Respuesta :

It's not clear whether you're talking about the bounded region to the left or right of the y-axis, but due to symmetry it doesn't really matter.

The two curves y = 8 and y = 8x ² intersect for

8 = 8x ²   →   x ² = 1   →   x = -1, x = 1

but we only care about one of these; take the positive solution, so the region of interest lies to the right of the y-axis.

Each washer has an outer radius (distance from the "top" curve to the axis of revolution) of 8 and an inner radius (distance from the "bottom" curve) of 8x ². Take the heights to be some small length ∆x . Then each has a volume of π (outer radius)² - π (inner radius)², or

π (64 - 64x ⁴) ∆x

Now take n such washers, take their sum, and let n → ∞ and ∆x → 0. This amounts to computing the integral,

∫₀¹ π (64 - 64x ⁴) dx = π (64x - 64/5 x ⁵) |₀¹

… = π (64 - 64/5)

… = 256π/5

Ver imagen LammettHash

The two curves y = 8 and y = 8x ² intersect for

8 = 8x ²   →   x ² = 1   →   x = -1, x = 1

but we only care about one of these; take the positive solution, so the region of interest lies to the right of the y-axis.

Each washer has an outer radius (distance from the "top" curve to the axis of revolution) of 8 and an inner radius (distance from the "bottom" curve) of 8x ². Take the heights to be some small length ∆x . Then each has a volume of π (outer radius)² - π (inner radius)², or

π (64 - 64x ⁴) ∆x

Now take n such washers, take their sum, and let n → ∞ and ∆x → 0. This amounts to computing the integral,

∫₀¹ π (64 - 64x ⁴) dx = π (64x - 64/5 x ⁵) |₀¹

… = π (64 - 64/5)

… = 256π/5

THANKS