Chapter 8.2, Problem 37E

To determine

To Find:

To find the area of the propeller shaped regions by curve $x-y^3=0$ and the line $x-y=0$ .

Answer to Problem 37E

The area of the propeller shaped regions by curve and line is $\frac{1}{2}$ .

Explanation of Solution

Given information:

The given function is $x-y^3=0$ and $x-y=0$ .

The red curve is the graph of $x-y^3=0$ and the blue line is the graph of $x-y=0$ .

  

Now we have to solve the equations for $x$ ,

  $\begin{array}{l}x=y^3\\ x=y\end{array}$

Now find the point of intersection by equating the above two equations,

  $\begin{array}{l}{y}^3=y\\ y^2=1\\ y=\pm 1\end{array}$

From the graph, the region below the x-axis has the same area as the region above the x-axis.

Hence, the integral runs from $0$ to $1$ and then multiplying by 2.

Therefore, the area of the propeller shaped regions are

  $\begin{array}{l}A=2{\displaystyle \underset{0}{\overset{1}{\int }}\left(y-y^3\right)dy}\\ A=2{\left[\frac{y^2}{2}-\frac{y^4}{4}\right]}_0^1\\ A=2\left[\frac{1}{2}-0-\frac{1}{4}+0\right]\\ A=2\left[\frac{1}{4}\right]\\ A=\frac{1}{2}\end{array}$

Therefore, the area of the propeller shaped regions by curve and line is $A=\frac{1}{2}$ .