Question

Let the area of the bounded region $ \{(x, y) : 0 \leq 9x \leq y^2, y \geq 3x - 6 \ be $ A $. Then  6A  is equal to:

Hint:

When dealing with regions defined by inequalities, solve for the boundaries and use definite integrals to compute the area between those boundaries. Remember to evaluate carefully.

Answer 1

Correct Answer: 81

1. Find the points of intersection. Intersection of \(9x = y^2\) and \(y = 3x - 6\): 
Substitute \(3x - 6\) for \(y\) in the first equation: \( 9x = (3x - 6)^2 \)
\( 9x = 9x^2 - 36x + 36 \)
\( 0 = 9x^2 - 45x + 36 \) 
\( 0 = x^2 - 5x + 4 \) 
\( 0 = (x - 1)(x - 4) \) 
So, \(x = 1\) or \(x = 4\). If \(x = 1\), \(y = 3(1) - 6 = -3\) If \(x = 4\), \(y = 3(4) - 6 = 6\) 
Therefore, the points of intersection are \((1, -3)\) and \((4, 6)\).
2. Express \(x\) in terms of \(y\) for both curves. From \(9x = y^2\), we get \(x = \frac{y^2}{9}\). From \(y = 3x - 6\), we get \(x = \frac{y + 6}{3}\).
3. Integrate to find the area.
The area \(A\) of the region is given by the integral: \[ A = \int_{-3}^{6} \left( \frac{y + 6}{3} - \frac{y^2}{9} \right) dy \] \[ A = \frac{1}{3} \int_{-3}^{6} (y + 6) dy - \frac{1}{9} \int_{-3}^{6} y^2 dy \] \[ A = \frac{1}{3} \left[ \frac{y^2}{2} + 6y \right]_{-3}^{6} - \frac{1}{9} \left[ \frac{y^3}{3} \right]_{-3}^{6} \] \[ A = \frac{1}{3} \left[ \left(\frac{36}{2} + 36\right) - \left(\frac{9}{2} - 18\right) \right] - \frac{1}{9} \left[ \left(\frac{216}{3}\right) - \left(\frac{-27}{3}\right) \right] \] \[ A = \frac{1}{3} \left[ 18 + 36 - \frac{9}{2} + 18 \right] - \frac{1}{9} \left[ 72 + 9 \right] \] \[ A = \frac{1}{3} \left[ 72 - \frac{9}{2} \right] - \frac{1}{9} \left[ 81 \right] \] \[ A = \frac{1}{3} \left[ \frac{144}{2} - \frac{9}{2} \right] - 9 \] \[ A = \frac{1}{3} \left[ \frac{135}{2} \right] - 9 \] \[ A = \frac{45}{2} - \frac{18}{2} \] \[ A = \frac{27}{2} \] 
4. Calculate \(6A\). \[ 6A = 6 \cdot \frac{27}{2} = 3 \cdot 27 = 81 \] Answer: \(6A\) is equal to 81.