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

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

Answer 2

Step 1: Given Equations
The given equations of the curves are: \[ y = ax^2 \quad \text{and} \quad x = ay^2 \] where \( a > 0 \).
Step 2: Find the Intersection Points
To find the points of intersection, substitute \( y = ax^2 \) into \( x = ay^2 \): \[ x = a(ax^2)^2 = a^3 x^4 \] Rearranging the equation gives: \[ a^3 x^4 - x = 0 \] Factor out \( x \): \[ x(a^3 x^3 - 1) = 0 \] Thus, \( x = 0 \) or \( a^3 x^3 = 1 \). Solving \( a^3 x^3 = 1 \) gives: \[ x^3 = \frac{1}{a^3} \quad \Rightarrow \quad x = \frac{1}{a} \] So, the points of intersection are \( x = 0 \) and \( x = \frac{1}{a} \).
Step 3: Set up the Integral for Area
The area between the curves is given by: \[ \text{Area} = \int_0^{\frac{1}{a}} \left( y_{\text{top}} - y_{\text{bottom}} \right) dx \] Here, the top curve is \( y = ax^2 \), and the bottom curve is \( y = \frac{x}{a} \) (from \( x = ay^2 \)). Thus, the area is: \[ \text{Area} = \int_0^{\frac{1}{a}} \left( ax^2 - \frac{x}{a} \right) dx \]
Step 4: Compute the Integral
Now, compute the integral: \[ \int_0^{\frac{1}{a}} \left( ax^2 - \frac{x}{a} \right) dx \] First, split the integral: \[ \text{Area} = \int_0^{\frac{1}{a}} ax^2 \, dx - \int_0^{\frac{1}{a}} \frac{x}{a} \, dx \] Calculate each integral separately: 1. For \( \int_0^{\frac{1}{a}} ax^2 \, dx \): \[ \int_0^{\frac{1}{a}} ax^2 \, dx = a \cdot \left[ \frac{x^3}{3} \right]_0^{\frac{1}{a}} = a \cdot \frac{1}{3a^3} = \frac{1}{3a^2} \] 2. For \( \int_0^{\frac{1}{a}} \frac{x}{a} \, dx \): \[ \int_0^{\frac{1}{a}} \frac{x}{a} \, dx = \frac{1}{a} \cdot \left[ \frac{x^2}{2} \right]_0^{\frac{1}{a}} = \frac{1}{a} \cdot \frac{1}{2a^2} = \frac{1}{2a^3} \] Thus, the total area is: \[ \text{Area} = \frac{1}{3a^2} - \frac{1}{2a^3} \]
Step 5: Set the Area Equal to 3
We are given that the area is 3 square units: \[ \frac{1}{3a^2} - \frac{1}{2a^3} = 3 \] To solve for \( a \), first find a common denominator: \[ \frac{2}{6a^3} - \frac{3}{6a^3} = 3 \quad \Rightarrow \quad \frac{-1}{6a^3} = 3 \] Now, multiply both sides by \( -6a^3 \): \[ 1 = -18a^3 \quad \Rightarrow \quad a^3 = \frac{1}{18} \] Thus, \( a = \frac{1}{3} \).
Step 6: Conclusion
The value of \( a \) is: \[ \boxed{\frac{1}{3}} \]