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Section 6-2 : Area Between Curves

10. Determine the area of the region bounded by \(x = {y^2} + 1\), \(x = 5\), \(y = - 3\) and \(y = 3\).

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Hint : It’s generally best to sketch the bounded region that we want to find the area of before starting the actual problem. Having the sketch of the graph will usually help with determining the right/left functions and the limits for the integral.
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Let’s start off with getting a sketch of the region we want to find the area of.

We are assuming that, at this point, you are capable of graphing most of the basic functions that we’re dealing with in these problems and so we won’t be showing any of the graphing work here.

Here is a sketch of the bounded region we want to find the area of.

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In the problem statement we were given two limits on \(y\). However, as seen in the sketch of the graph above the curves intersect in this region and the right/left functions differ depending on what range of \(y\)’s we are looking for.

Therefore, we’ll need to find the intersection points. However, we should never just assume that our graph is accurate or that we will be able to read it accurately enough to guess the coordinates from the graph. This is especially true when the intersection points of the two curves do not occur on an axis (as they don’t in this case).

So, to determine the intersection points correctly we’ll need to find them directly. The intersection points are where the two curves intersect and so all we need to do is set the two equations equal and solve. Doing this gives,

\[{y^2} + 1 = 5\hspace{0.25in} \to \hspace{0.25in}{y^2} = 4\hspace{0.25in} \to \hspace{0.25in}y = - 2,\,\,\,\,y = 2\]

Note that you may well have found the intersection points in the first step to help with the graph if you were graphing by hand which is not a bad idea with faced with graphing this kind of region.

So, from the graph then it looks like we’ll need three integrals since there are three ranges of \(x\) (\( - 3 \le x \le - 2\), \( - 2 \le x \le 2\) and \(2 \le x \le 3\)) for which the right/left functions are different.

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At this point there isn’t much to do other than step up the integrals and evaluate them.

We are assuming that you are comfortable with basic integration techniques so we’ll not be including any discussion of the actual integration process here and we will be skipping some of the intermediate steps.

The area is,

\[\begin{align*}A & = \int_{{ - 3}}^{{ - 2}}{{{y^2} + 1 - 5\,dy}} + \int_{{ - 2}}^{2}{{5 - \left( {{y^2} + 1} \right)\,dy}} + \int_{2}^{3}{{{y^2} + 1 - 5\,dy}}\\ & = \int_{{ - 3}}^{{ - 2}}{{{y^2} - 4\,dy}} + \int_{{ - 2}}^{2}{{4 - {y^2}\,dy}} + \int_{2}^{3}{{{y^2} - 4\,dy}}\\ & = \left. {\left( {\frac{1}{3}{y^3} - 4y} \right)} \right|_{ - 3}^{ - 2} + \left. {\left( {4y - \frac{1}{3}{y^3}} \right)} \right|_{ - 2}^{ 2} + \left. {\left( {\frac{1}{3}{y^3} - 4y} \right)} \right|_2^3\\ & = \frac{7}{3} + \frac{{32}}{3} + \frac{7}{3} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{46}}{3}}}\end{align*}\]