Question:
A solid lies between planes perpendicular to the x-axis at x=0 and x=12. The cross-sections perpendicular to the axis on the interval 0≤x≤12 are squares with diagonals that run from the parabola y=-2√x to the parabola y=2√x. Find the volume of the solid.
Answer:
576
Step-by-step explanation:
Given:
Length of diagonal square:
[tex] D = 2\sqrt{x} - (-2\sqrt{x}) [/tex]
[tex] D = 4\sqrt{x} [/tex]
Here, the diagonal is the hypotenus of a right angle triangle, with leg S, where the square has a side of length S.
Using Pythagoras theorem:
[tex] S^2 + S^2 = D^2 [/tex]
[tex] S^2 + S^2 = (4\sqrt{x})^2 [/tex]
[tex] 2S^2 = 16x [/tex]
Divide both sides by 2
[tex] S^2 = 8x [/tex]
Thus,
Area, A = S² = 8x
Take differential volume, dx =
dV = Axdx
dV = 8xdx
Where limit of solid= 0≤x≤12
Volume of solid, V:
V =∫₀¹² dV
V = 8 ∫₀¹² xdx
V = [4x²]₀¹²
V = 4 (12)²
V = 12 * 144
= 576
Volume of solid = 576
