y = x² has a lower bound at the vertex (0, 0); y = 0
y = 9 has an upper bound at y = 9
x² = 9 ⇒ x = +/- 3 the region is bounded on the left at x = -3 and on the right at x = 3
Solve the integral from -3 to 3 to find the area under the curve, then multiply by 1/3 to calculate 1/3 of the area.
[tex]\int\limits^3_3 (9 - {x^{2}}) \, dx[/tex] = (9x - x³/3) | from -3 to 3
= (9(3) - 3³/3) - (9(-3) - (-3)³/3)
= (27 - 9) - (- 27 - (-9))
= 16 - (-16)
= 32
*** 32/3 is 1/3 of the area ****
Now that we know what 1/3 of the area is, we can find the height at y = a which will satisfy 1/3 of the area. We do this by solving the integral from "0 to a" and setting it equal to 1/3 of the area (32/3).
[tex]\int\limits^a_0 (9 - {x^{2}}) \, dx[/tex] - a(9 - a²)
= (9x - x³/3) | from 0 to a - (9a - a³)
= (9a - a³/3) - (9(0) - (0)³/3) - 9a + a³
= 9a - a³/3 - 9a + a³
= a³ - a³/3
= 2a³/3
32/3 = 2a³/3
32 = 2a³
16 = a³
∛16 = a
2∛2 = a (This is the distance from the upper bound (y = 9)
Answer: y = 9 - 2∛2
Follow the same steps using 2/3 of the area (64/3) to find "b".
64 = 2b³
32 = b³
∛32 = b 3.17
2∛4 = b (This is the distance from the upper bound (y = 9)
Answer: y = 9 - 2∛4
