Question

If the area enclosed between the curves $y = a{x^2}$ and $x = a{y^2}$ is $1$ sq. unit, then what is the value of $a$ ?
A. $\dfrac{1}{{\sqrt 3 }}$
B. $\dfrac{1}{2}$
C. $1$
D. $\dfrac{1}{3}$

Answer 1

Hint:To solve the given question proceed by finding the intersection points of the two curves. Shade the area enclosed within the two curves above the x-axis and then, calculate its value using integration. Find the value of $a$ from the obtained equations.

Formula used: For two curves, $f(x)$ and $g(x)$ , the area enclosed within them, above x-axis, between two points $x = a$ and $x = b$ is given by: $A = \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]dx} $

Complete step by step Solution:
Given curves:
$y = a{x^2}$ … (1)
And
$x = a{y^2}$ … (2)
Area enclosed within these curves $ = 1$ sq. unit
Let us proceed with finding the intersection points of the two curves.
Squaring equation (1),
${\left( {a{x^2}} \right)^2} = {y^2}$
Substituting the values of ${y^2}$ from equation (2) in equation (1),
${a^3}{x^4} = x$
Simplifying further,
$x\left( {{a^3}{x^3} - 1} \right) = 0$
As two of the roots of the cubic equation will result in imaginary roots, therefore,
The equation gives the intersection points as: $x = 0$ and $x = \dfrac{1}{a}$ .
We know that for two curves, $f(x)$ and $g(x)$ , the area enclosed within them, above x-axis, between two points $x = p$ and $x = q$ is given by:
$A = \int_p^q {\left[ {f\left( x \right) - g\left( x \right)} \right]dx} $
For the given question,
$p = 0$
$q = \dfrac{1}{a}$
$f\left( x \right) = \sqrt {\dfrac{x}{a}} $ (Considering only the part of the curve above x-axis, as only that involves the shaded area)
$g\left( x \right) = a{x^2}$

Thus, the value of the area enclosed within the curve will be:
$A = \int\limits_0^{\dfrac{1}{a}} {\left( {\sqrt {\dfrac{x}{a}} - a{x^2}} \right)dx} $
Performing integration,
\[A = {\left[ {\frac{2}{3} \times \frac{{{x^{\frac{3}{2}}}}}{{\sqrt a }} - \frac{{a{x^3}}}{3}} \right]_0}^{\frac{1}{a}}\]
Putting the limits,
\[A = \left[ {\frac{2}{3} \times \frac{{{{\left( {\frac{1}{a}} \right)}^{\frac{3}{2}}}}}{{\sqrt a }} - \frac{{a{{\left( {\frac{1}{a}} \right)}^3}}}{3} - 0 + 0} \right]\]
Simplifying further,
$A = \left( {\dfrac{2}{{3{a^2}}} - \dfrac{1}{{3{a^2}}}} \right) = \dfrac{1}{{3{a^2}}}$
We are also given that $A = 1$ . Thus,
$\dfrac{1}{{3{a^2}}} = 1$
This gives: $a = \pm \sqrt {\dfrac{1}{3}} $
As $a > 0$ , therefore, $a = \sqrt {\dfrac{1}{3}} $ .

Therefore, the correct option is (A).

Note: While calculating the value of the area enclosed within the curves, we consider the equation of the curves. In the case of the above solution, for $a{y^2} = x$ , there are two values of $y$ for a corresponding value of $x$ . Yet, we considered only the positive one as that is the one contributing to the shaded area.