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Section 6-2 : Area Between Curves

2. Determine the area to the left of \(g\left( y \right) = 3 - {y^2}\) and to the right of \(x = - 1\).

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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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It should be clear from the graph that the right function is the parabola (i.e. \(x = 3 - {y^2}\)) and the left function is the line \(x = - 1\).

Since we weren’t given any limits on \(y\) in the problem statement we’ll need to get those. 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 limits from the graph. This is especially true when the intersection points of the two curves (i.e. the limits on \(y\) that we need) 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,

\[3 - {y^2} = - 1\hspace{0.25in} \to \hspace{0.25in}{y^2} = 4\hspace{0.25in} \to \hspace{0.25in}y = - 2,\,\,\,\,y = 2\]

So, the limits on \(y\) are : \( - 2 \le y \le 2\).

Show Step 3

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

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,

\[A = \int_{{ - 2}}^{2}{{3 - {y^2} - \left( { - 1} \right)\,dy}} = \int_{{ - 2}}^{2}{{4 - {y^2}dy}} = \left. {\left( {4y - \frac{1}{3}{y^3}} \right)} \right|_{ - 2}^2 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{32}}{3}}}\]