Regions Defined with Respect to y

Learning Outcomes

Determine the area of a region between two curves by integrating with respect to the dependent variable
Recognize when it is easier to integrate with respect to the independent or dependent variable

In the previous example, 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 the last example. 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 figure is has two graphs in the first quadrant. They are the functions v(y) and u(y). In between these graphs is a shaded region, bounded to the left by v(y) and to the right by u(y). The region is labeled R. The shaded area is between the horizontal boundaries of y=c and y=d.

Figure 8. We can find the area between the graphs of two functions, $u(y)$ and $v(y).$

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 $v({y}_{i}^{*})$ to $u({y}_{i}^{*}).$ Figure 9(a) shows the rectangles when ${y}_{i}^{*}$ is selected to be the lower endpoint of the interval and $n=10.$ Figure 9(b) shows a representative rectangle in detail.

This figure is has three graphs. The first figure has two curves. They are the functions v(y*) and u(y*). In between these curves is a horizontal rectangle. The second figure labeled “(a)”, is a shaded region, bounded to the left by v(y) and to the right by u(y). The shaded area is between the horizontal boundaries of y=c and y=d. This shaded area is broken into rectangles between the curves. The third figure, labeled “(b)”, is the two curves v(y*) and u(y*). In between the curves is a horizontal rectangle with width delta y.

Figure 9. (a) Approximating the area between the graphs of two functions, $u(y)$ and $v(y),$ with rectangles. (b) The area of a typical rectangle.

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

A \approx \underset{i = 1}{∑^{n}} [u (y_{i}^{*}) - v (y_{i}^{*})] Δ y .

$A\approx \underset{i=1}{\overset{n}{\text{∑}}}\left[u({y}_{i}^{*})-v({y}_{i}^{*})\right]\text{Δ}y.$

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

A = lim_{n \to \infty} \underset{i = 1}{∑^{n}} [u (y_{i}^{*}) - v (y_{i}^{*})] Δ y = \int_{c}^{d} [u (y) - v (y)] d y .

$A=\underset{n\to \infty }{\text{lim}}\underset{i=1}{\overset{n}{\text{∑}}}\left[u({y}_{i}^{*})-v({y}_{i}^{*})\right]\text{Δ}y={\displaystyle\int }_{c}^{d}\left[u(y)-v(y)\right]dy.$

These findings are summarized in the following theorem.

Finding the Area between Two Curves, Integrating along the $y$ -axis

Let $u(y)$ and $v(y)$ be continuous functions such that $u(y)\ge v(y)$ for all $y\in \left[c,d\right].$ Let $R$ denote the region bounded on the right by the graph of $u(y),$ on the left by the graph of $v(y),$ and above and below by the lines $y=d$ and $y=c,$ respectively. Then, the area of $R$ is given by

A = \int_{c}^{d} [u (y) - v (y)] d y .

$A={\displaystyle\int }_{c}^{d}\left[u(y)-v(y)\right]dy.$

Example: Integrating with Respect to $y$

Let’s revisit finding the area of a complex region, only this time let’s integrate with respect to $y.$ Let $R$ be the region depicted in Figure 10. Find the area of $R$ by integrating with respect to $y.$

This figure is has two graphs in the first quadrant. They are the functions f(x) = x^2 and g(x)= 2-x. In between these graphs is a shaded region, bounded to the left by f(x) and to the right by g(x). All of which is above the x-axis. The region is labeled R. The shaded area is between x=0 and x=2.

Figure 10. The area of region $R$ can be calculated using one integral only when the curves are treated as functions of $y.$

Show Solution

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 $(1,1),$ 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} A & = \int_{c}^{d} [u (y) - v (y)] d y \\ = \int_{0}^{1} [(2 - y) - \sqrt{y}] d y = {[2 y - \frac{y^{2}}{2} - \frac{2}{3} y^{3 / 2}] |}_{0}^{1} \\ = \frac{5}{6} . \end{array}

$\begin{array}{cc}\hfill A& ={\displaystyle\int }_{c}^{d}\left[u(y)-v(y)\right]dy\hfill \\ & ={\displaystyle\int }_{0}^{1}\left[(2-y)-\sqrt{y}\right]dy={\left[2y-\frac{{y}^{2}}{2}-\frac{2}{3}{y}^{3\text{/}2}\right]|}_{0}^{1}\hfill \\ & =\frac{5}{6}.\hfill \end{array}$

The area of the region is $\frac{5}{6}$ units².

Try It

Let’s revisit the checkpoint associated with finding the area of a complex region, only this time, let’s integrate with respect to $y$ . Let $R$ be the region depicted in the following figure. Find the area of $R$ by integrating with respect to $y.$

This figure is has two graphs in the first quadrant. They are the functions f(x) = squareroot of x and g(x)= 3/2 – x/2. In between these graphs is a shaded region, bounded to the left by f(x) and to the right by g(x). All of which is above the x-axis. The shaded area is between x=0 and x=3.

Figure 11.

Show Solution

$\frac{5}{3}$ units²

Hint

Follow the process from the previous example.

Watch the following video to see the worked solution to the above Try It.

Closed Captioning and Transcript Information for Video

Module 6: Applications of Integration