Question

The area enclosed by the curve \[{{y}^{2}}+{{x}^{4}}={{x}^{2}}\] is:
 \[\begin{align}
  & (\text{A})\text{ }\dfrac{2}{3} \\
 & (B)\text{ }\dfrac{4}{3} \\
 & (C)\text{ }\dfrac{8}{3} \\
 & (D)\text{ }\dfrac{10}{3} \\
\end{align}\]

Answer 1

Hint: First, try to make a rough sketch. Then find out the symmetry. Then find the area under the curve using the formula; Finding the area enclosed by $f(x)$ between $x=a$ and $x=b$ can be written as $=|\int\limits_{a}^{b}{f(x)dx|}$.

Complete step by step answer:
Consider the given curve,
\[{{y}^{2}}+{{x}^{4}}={{x}^{2}}\]
Put \[y=-y\]and note that the expression remains unchanged. So the curve is symmetric about the x-axis.
Similarly, the curve is symmetric about the y-axis.
So, we can write the curve as,
\[\begin{align}
  & {{y}^{2}}+{{x}^{4}}={{x}^{2}} \\
 & \Rightarrow {{y}^{2}}={{x}^{2}}-{{x}^{4}} \\
 & \Rightarrow {{y}^{2}}={{x}^{2}}(1-{{x}^{2}}) \\
\end{align}\]
Squaring on both sides, we get
\[y=x\sqrt{1-{{x}^{2}}}\]
So, first of all let’s try plotting the curve considering various values,
Note that at\[x=0,y=0\]
And at \[y=0,\left( 1-x \right)\left( 1+x \right)=0\]or \[x=1,-1\]
We will consider one more point,
\[\begin{align}
  & x=0.5 \\
 & \Rightarrow y=\sqrt{{{\left( 0.5 \right)}^{2}}\left( 1-{{(0.5)}^{2}} \right)} \\
 & \Rightarrow y=\pm 0.43 \\
\end{align}\]
Similarly,
\[\begin{align}
  & x=-0.5 \\
 & \Rightarrow y=\sqrt{{{\left( -0.5 \right)}^{2}}\left( 1-{{(-0.5)}^{2}} \right)} \\
 & \Rightarrow y=\pm 0.43 \\
\end{align}\]
So, the points calculated are,

Use the above observations to plot the graph.

We can see that the area is symmetrical about x-axis and y-axis. So the curve can be divided into four parts. So we can now find the area by integrating the function from $x=0$to $x=1$. Then multiply it by four to get the total area under the curve.
The formula of finding the area enclosed by $f(x)$ between $x=a$ and $x=b$ can be written as $=|\int\limits_{a}^{b}{f(x)dx|}$.
So the required area under the curve is,
\[\text{Area}=\int\limits_{0}^{1}{x\sqrt{1-{{x}^{2}}}dx}.........(i)\]
Put \[1-{{x}^{2}}={{u}^{2}}\]
Differentiating on both sides, we get
\[\Rightarrow -2x\text{ }dx=2u\text{ }du\]
\[\Rightarrow x\text{ }dx=-u\text{ }du\]
Now we will find the limits, when this value is substituted.
When
 $\begin{align}
  & x=0 \\
 & \Rightarrow (1-{{0}^{2}})={{u}^{2}} \\
 & \Rightarrow u=1 \\
\end{align}$
When
$\begin{align}
  & x=1 \\
 & \Rightarrow (1-{{1}^{2}})={{u}^{2}} \\
 & \Rightarrow u=0 \\
\end{align}$
Substituting these values in equation (i), we get
\[\text{Area}=\int\limits_{1}^{0}{\sqrt{{{u}^{2}}}\left( -udu \right)}\]
\[\text{Area }=-\int\limits_{1}^{0}{{{u}^{2}}\text{ }du}\]
On integrating, we get
\[Area=-\left[ \dfrac{{{u}^{3}}}{3} \right]_{1}^{0}\]
Applying the limits, we get
\[Area=-\left[ \dfrac{{{0}^{3}}}{3}-\dfrac{{{1}^{3}}}{3} \right]\]
\[Area=\dfrac{1}{3}\text{ sq}\text{. units}\]
Note that the above area is for the first quadrant. As we need the entire area enclosed by the curve, so multiply the area by \[4\], we get
\[\text{Total Area}=4\times \dfrac{1}{3}\text{=}\dfrac{4}{3}\text{ sq}\text{. units}\]

So, the correct answer is “$\dfrac{4}{3}$”.

Note: The possibility of mistake is that the student might not multiply the result by \[4\]. One more possibility is the students forget to change the limits when substituting with ‘u’.
Instead of the is\[\text{Area }=-\int\limits_{1}^{0}{{{u}^{2}}\text{ }du}\], they will write \[\text{Area }=-\int\limits_{0}^{1}{{{u}^{2}}\text{ }du}\]. They will get a negative answer.