Question

If the area bounded by the curves \( y = ax^2 \) and \( x = ay^2 \) (where \( a>0 \)) is 3 sq. units, then the value of \( a \) is:

• A. \( \frac{2}{3} \)
• B. \( \frac{1}{3} \)
• C. 1
• D. 4

Hint:

When calculating the area between curves, find the points of intersection first, then set up the integral with the difference of the functions. Solve for the unknown constant by using the given area.

Answer 1

The Correct Option is B

\[ The curves are given by \( y = ax^2 \) and \( x = ay^2 \). \] \[ When \( x=0 \), \( y=0 \). When \( x=\frac{1}{a} \), \( y=\frac{1}{a} \). \] \[ The points of intersection for \( y=ax^2 \) and \( x=ay^2 \) are \( (0,0) \) and \( \left(\frac{1}{a}, \frac{1}{a}\right) \). \] \[ The required area is given by: \] \[ A = \int_{x=a}^{x=b} [f_2(x) - f_1(x)] \, dx \] \[ Given that the area is 3, we have: \] \[ 3 = \int_{0}^{1/a} \left(\frac{\sqrt{x}}{\sqrt{a}} - ax^2\right) \, dx \] \[ Evaluating the integral: \] \[ 3 = \left[\frac{2}{3\sqrt{a}} x^{3/2} - \frac{ax^3}{3}\right]_0^{1/a} \] \[ Substituting the limits: \] \[ 3 = \frac{2}{3\sqrt{a}} \times \frac{1}{a\sqrt{a}} - \frac{a}{3} \times \frac{1}{a^3} \] \[ Simplifying the expression: \] \[ 3 = \frac{2}{3a^2} - \frac{1}{3a^2} \] \[ 3 = \frac{1}{3a^2} \] \[ Rearranging the equation: \] \[ 9a^2 = 1 \] \[ Solving for \( a^2 \): \] \[ a^2 = \frac{1}{9} \Rightarrow a = \frac{1}{3} \] \[ The value of \( a \) is \( \frac{1}{3} \). \]