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zlopas [31]
3 years ago
14

The volume V of a sphere with radius R is given by formula V=4/3 x 3.14 x r^3

Mathematics
1 answer:
Olin [163]3 years ago
5 0
36(3.14) = 4/3 x 3.14 x r^3
113.04 = 4.187(r^3)
113.04/4.187 = r^3
26.997 = r^3
3√26.997 = r
r = ~3

Check: 3^3(\pi)(4/3) = 27(3.14)(4/3) = 113.04
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(a) See the attached sketch. Each shell will have a radius <em>y</em> chosen from the interval [2, 4], a height of <em>x</em> = 2/<em>y</em>, and thickness ∆<em>y</em>. For infinitely many shells, we have ∆<em>y</em> converging to 0, and each super-thin shell contributes an infinitesimal volume of

2<em>π</em> (radius)² (height) = 4<em>πy</em>

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\displaystyle 4\pi \int_2^4 y\,\mathrm dy = 2\pi y^2\bigg|_{y=2}^{y=4} = 2\pi (4^2-2^2) = \boxed{24\pi}

(b) See the other attached sketch. (The text is a bit cluttered, but hopefully you'll understand what is drawn.) Each shell has a radius 9 - <em>x</em> (this is the distance between a given <em>x</em> value in the orange shaded region to the axis of revolution) and a height of 8 - <em>x</em> ³ (and this is the distance between the line <em>y</em> = 8 and the curve <em>y</em> = <em>x</em> ³). Then each shell has a volume of

2<em>π</em> (9 - <em>x</em>)² (8 - <em>x</em> ³) = 2<em>π</em> (648 - 144<em>x</em> + 8<em>x</em> ² - 81<em>x</em> ³ + 18<em>x</em> ⁴ - <em>x</em> ⁵)

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\displaystyle 2\pi \int_0^2 (648-144x+8x^2-81x^3+18x^4-x^5)\,\mathrm dx = \boxed{\frac{24296\pi}{15}}

I leave the details of integrating to you.

3 0
2 years ago
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Elan Coil [88]

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