Section 6.2 : Area Between Curves
5. Determine the area of the region bounded by \(x = 3 + {y^2}\), \(x = 2 - {y^2}\), \(y = 1\) and \(y = - 2\).
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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.
It should be clear from the graph that the right function is \(x = 3 + {y^2}\) and the left function is \(x = 2 - {y^2}\).
Next, we were given limits on \(y\) in the problem statement and we can see that the two curves do not intersect in that range. Note that this is something that we can’t always guarantee and so we need the graph to verify if the curves intersect or not. We should never just assume that because limits on \(y\) were given in the problem statement that the curves will not intersect anywhere between the given limits.
So, because the curves do not intersect we will be able to find the area with a single integral using the limits : \( - 2 \le y \le 1\).
Show Step 3At 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,
\[A = \int_{{ - 2}}^{1}{{3 + {y^2} - \left( {2 - {y^2}} \right)\,dy}} = \int_{{ - 2}}^{1}{{1 + 2{y^2}\,dy}} = \left. {\left( {y + \frac{2}{3}{y^3}} \right)} \right|_{ - 2}^1 = \require{bbox} \bbox[2pt,border:1px solid black]{9}\]