Mayr won the Olympic gold metal in 2000 with the time of 9.87 his time was 0.93 seconds faster than jermeys who won in the 1900
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Step-by-step explanation:
First, graph the region. The first equation is x = 3y² − 2, which has a vertex at (-2,0). The second equation is x = y², which has a vertex at (0, 0). The two curves meet at the point (1, 1). The region should look kind of like a shark fin.
(a) Rotate the region about y = -1. Make vertical cuts and divide the volume into a stack of hollow disks (washers).
Between x=-2 and x=0, the outside radius of each washer is y₁ + 1, and the inside radius is 1. Between x=0 and x=1, the outside radius of each washer is y₁ + 1, and the inside radius is y₂ + 1.
The thickness of each washer is dx.
Solve for y in each equation:
y₁ = √(⅓(x + 2))
y₂ = √x
The volume is therefore:
∫₋₂⁰ {π[√(⅓(x+2)) + 1]² − π 1²} dx + ∫₀¹ {π[√(⅓(x+2)) + 1]² − π[√x + 1]²} dx
∫₋₂⁰ π[⅓(x+2) + 2√(⅓(x+2))] dx + ∫₀¹ π[⅓(x+2) + 2√(⅓(x+2)) − x − 2√x] dx
∫₋₂¹ π[⅓(x+2) + 2√(⅓(x+2))] dx − ∫₀¹ π(x + 2√x) dx
π[⅙(x+2)² + 4 (⅓(x+2))^(3/2)] |₋₂¹ − π[½x² + 4/3 x^(3/2)] |₀¹
π(3/2 + 4) − π(½ + 4/3)
11π/3
(b) This time, instead of slicing vertically, we'll divide the volume into concentric shells. The radius of each shell y + 1. The width of each shell is x₂ − x₁.
The thickness of each shell is dy.
The volume is therefore:
∫₀¹ 2π (y + 1) (x₂ − x₁) dy
∫₀¹ 2π (y + 1) (y² − (3y² − 2)) dy
∫₀¹ 2π (y + 1) (2 − 2y²) dy
4π ∫₀¹ (y + 1) (1 − y²) dy
4π ∫₀¹ (y − y³ + 1 − y²) dy
4π (½y² − ¼y⁴ + y − ⅓y³) |₀¹
4π (½ − ¼ + 1 − ⅓)
11π/3
As you can see, when given x = f(y) and a rotation axis of y = -1, it's easier to use shell method.
(c) Since we're given x = f(y), and the rotation axis is x = -4, we should use washer method.
Make horizontal slices and divide the volume into a stack of washers. The inside radius is 4 + x₁, and the outside radius is 4 + x₂.
The thickness of each washer is dy.
The volume is therefore:
∫₀¹ π [(4 + x₂)² − (4 + x₁)²] dy
∫₀¹ π [(4 + y²)² − (3y² + 2)²] dy
∫₀¹ π [(y⁴ + 8y² + 16) − (9y⁴ + 12y² + 4)] dy
∫₀¹ π (-8y⁴ − 4y² + 12) dy
-4π ∫₀¹ (2y⁴ + y² − 3) dy
-4π (⅖y⁵ + ⅓y³ − 3y) |₀¹
-4π (⅖ + ⅓ − 3)
136π/15
