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3. The brilliant Greek engineer Archimedes (287 B.C. – 212 B. C.) was concerned with using a plane to cut a sphere into two pieces, one of which had a volume of twice the other. He found a formula for the volume of a section of a given height: $V = frac{1}{3}pi h^2 (3r - h)$ where $h$ is the height and $r$ is the radius of the sphere. (a) If a plane at a distance $x = r - h$ from the center of the unit sphere cuts it into two segments, one with twice the volume of the other, show that $x$ must satisfy the equation $3x^3 - 9x + 2 = 0$. (b) Now use Newton's method to find a solution $x$ to the problem of Archimedes, accurate to 8 decimal places.

          3. The brilliant Greek engineer Archimedes (287 B.C. – 212 B. C.) was concerned with using a plane to
cut a sphere into two pieces, one of which had a volume of twice the other. He found a formula for the
volume of a section of a given height: $V = frac{1}{3}pi h^2 (3r - h)$ where $h$ is the height and $r$ is the radius of
the sphere.
(a) If a plane at a distance $x = r - h$ from the center of the unit sphere cuts it into two segments, one
with twice the volume of the other, show that $x$ must satisfy the equation $3x^3 - 9x + 2 = 0$.
(b) Now use Newton's method to find a solution $x$ to the problem of Archimedes, accurate to 8
decimal places.

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$3. The brilliant Greek engineer Archimedes (287 B.C. – 212 B. C.) was concerned with using a plane to cut a sphere into two pieces, one of which had a volume of twice the other. He found a formula for the volume of a section of a given height: V = frac13pi h^2 (3r - h) where h is the height and r is the radius of the sphere. (a) If a plane at a distance x = r - h from the center of the unit sphere cuts it into two segments, one with twice the volume of the other, show that x must satisfy the equation 3x^3 - 9x + 2 = 0. (b) Now use Newton's method to find a solution x to the problem of Archimedes, accurate to 8 decimal places.$

Added by Kimberly P.

Calculus: Early Transcendentals

James Stewart 8th Edition

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04/22/2023

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Let V1 be the volume of the smaller segment and V2 be the volume of the larger segment. We are given that V2 = 2V1. We can use Archimedes' formula for the volume of a section of given height to find the volumes of the two segments. The height of the smaller Show more…

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The brilliant Greek engineer Archimedes (287 B.C. – 212 B. C.) was concerned with using a plane to cut a sphere into two pieces, one of which had a volume of twice the other. He found a formula for the volume of a section of a given height: V = 1/3 ̇ π ̇ h²(3r - h) where h is the height and r is the radius of the sphere. (a) If a plane at a distance x = r - h from the center of the unit sphere cuts it into two segments, one with twice the volume of the other, show that x must satisfy the equation 3x³ - 9x + 2 = 0. (b) Now use Newton's method to find a solution x to the problem of Archimedes, accurate to 8 decimal places.

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00:01 In this problem, we have a sphere like this and the radius is r.

00:09 And we are cutting this sphere by a plane, something like this.

00:16 And let's say the height of this piece is h.

00:22 Then the volume of this portion turns out to be, and the formula is given to be, okay, so this is the setup.

00:40 Now we want to cut this sphere with a plane so that the volume of this portion, let's say v1, is equal to 2 times the volume of this portion, let's say v2.

01:00 Okay, now we are told to assume that this distance is x.

01:08 So we want to find an equation that is satisfied by x according to this condition that says v1 equal to 2v2.

01:20 Okay, now that we know some part of the height, let us compute the entire height in terms of x.

01:28 So here is, okay we have r here, r plus x.

01:36 And this part, okay, noting that this is r, the smaller part is r minus x.

01:44 So now we know the height of each section, each portion, so we can write down their individual volumes.

01:53 V1 is equal to one third pi.

01:57 Okay, the height of v1 is r plus x minus 3r minus.

02:05 We have height again, r plus x.

02:12 Okay, before maybe v2, let us expand this expression.

02:16 If you carry out the multiplication and squaring, you will get 2 pi r cubed over 3 plus pi r squared x minus pi x cubed over 3.

02:34 Okay, now let us write down v2.

02:36 We have one third pi.

02:39 Height again, height is r minus x squared 3r minus r minus x.

02:49 And if you expand this, you will get 2 pi r cubed over 3 plus pi r squared x plus pi x cubed over 3.

03:05 So we want v1 equal to 2v2.

03:10 And we have this other requirement that says this is a unit sphere.

03:15 So r is equal to 1.

03:17 But i will do that at the end.

03:20 Okay, let's write v1.

03:22 Let's just copy it actually from here.

03:29 So this is v1...

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