Chapter 8.2, Problem 4E

To determine

To Find: The area of the shaded region analytically

Answer to Problem 4E

The area of the shaded region $=\frac{4}{3}$

Explanation of Solution

Given information: The shaded region between $x=12y^2-12y^3$ and $x=2y^2-2y$

  

Concept used: If $f(x)$ and $g(x)$ are continuous with $f(x)\ge g(x)$ throughout $[a,b]$ , then the area between the curves $y=f(x)$ and $y=g(x)$ from a to b is the integral of $[f(x)-g(x)]$ from a to b

  $A={\displaystyle \underset{a}{\overset{b}{\int }}[f(x)-g(x)]}dx$

Calculation: The shaded region between $x=12y^2-12y^3$ and $x=2y^2-2y$

Area of the shaded region will be the integral of the line on the right minus the line on the left with the limits going from $0$ to $1$

  $\begin{array}{l}A={\displaystyle \underset{0}{\overset{1}{\int }}\left[\left(12y^2-12y^3\right)-\left(2y^2-2y\right)\right]}dy\\ A={\displaystyle \underset{0}{\overset{1}{\int }}\left[12y^2-12y^3-2y^2+2y\right]}dy\\ A={\displaystyle \underset{0}{\overset{1}{\int }}\left[-12y^3+10y^2+2y\right]}dy\\ A={\left[\frac{-12y^4}{4}+\frac{10y^3}{3}+\frac{2y^2}{2}\right]}_0^1\\ A=\frac{-12}{4}+\frac{10}{3}+1-0\\ A=\frac{4}{3}\end{array}$

The area of the shaded region $=\frac{4}{3}$