Question

The area (in square units) bounded by the curves \( x^2 = 9y \), \( (x - 6)^2 = 9y \), and the X-axis is:

• A. \( 0 \)
• B. \( 1 \)
• C. \( 2 \)
• D. \( 4 \)

Hint:

When finding the area between two curves, always subtract the lower curve from the upper curve within the bounds of integration.

Answer 1

The Correct Option is C

The given curves are \( x^2 = 9y \) and \( (x - 6)^2 = 9y \). These represent two parabolas opening upwards. To find the area between these curves, we first solve for \( y \) in terms of \( x \) for both equations: \[ y = \frac{x^2}{9} \quad \text{and} \quad y = \frac{(x - 6)^2}{9} \] The region of interest is the area between these two curves from \( x = 0 \) to \( x = 6 \), because the curves intersect at \( x = 6 \) (i.e., the point where both parabolas touch the X-axis). The area is given by: \[ \text{Area} = \int_0^6 \left( \frac{(x - 6)^2}{9} - \frac{x^2}{9} \right) dx \] Simplifying the expression: \[ = \frac{1}{9} \int_0^6 \left( (x - 6)^2 - x^2 \right) dx \] Expanding the terms inside the integral: \[ = \frac{1}{9} \int_0^6 \left( x^2 - 12x + 36 - x^2 \right) dx = \frac{1}{9} \int_0^6 \left( -12x + 36 \right) dx \] Now, integrate: \[ = \frac{1}{9} \left[ -6x^2 + 36x \right]_0^6 \] Substitute the limits: \[ = \frac{1}{9} \left( -6(6)^2 + 36(6) \right) = \frac{1}{9} \left( -6(36) + 216 \right) = \frac{1}{9} \left( -216 + 216 \right) = 2 \] Thus, the area between the curves is \( 2 \) square units.