Integration Application

Question

Use the slicing method to find the volume of a cone with radius 8 units and height 10 units. Use a sketch to assist in the explanation.

Answer 1

For the slicing method, we want to determine a consistent cross-sectional area we can use to integrate over some dimension of the volume. For a cone, we can use its circular cross-section. We know that the radius along the cone's height, starting from the tip, goes from r = 0 to r = 8. This sounds like something we could integrate over, but it doesn't account for the height of the cone.

We need some way to relate the radius, r, to the height.

If you can picture the cone's height to lie along an axis, the x-axis on a plot, and picture the cone sliced in half, you get a triangle that is mirrored over the x-axis, where r = 0 at the origin and r = 8 at x = 10. I've done my best to mock it up here: https://www.desmos.com/calculator/sv82aegsir

But the relationship between the radius and the height looks like a linear equation. Well, let's treat it like one - if radius is our "y" and height is our "x", we can use our old friend y = mx + b

where,

b = 0

and,

m = (y2-y1)/(x2-x) = (8-0)/(10-0) = 4/5

so,

y = 4/5 * x

or,

r = 4/5 * h

Great! So as we go along the cone's height, we can calculate the radius, and then use the radius to calculate the area of the cross-section at each point. To get volume, we take infinite cross-sections and add them together - we integrate!

So, V = ∫ A(h) dh [0, 10]

Or, V = ∫ (π r²) dh [0, 10] = ∫ (π (h * 4/5)²) dh [0, 10]

Moving out constants, we get:

V = π*16/25 (∫ h² dh [0, 10])

And integrating:

V = π*16/25 (h³ / 3) [0, 10] = π*16/25 (10³ / 3 - 0³ / 3)

Multiplying out:

V = π *16 * (1000 / 75) ≈ 670.21

To test, just plug in r = 8 and h = 10 into the volume equation for a cone:

V = π/3 r²h = π/3 (8)² (10) ≈ 670.21

I hope this was clear enough and helped clarify the approach as much as it answered the question.

Answer 2

let us assume the cone has radius of R and height of H, at distance h from the base, the cross section is a circle with radius of r. Based on the knowledge of similar triangles, we have

r/R = (H-h)/h,

therefore, r = R(H-h) / H.

area of the layer at h is s = πr², and the volume of the layer is dv = sdh = πr² dh

so the whole volume of the cone is:

V = ∫₀^H πr² dh

= ∫₀^H π (R(H-h)/H)² dh

= πR²/H² ∫₀^H (H-h)² dh

= πR²/(3H²) [h-H]³|₀^H

= (1/3)πR²H

plug in the value R and H you will get your answer.

Answer 3

Answer 4

As an engineer the way I'd do this is bh/3.

64pi/3=67.02