I am having difficulties understanding the set-up (of the integrals) in evaluating the area between these two curves given below.

Sketch of problem

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."


(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.)

  • $\begingroup$ Great explanation, +1! One thing I did not understand however was your last line: $$dA=(8-y-y^2)\,dy\;.$$ Where did you get the $8?$ Once expanded, shouldn't it be $$(6 - y^{2} - y) dy ?$$ $\endgroup$ – Joe Apr 13 '12 at 23:35
  • 1
    $\begingroup$ @Jay: It certainly should; fixed. Thanks. Apparently my fingers decided to multiply instead of add. $\endgroup$ – Brian M. Scott Apr 13 '12 at 23:40

For the area by vertical slices, the first part of the question, the slices do not always extend from $\sqrt{4-x}$ to $x+2$. They do from $x=-5$ to $x=0$, but then from $x=0$ to $x=4$ the vertical slice is between the two arms of the parabola. You need to split the integral into two pieces and integrate each one separately. I don't know where your limits of $0$ to $5$ come from. You should have $\int_{-5}^{0} x+2-(\sqrt{4-x}) dx + \int_0^4 (\sqrt{4-x}-(-\sqrt{4-x}))dx$

For the second, you want the horizontal distance between the curves. In this case the lower (left-hand) curve is always the straight line and the higher (right hand) curve is always the parabola. Your drawing captures this nicely and your integral is correct. The "small piece" of area is the rectangle from the straight line to the parabola of height $dy$

  • $\begingroup$ Ah, my upper and lower limits on 1) were simply a typo - sorry (I have now fixed this.) Nonetheless, nice explanation; +1. Thanks. $\endgroup$ – Joe Apr 13 '12 at 23:37

I'm guessing for the first one you mean $\int_{-5}^0x+2-\sqrt{4-x^2}dx$. I see you have an illustration of the 2 curves and the area between them. There are 2 problems with this.

  1. The area extends past the y-axis x=0.

  2. The lower boundary is not $\sqrt{4-x^2}$.

First we'll correct problem number 2. You'll notice that the values of y here are negative. The lower curve is not $\sqrt{4-x^2}$, it's $-\sqrt{4-x^2}$. So the area to the left of the y axis is $\int_{-5}^0x+2+\sqrt{4-x^2}dx$.

Now to correct the first problem. The issue here is the top curve is no longer $x+2$, it's half of the parabola. So this area is $\int_0^4\sqrt{4-x^2}-(-\sqrt{4-x^2})dx=2\int_0^4 \sqrt{4-x^2}dx$.

This makes the total area

$\int_{-5}^0x+2+\sqrt{4-x^2}dx+2\int_0^4 \sqrt{4-x^2}dx$

As for the 2nd question, draw a horizontal line. Or in this case, you can simply look at the x axis. At y=0, one function is -2 and the other is 4. The one with the highest x value (the parabola) is the "rightmost".


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