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(1 point) A cone is to be built from stone so that it has height 10 feet and a base with radius 4. (The base of the cone will lie on the floor.) Suppose the density of the stone used to make the cone is 70 lb/ft3. Calculate the work required to build the cone from the ground up.

Respuesta :

Answer:

29260 lb.ft

Explanation:

Elemental Volume of the stone is

dV = πr²dh

r varies linearly from 4 at the base (h=0) to 0 at h=10

Using the concept of similar triangles,

(r/4) = (10-h/10)

r = 4 - 0.4h

dV = π(4 - 0.4h)² dh

dV = π (16 - 3.2h + 0.16h²) dh

dweight = ρdV = 70 dV

dweight = 70π (0.16h² - 3.2h + 16) dh

dWork = dweight x distance

dW = 70π (0.16h² - 3.2h + 16)h × dh)

dW = 70π (0.16h³ - 3.2h² + 16h) dh

dW = (70π) ∫¹⁰₀ (0.16h³ - 3.2h² + 16h) dh

W = (70π) [0.04h⁴ - 1.067h³ + 8h²]¹⁰₀

W = 29260 lb.ft

The work required to build the cone from the ground up is 29260 lb.ft

Calculation of the work required:

Since

Elemental Volume of the stone is

dV = πr²dh

here,

r varies linearly from 4 at the base (h=0) to 0 at h=10

Also, we used the similar triangles triangle.

(r/4) = (10-h/10)

r = 4 - 0.4h

Now

dV = π(4 - 0.4h)² dh

dV = π (16 - 3.2h + 0.16h²) dh

And,

dweight = ρdV = 70 dV

dweight = 70π (0.16h² - 3.2h + 16) dh

Now

dWork = dweight x distance

dW = 70π (0.16h² - 3.2h + 16)h × dh)

dW = 70π (0.16h³ - 3.2h² + 16h) dh

dW = (70π) ∫¹⁰₀ (0.16h³ - 3.2h² + 16h) dh

W = (70π) [0.04h⁴ - 1.067h³ + 8h²]¹⁰₀

W = 29260 lb.ft

learn more about density here; https://brainly.com/question/18997330

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