Let the area of the bounded region {(x,y):0≤ 9x≤ y²,y≥ 3x-6} be A. Then 6A is equal to _________.

Solution

Maths sir · 4 steps

Read the inequalities as left of the parabola, right of the y-axis, above the line, then slice horizontally.

1. Read the region. 0 ≤ 9x ≤ y² means 0 ≤ x ≤ y²/9: right of the y-axis and left of the parabola x = y²/9, not inside it. y ≥ 3x - 6 means x ≤ (y+6)/3, the side of the line that contains the origin.
2. Find the bounded piece. At y = 0 the parabola meets the y-axis, so the region pinches to a single point there. For y > 0 it runs upward forever, so the bounded part is the piece with y < 0. The line meets x = 0 at y = -6, and y²/9 = (y+6)/3 gives y² - 3y - 18 = 0, so the line meets the parabola at y = -3 (and at y = 6, which lies in the unbounded part).
3. Pick the right boundary. Use horizontal strips from x = 0 to the nearer curve. For -6 ≤ y ≤ -3 the line is nearer, and for -3 ≤ y ≤ 0 the parabola is nearer.

   A = ∫₋₆⁻³ (y+6)/3 dy + ∫₋₃⁰ y²/9 dy = [(y+6)²/6]₋₆⁻³ + [y³/27]₋₃⁰ = 3/2 + 1

4. So. A = 5/2, and 6A = 15.

The common slip: 81. Taking the area enclosed between the line and the parabola from y = -3 to y = 6 gives A = 27/2 and 6A = 81. But 9x ≤ y² means left of the parabola, not inside it.

The idea to keep. Turn each inequality into a side of a curve before integrating. If x is given in terms of y, slice horizontally and switch the boundary wherever the curves cross.

Answer from the NTA final answer key. Solution by PrepWiser.