Question

The area enclosed by the curve ${{y}^{2}}+{{x}^{4}}={{x}^{2}}$ is:

& (\text{A})\text{ }\dfrac{2}{3} \\\ & (B)\text{ }\dfrac{4}{3} \\\ & (C)\text{ }\dfrac{8}{3} \\\ & (D)\text{ }\dfrac{10}{3} \\\ \end{aligned}$$

Answer 1

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,

& {{y}^{2}}+{{x}^{4}}={{x}^{2}} \\\ & \Rightarrow {{y}^{2}}={{x}^{2}}-{{x}^{4}} \\\ & \Rightarrow {{y}^{2}}={{x}^{2}}(1-{{x}^{2}}) \\\ \end{aligned}$$ Squaring on both sides, we get $$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{aligned} & x=0.5 \\\ & \Rightarrow y=\sqrt{{{\left( 0.5 \right)}^{2}}\left( 1-{{(0.5)}^{2}} \right)} \\\ & \Rightarrow y=\pm 0.43 \\\ \end{aligned}$$ Similarly, $$\begin{aligned} & x=-0.5 \\\ & \Rightarrow y=\sqrt{{{\left( -0.5 \right)}^{2}}\left( 1-{{(-0.5)}^{2}} \right)} \\\ & \Rightarrow y=\pm 0.43 \\\ \end{aligned}$$ So, the points calculated are,  Use the above observations to plot the graph.  We can see that the area is symmetrical about x-axis and y-axis. So the curve can be divided into four parts. So we can now find the area by integrating the function from $x=0$to $x=1$. Then multiply it by four to get the total area under the curve. The formula of finding the area enclosed by $f(x)$ between $x=a$ and $x=b$ can be written as $=|\int\limits_{a}^{b}{f(x)dx|}$. So the required area under the curve is, $$\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$$ Now we will find the limits, when this value is substituted. When $\begin{aligned} & x=0 \\\ & \Rightarrow (1-{{0}^{2}})={{u}^{2}} \\\ & \Rightarrow u=1 \\\ \end{aligned}$ When $\begin{aligned} & x=1 \\\ & \Rightarrow (1-{{1}^{2}})={{u}^{2}} \\\ & \Rightarrow u=0 \\\ \end{aligned}$ Substituting these values in equation (i), we get $$\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.