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

\[ We\ are\ given\ the\ curves\ y = ax^2 \ and\ x = ay^2. \] \[ {When, } x=0 \Rightarrow y=0 { and } x=\frac{1}{a} \Rightarrow y=\frac{1}{a} \] \[ {Here, points of intersection of curves } y=ax^2 { and } x=ay^2 { are } (0,0) { and } \left(\frac{1}{a}, \frac{1}{a}\right) \] \[ \therefore { Required area } \] \[ A = \int_{x=a}^{x=b} [f_2(x) - f_1(x)] \, dx \] \[ 3 = \int_{0}^{1/a} \left(\frac{\sqrt{x}}{\sqrt{a}} - ax^2\right) \, dx \] \[ 3 = \left[\frac{2}{3\sqrt{a}} x^{3/2} - \frac{ax^3}{3}\right]_0^{1/a} \] \[ 3 = \frac{2}{3\sqrt{a}} \times \frac{1}{a\sqrt{a}} - \frac{a}{3} \times \frac{1}{a^3} \] \[ 3 = \frac{2}{3a^2} - \frac{1}{3a^2} \] \[ 3 = \frac{1}{3a^2} \] \[ 9a^2 = 1 \] \[ a^2 = \frac{1}{9} \Rightarrow a = \frac{1}{3} \] \[ Solving\ for\ a, \ we\ get\ a = \frac{1}{3}. \]