $-5$ to $4$, so if you could do the calculation as a single integral, it would be $\int_{-5}^4\text{ something }dx$. But when $x$ is between $-5$ and $0$, vertical slices from from $y=x+2$ at the top down to $y=-\sqrt{x-4}$ at the bottom, while for $x$ between $0$ and $4$ they run from $\sqrt{x-4}$ at the top down to $-\sqrt{x-4}$ at the bottom. In other words, for $-5\le x\le 0$ the slice at $x$ has length
$$\begin{align*}\text{top}-\text{bottom}&=(x+2)-(-\sqrt{x-4})\\ &=x+2+\sqrt{x-4}\;, \end{align*}$$
but for $0\le x\le 4$ it has length $$\begin{align*}\text{top}-\text{bottom}&=\sqrt{x-4}-(-\sqrt{x-4})\\ &=2\sqrt{x-4}\;. \end{align*}$$
Since these are not the same, you have to break the calculation into two parts: the area is
$$\int_{-5}^0(x+2+\sqrt{4-x})\,dx+\int_0^4 2\sqrt{4-x}\,dx\;.$$
Notice that I can’t simply say that $y=\sqrt{4-x}$: $y$ is $\sqrt{4-x}$ for points on the upper branch of the parabola, but for points on the lower branch $y=-\sqrt{4-x}$.
(2) Again, look at the shaded region: when you cut a horizontal slice across it, the slice starts on the left at a point on the straight line and ends at a point on the right on the parabola. For a given $y$, the $x$-coordinate on the parabola is given by $x=4-y^2$, so that’s the $x$-coordinate at the righthand end of the slice; the $x$-coordinate on the straight line is given by $x=y-2$, so that’s the $x$-coordinate at the lefthand end of the slice. The length of the slice is therefore $$\text{right}-\text{left}=(4-y^2)-(y-2)=6-y-y^2\;,$$ and the infinitesimal bit of area contributed by it is its length times its width $dy$:
$$dA=(6-y-y^2)\,dy\;.$$ (For this one you do have the correct limits of integration.)