Single Variable Calculus, Chapter 6, 6.1, Section 6.1, Problem 28

Sketch the region enclosed by the curves $\displaystyle y = 3x^2, y = 8x^2, 4x + y = 4$ and $x \geq 0$. Then find the area of the region.







Let $A_1, A_2$ and $A_3$ be the area in the left, middle and the right part respectively.

Notice that orientation of the curve changes at the point of intersection. Let $A_1$ and $A_2$ be the area in left part respectively. Thus,

$\displaystyle A_1 = \int^{x_2}_{x_1} (y_{\text{upper}} - y_{\text{lower}}) dx$

To determine the values of the upper and lower limit, we must get the point of intersection of $y = 8x^2$ and $4x + y = 4$ So..


$
\begin{equation}
\begin{aligned}

& 8x^2 = 4 - 4x
\\
\\
& 8x^2 + 4x - 4 = 0

\end{aligned}
\end{equation}
$


By applying Quadratic Formula,

$x = 0.5$ and $x = -1$

Since the functions are defined for $x \geq 0$, we have $x = 0.5$


$
\begin{equation}
\begin{aligned}

A_1 =& \int^{0.5}_0 [4 \cdot 4x - 8x^2] dx
\\
\\
A_1 =& \left[ 4x - \frac{4x^2}{2} - \frac{8x^3}{3} \right] ^{0.5}_0
\\
\\
A_1 =& \frac{7}{6} \text{ square units}

\end{aligned}
\end{equation}
$


For the middle part, notice that we can't use a vertical strip for the entire region, we must divide the region into two sub region. Let it be $A_{2A}$ and $A_{2B}$. By referring to the graph we can determine the upper unit of $A_{2A}$ from the point of intersection of $y = 8x^2$ and $4x + y = 4$ that is at $x = 0.5$. So..


$
\begin{equation}
\begin{aligned}

A_{2A} =& \int^{0.5}_0 (8x^2 - 3x^2) dx
\\
\\
A_{2A} =& \left[ \frac{5x^3}{3} \right] ^{0.5} _0
\\
\\
A_{2A} =& \frac{5}{24} \text{ square units}

\end{aligned}
\end{equation}
$



For the other sub region, we can determine the upper limit by getting the point of intersection of $y = 3x^2$ and $4x + y = 4$. So..


$
\begin{equation}
\begin{aligned}

& 3x^2 = 4 - 4x
\\
\\
& 3x^2 + 4x - 4 = 0

\end{aligned}
\end{equation}
$


by applying Quadratic Formula,

$\displaystyle x = \frac{2}{3}$ and $x = -2$

Since the function is defined on the interval $x \geq 0$, we have $\displaystyle x = \frac{2}{3}$

Then


$
\begin{equation}
\begin{aligned}

A_{2B} =& \int^{\frac{2}{3}}_0.5 [4 - 4x - 3x^2] dx
\\
\\
A_{2B} =& \left[ 4x - \frac{4x^2}{2} - \frac{3x^3}{3} \right]^{\frac{2}{3}}_{0.5}
\\
\\
A_{2B} =& \frac{23}{216} \text{ square units}

\end{aligned}
\end{equation}
$


Hence, the total area for the middle part is

$\displaystyle A_2 = A_{2A}+ A_{2B} = \frac{17}{54}$ square units

Lastly, for the right part, we can use a horizontal strip to be able to get the area without dividing the area into two sub region. So..

$\displaystyle A_3 = \int^{y_2}_{y_1} (x_{\text{right}}, x_{\text{left}}) dy$

To get the upper limit, we know that the point of intersection of the curve $y = 3x^2$ and $4x + y = 4$ is at $\displaystyle x = \frac{2}{3}$

So if $\displaystyle x = \frac{2}{3}$, then

$\displaystyle y = 3 \left( \frac{2}{3} \right)^2 = \frac{4}{3}$


$
\begin{equation}
\begin{aligned}

A_3 =& \int^{\frac{4}{3}}_0 \left( \frac{4 - 4}{4} - \sqrt{\frac{4}{3}} \right) dy
\\
\\
A_3 =& 0.5185 \text{ square units}

\end{aligned}
\end{equation}
$


Therefore, the total area of the entire region is..

$A_T = A_1 + A_2 + A_3 = 2$ units