The problem I am working on is, "In Exercises 17 and 18, find the area of the region by integrating (a) with respect to and x (b) with respect to y."

The two functions: $g(y)=4-y^2$, and $f(y)=y-2$

I was able to solve part (b), but (a) proved to be a bit more complicated.

Here is my work, for part (a).

So, the region bounded by the two functions is found by determining where the intersect:


Rewriting the the functions in terms of x being the independent variable:

$g(x)=\pm\sqrt{4-x}$ and $f(x)=x+2$

The interval of integration being, $[-5,0]$; but I have a feeling this is wrong. Another part I felt insecure about is what to do with the $\pm$.

I would appreciate the help.


The first step in any problem like this is to draw a picture. Then you can find the limits a bit easier... Here's a plot: http://www.wolframalpha.com/input/?i=plot{y%3Dx%2B2%2C+x%3D4-y^2}

(Note: You rewrote $f(y)$ incorrectly... it should change into $f(x) = x + 2$)

This problem is really just trying to emphasize how integrating w.r.t. y or x can make a problem a lot harder or easier.

Basically, when integrating w.r.t. x, you're going to have to split the region in two: Region one ($R_1$) is from x=-5 to x=0, and region two ($R_2$) from x=0 to x=4.

This is because your "top" curve changes at x=0. On $R_1$, your area is bounded above by y=x+2, below by $y=-\sqrt{4-x}$. On $R_2$, you are bounded above by $y=\sqrt{4-x}$, below by $y=-\sqrt{4-x}$.

Thus, your area is:

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

Simplifies to:

$$\int_{-5}^0 [(x+2)+\sqrt{4-x})]dx + \int_0^42\sqrt{4-x}dx$$

EDIT: To determine analytically, without graph.

It is possible, but I wouldn't recommend it, as it could (and typically does) involve more work. There is also more of a chance of making a mistake.

First, find all intersections of the functions. In this case, there are three functions: $f_1(x)=x+2$, $f_2(x)=\sqrt{4-x}$, $f_3(x)=-\sqrt{4-x}$. Now find the intersections of these graphs, which are: $$(-5, -3),(0, 2),(4, 0)$$ This splits the x axis into five regions: $$(-\infty, -3), (-3, 0), (0, 4), (4, \infty)$$ (Note: the above are not points, but rather ranges of x. I have made the endpoints excluded really because it doesn't matter so much...)

Now, determine the top and bottom curves for each region. This means you must determine, for example, the function with the greatest y value for the region $(-\infty, -3)$. Do this for all ranges. This basically involves a lot of manipulating inequalities.

Now you know your upper and lower limits for each of those regions. Now you need to know what regions you need to find the area of. Do this by a similar process as above for the y-axis.

Now you know your limits of integration as well as your integrands.

I think you will agree that drawing a graph is a better approach... :P

  • $\begingroup$ beat me by 5 seconds. :) $\endgroup$ – Mike Nov 25 '12 at 22:43
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    $\begingroup$ Don't you just hate it when that happens??? I've been waiting around on this site for a while now just so I could finally answer a question first today... and then amWhy beats me. :P On the bright side, our answers match. :) $\endgroup$ – apnorton Nov 25 '12 at 22:45
  • $\begingroup$ I am curious: is there a way to solve this purely analytically, without the aid of the graph? For instance, is there an analytically way of showing that the vertex of the parabola is a bound for the region? $\endgroup$ – Mack Nov 26 '12 at 14:24
  • $\begingroup$ I highly recommend against it. I've updated my answer to outline how one could go about doing so, but I think that process is very much error prone. Sketching a graph isn't that much work, anyway. When you get to multivariable calculus, too, you'd much rather sketch a graph than manipulate inequalities... $\endgroup$ – apnorton Nov 26 '12 at 23:49

I think you want $g(x) = \pm \sqrt{4 - x} \text{ and}\; f(x) = x + 2$.

Yes, you will need to consider both positive values for $g(x)$ and negative values:

That is, you need to consider both $g(x) = \sqrt{4-x}\;$ AND $\;g(x) = -\sqrt{4 - x}$).

Have you tried graphing the functions? That will help you check your intervals of integration.

$\quad\quad\quad\quad$ enter image description here

This helps confirm your doubts about the bounds of integration.

On the interval $-5 \le x \leq 0:\; f(x) \ge g(x)$, i.e., $x+2 \ge -\sqrt{4-x}$

But on the interval $0 \le x \le 4:\; \sqrt{4 - x} \ge -\sqrt{4 - x}$

  • $\begingroup$ Thank you very much for your help. $\endgroup$ – Mack Nov 26 '12 at 14:22

It might have helped if you tried graphing the 2 functions. One is a leftward facing parabola with vertex at $(4,0)$. The other, of course, is a line of slope $1$ with a $y$-intercept at $2$. I believe where $0<x<5$, the topmost function is the line $y=x+2$ (if $x=f(y)=y-2$, then $y=x+2$, not $x-2$) and the bottommost function is $y=-\sqrt{4-x}$.

But this is not the complete area as the parabola extends past the $y$-axis. The bottommost function is still $y=-\sqrt{4-x}$, but the topmost function is now $y=\sqrt{4-x}$. This makes the area you're looking for

$$\int\limits_{-5}^0x+2-(-\sqrt{4-x})dx+\int\limits_0^4\sqrt{4-x}-(-\sqrt{4-x})dx=$$ $$\int\limits_{-5}^0x+2+\sqrt{4-x}dx+\int\limits_0^42\sqrt{4-x}dx$$

  • $\begingroup$ Thank you very much for your help. $\endgroup$ – Mack Nov 26 '12 at 14:23

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