Figure

Finding the Area Between the Graphs of Two Functions

Find the area $A$ of the region enclosed by the graphs of $y=\sqrt{4-4x}, y=\sqrt{4-x}$, and the $x$-axis:

by partitioning the $x$-axis.
by partitioning the $y$-axis.

Solution (a) We begin by graphing the two equations and identifying the region whose area we seek. See Figure 11.

Figure 11 The area $A$ is the sum of the areas $A_{1}$ and $A_{2}$.

If we partition the $x$-axis, the area $A$ of the region we seek must be expressed as the sum of the two areas $A_{1}$ and $A_{2}$ marked in the figure. [Do you see why? The bottom graph changes at $x=1$ from $y=\sqrt{4-4x}$ to $y=0$ (the $x$-axis)].

Area $A_{1}$ is the region enclosed by the graphs of $y=\sqrt{4-x}$, and $y=\sqrt{4-4x}$, and the line $x=1$. Area $A_{2}$ is the region enclosed by the graph $y=\sqrt{4-x}$, the $x$-axis,

410

and the line $x=1$. Then \begin{eqnarray*} A &=&A_{1}+A_{2}=\int_{0}^{1}( \sqrt{4-x}-\sqrt{4-4x})~{\it dx}+\int_{1}^{4}\sqrt{4-x}~{\it dx} \\[4pt] &=&\int_{0}^{1}\sqrt{4-x}~{\it dx}-\int_{0}^{1}\sqrt{4-4x}~{\it dx}+\int_{1}^{4}\sqrt{4-x}~{\it dx} \\[4pt] &=&\int_{0}^{4}\sqrt{4-x}~{\it dx}-\int_{0}^{1}\sqrt{4-4x}~{dx}\!\qquad {\color{#0066A7}{\hbox{$\int_{0}^{1}\sqrt{ {4-x}}{~{dx}}+\int_{1}^{4}\sqrt{{4-x}}{~{dx}}=\int_{0}^{4}\sqrt{{4-x}}{~{dx}}$}}} \end{eqnarray*}

To find the first integral, we use the substitution $u=4-x$. Then ${\it du}=-{\it dx}$, and \begin{equation*} \int_{0}^{4}\sqrt{4-x}~{\it dx}=-\int_{4}^{0}u^{1/2}{\it du}=\int_{0}^{4}u^{1/2}~{\it du}= \left[ \dfrac{2}{3}u^{3/2}\right] _{0}^{4}=\frac{2}{3}( 8-0) = \frac{16}{3} \end{equation*}

For the other integral, we use the substitution $u=4-4x$. Then ${\it du}=-4\,{\it dx}$, or equivalently, ${\it dx}=-\dfrac{{\it du}}{4}$, and \begin{equation*} \int_{0}^{1}\sqrt{4-4x}~{\it dx}=-\frac{1}{4}\int_{4}^{0}u^{1/2}~{\it du}=\dfrac{1}{4} \int_{0}^{4}u^{1/2}~{\it du}=\dfrac{1}{4}\left[ \dfrac{2}{3}u^{3/2}\right] _{0}^{4}= \frac{1}{6}( 8-0) =\frac{4}{3} \end{equation*}

The area $A=\dfrac{16}{3}-\dfrac{4}{3}=4$ square units.

Figure 12

(b) Figure 12 shows the graphs of $y=\sqrt{4-4x}$ and $y=\sqrt{4-x}$ and the region whose area $A$ we seek. Since the graphs of $y=\sqrt{4-4x}$ and $y=\sqrt{4-x}$ satisfy the Horizontal-line Test for $0\leq y\leq 2$, we can express $y=\sqrt{4-4x}$ as a function $x=f(y)$ and $y=\sqrt{4-x}$ as a function $x=g(y)$. To find $x=f(y)$, we solve $y=\sqrt{4-4x}$ for $x$, where $x\geq 0$: \begin{eqnarray*} y&=&\sqrt{~4-4x} \\[4pt] y^{2}&=&4-4x \\[4pt] x&=&\dfrac{4-y^{2}}{4} \\[4pt] x&=&f(y)=1-\dfrac{y^{2}}{4} \end{eqnarray*}

To express $y=\sqrt{4-x}$ as a function $x=g(y)$, we solve for $x$, where $x\geq 0$: \begin{eqnarray*} y &=&\sqrt{~4-x} \\[4pt] y^{2} &=&4-x \\[4pt] x &=&g(y)=4-y^{2} \end{eqnarray*}

The graph of $x=g(y)=4-y^{2}$ is to the right of the graph of $x=f(y)=1-\dfrac{y^{2}}{4}, 0\leq y \leq 2$. So, $g(y)\geq f(y)$. Then \begin{eqnarray*} A&=&\int_{0}^{2}[ g(y)-f(y)]\, {\it dy}=\int_{0}^{2}\left[ (4-y^{2})-\left( 1-\dfrac{y^{2}}{4}\right) \right] {\it dy}=\int_{0}^{2}\left( 3-\dfrac{3y^{2}}{4}\right) {\it dy}\\[4pt] &=&\left[ 3y-\dfrac{y^{3}}{4}\right] _{0}^{2}=6-\dfrac{8}{4}=4\hbox{ square units} \end{eqnarray*}