Introduction

Welcome back! Like we promised last time, today’s topic is finding the area between curves, but with functions of $y$ instead of functions of $x$. The method for this is identical to the method for finding the area using functions of $x$, with the only difference being that the higher and lower function is determined by their distance from the $y$-axis instead of the $x$-axis. The main challenge with this topic is that you may be given functions of $x$ that have to be turned into functions of $y$. However, this can be done with some simple algebra. 

Changing Variables

So why do we change our variables? Although most of the time finding the area by integrating with respect to either $x$ or $y$ results in the same answer, there might be situations where integrating with respect to one variable is simpler than integrating for the other variable. Let’s go over one example problem that deals with changing the variable of the function, and then we’ll jump right into the practice section. 

Example Problem

Changing Variables

First, we’ll use basic operations to switch the functions to functions of $y$.

$y=\sqrt{x}$

$x=y^2$

$y=x-2$

$x=y+2$

Finding Bounds

Now that we have our equations in terms of $y$, we’ll set them equal to each other and solve to find the bounds of the integral for area.

$y^2=y+2$

$y^2-y-2=0$

$(y-2)(y+1)=0$

$y=-1,2$

Therefore, the bounds of our integral are $-1$ and $2$.

Checking top-bottom

Now that we have our bounds, we’re going to check which function is farther from the y-axis on the interval $-1\le y\le 2$. To do this, we’ll check $y=0$.

$x=(0{)}^2$

$x=0$

$x=(0)+2$

$x=2$

Therefore, $x=y+2$ is higher on the interval $-1\le y\le 2$

Plugging in

Now, we can take our integral and find the area.

$A={\int }_{-1}^2(y+2-y^2)dy$

$A=[\frac{y^2}{2}+2y-\frac{y^3}{3}{]}_{-1}^2$

$A=[\frac{(2{)}^2}{2}+2(2)-\frac{(2{)}^3}{3}]-[\frac{(-1{)}^2}{2}+2(-1)-\frac{(-1{)}^3}{3}]$

$A=[2+4-\frac{8}{3}]-[\frac{1}{2}-2-\frac{-1}{3}]$

$A=\left[\frac{18}{3}-\frac{8}{3}\right]-\left[\frac{-9}{6}-\frac{-2}{6}\right]$

$A=\frac{20}{6}+\frac{7}{6}$

$A=\frac{9}{2}$

Therefore, the area between the curves generated by the functions $y=\sqrt{x}$ and $y=x-2$ is $\frac{9}{2}$ or $4.5$.

Now that we’ve covered an example of how to change the variable of a function, let’s jump into the practice section.

Practice

Practice Answers

1)

First, we’ll find the bounds by setting the functions equal to each other.

$y^2=4y$

$y^2-4y=0$

$y(y-4)=0$

$y=0,4$

Therefore, our bounds are 0 and 4. Next, we’ll see which function is farther from the y-axis by testing $y=2$

$(2{)}^2=4$

$4(2)=8$

Therefore $x=4y$ is farther from the y-axis on the interval $0\le y\le 4$. Finally, we can plug in the bounds and the values into our integral and solve.

$A={\int }_0^4(4y-y^2)dy$

$A={\left[2y^2-\frac{y^3}{3}\right]}_0^4$

$A=\left[2(4{)}^2-\frac{(4{)}^3}{3}\right]-\left[2(0{)}^2-\frac{(0{)}^3}{3}\right]$

$A=\left[32-\frac{64}{3}\right]-[0-0]$

$A=\frac{128}{3}-\frac{64}{3}$

$A=\frac{64}{3}$

Therefore, the area of the region enclosed by the curves $x=y^2$ and $x=4y$ is $\frac{64}{3}$.

2)

First, we’ll have to change the variable of our functions.

Doing so gives us $x=\sqrt{y}$ and $x=\frac{y+3}{4}$. Now, we’ll set them equal to each other to get our bounds. Since the region is bounded by the x-axis, our lowest bound will automatically be 0.

$\sqrt{y}=\frac{y+3}{4}$

$4y=y^2+6y+9$

$y^2-2y+9=0$

Using the quadratic formula, we get y = 1 for our second bound. Now, let’s see which is farther from the y-axis on the interval $0\le y\le 1$. Let’s test $y=\frac{1}{4}$.

$\sqrt{\frac{1}{4}}=\frac{1}{2}$

$\frac{\frac{1}{4}+3}{4}=\left(\frac{13}{4}\right)\left(\frac{1}{4}\right)$

$=\frac{13}{16}$

Therefore, $x=\frac{y+3}{4}$ is farther from the y-axis on the interval $0\le y\le 1$. Now, let’s plug in our bounds and our functions into the formula and solve.

$A={\int }_0^1(\frac{1}{4}(y+3)-\sqrt{y})dy$

$A={\int }_0^1(\frac{1}{4}(y+3)dy-{\int }_0^1(\sqrt{y})dy$

$A=\frac{1}{4}{\left[\frac{y^2}{2}+3y\right]}_0^1-{\left[\frac{2y^{\frac{3}{2}}}{3}\right]}_0^1$

$A=\frac{1}{4}\left(\left[\frac{(1{)}^2}{2}+3(1)\right]-\left[\frac{(0{)}^2}{2}+3(0)\right]\right)-\left(\left[\frac{2(1{)}^{\frac{3}{2}}}{3}\right]-\left[\frac{2(0{)}^{\frac{3}{2}}}{3}\right]\right)$

$A=\frac{1}{4}\left(\left[\frac{1}{2}+3\right]-\left[\frac{2(1)}{3}\right]\right)$

$A=\frac{1}{4}\left(\frac{7}{2}\right)-\frac{16}{24}$

$A=\frac{7}{8}-\frac{16}{24}$

$A=\frac{21}{24}-\frac{16}{24}$

$A=\frac{5}{24}$

Therefore, the area of the region enclosed by the curves $y=x^2$ and $y=4x-3$, as well as the x-axis is $\frac{5}{24}$.

3)

First, we’ll find the bounds.

$\sqrt{y}=2-y$

$y=4-4y+y^2$

$0=4-5y+y^2$

$(y-4)(y-1)=0$

$y=1,4$

Therefore, our bounds 1 and 4. Next, we’ll test $y=3$ to see which function is farther from the y-axis on the interval $1\le y\le 4$.

$\sqrt{y}$

$\sqrt{3}$

$2-y$

$2-3$

$-1$

Therefore, the function $x=\sqrt{y}$ is farther from the y-axis on the interval $1\le y\le 4$. Lastly, let’s plug everything into our integral and solve.

$A={\int }_1^4(\sqrt{y}-2+y)dy$

$A={\left[\frac{2y^{\frac{3}{2}}}{3}-2y+\frac{y^2}{2}\right]}_1^4$

$A=\left[\frac{2(4{)}^{\frac{3}{2}}}{3}-2(4)+\frac{(4{)}^2}{2}\right]-\left[\frac{2(1{)}^{\frac{3}{2}}}{3}-2(1)+\frac{(1{)}^2}{2}\right]$

$A=\left[\frac{2(8)}{3}-8+8\right]-\left[\frac{2(1)}{3}-2+\frac{1}{2}\right]$

$A=\left[\frac{16}{3}\right]-\left[\frac{4}{6}-\frac{12}{6}+\frac{3}{6}\right]$

$A=\left[\frac{16}{3}\right]-\left[\frac{-5}{6}\right]$

$A=\frac{21}{3}$

$A=7$

Therefore, the area of the region enclosed by the curves $x=\sqrt{y}$ and $x=2-y$ is $7$.