6.1 Areas between curves  (Page 3/4)

Consider the region depicted in the following figure. Find the area of $R.$

$\frac{5}{3}$ units ²

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Regions defined with respect to y

In [link] , we had to evaluate two separate integrals to calculate the area of the region. However, there is another approach that requires only one integral. What if we treat the curves as functions of $y,$ instead of as functions of $x?$ Review [link] . Note that the left graph, shown in red, is represented by the function $y=f(x)=x^2.$ We could just as easily solve this for $x$ and represent the curve by the function $x=v(y)=\sqrt{y}.$ (Note that $x=\text{−}\sqrt{y}$ is also a valid representation of the function $y=f(x)=x^2$ as a function of $y.$ However, based on the graph, it is clear we are interested in the positive square root.) Similarly, the right graph is represented by the function $y=g(x)=2-x,$ but could just as easily be represented by the function $x=u(y)=2-y.$ When the graphs are represented as functions of $y,$ we see the region is bounded on the left by the graph of one function and on the right by the graph of the other function. Therefore, if we integrate with respect to $y,$ we need to evaluate one integral only. Let’s develop a formula for this type of integration.

Let $u(y)$ and $v(y)$ be continuous functions over an interval $\left[c,d\right]$ such that $u(y)\ge v(y)$ for all $y\in \left[c,d\right].$ We want to find the area between the graphs of the functions, as shown in the following figure.

This time, we are going to partition the interval on the $y\text{-axis}$ and use horizontal rectangles to approximate the area between the functions. So, for $i=0,1,2\text{,…},n,$ let $Q=\left\{y_i\right\}$ be a regular partition of $\left[c,d\right].$ Then, for $i=1,2\text{,…},n,$ choose a point $y_i^{*}\in \left[y_{i-1},y_i\right],$ then over each interval $\left[y_{i-1},y_i\right]$ construct a rectangle that extends horizontally from $v\left(y_i^{*}\right)$ to $u\left(y_i^{*}\right).$ [link] (a) shows the rectangles when $y_i^{*}$ is selected to be the lower endpoint of the interval and $n=10.$ [link] (b) shows a representative rectangle in detail.

The height of each individual rectangle is $\text{Δ}y$ and the width of each rectangle is $u\left(y_i^{*}\right)-v\left(y_i^{*}\right).$ Therefore, the area between the curves is approximately

This is a Riemann sum, so we take the limit as $n\to \infty ,$ obtaining

These findings are summarized in the following theorem.

Integrating with respect to y

Let’s revisit [link] , only this time let’s integrate with respect to $y.$ Let $R$ be the region depicted in [link] . Find the area of $R$ by integrating with respect to $y.$

We must first express the graphs as functions of $y.$ As we saw at the beginning of this section, the curve on the left can be represented by the function $x=v(y)=\sqrt{y},$ and the curve on the right can be represented by the function $x=u(y)=2-y.$

Now we have to determine the limits of integration. The region is bounded below by the x -axis, so the lower limit of integration is $y=0.$ The upper limit of integration is determined by the point where the two graphs intersect, which is the point $\left(1,1\right),$ so the upper limit of integration is $y=1.$ Thus, we have $\left[c,d\right]=\left[0,1\right].$

Calculating the area of the region, we get

$\begin{array}{cc}\hfill A& ={{\displaystyle \int }}_c^d\left[u(y)-v(y)\right]dy\hfill \\ & ={{\displaystyle \int }}_0^1\left[\left(2-y\right)-\sqrt{y}\right]dy={\left[2y-\frac{y^2}{2}-\frac{2}{3}{y}^{3\text{/}2}\right]|}_0^1\hfill \\ & =\frac{5}{6}.\hfill \end{array}$

The area of the region is $5\text{/}6$ units ² .

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