Given information:

The segment of a sphere with radius r and height h.

Consider that the Equation of circle as x2+y2=r2 (1)

Rearrange Equation (1).

x2=r2−y2x=r2−y2

The dimensions of the sphere as shown in Figure 1.

Refer to Figure 1.

Consider that the sphere is obtained by rotating the circle x2+y2=r2 about y-axis.

The region lies between a=r−h and b=r.

Calculation:

The expression to find the volume of the segment of a sphere as shown below:

V=∫abA(y)dy (2)

Find the area of the segment of a sphere as shown below.

A(y)=πr2=π(r2−y2)2=π(r2−y2)

Substitute (r−h) for a, r for b, and π(r2−y2) for A(y) in Equation (2).

V=∫r−hrπ(r2−y2)dy=π(r2y−y33)r−hr=π(r2×r−r33−(r2(r−h)−(r−h)33))=π(r3−r33−(r−h)(r2−(r−h)23))

V=π(2r33−(r−h)3(3r2−(r2+h2−2rh)))=13π(2r3−(r−h)(2r2+2rh−h2))=13π(2r3−2r3−2r2h+rh2+2r2h+2rh2−h3)=13π(3rh2−h3)

=13πh2(3r−h)

Therefore, the volume of the segment of a sphere is V=13πh2(3r−h)_.

Given information:

A sphere of radius 1 is sliced by a plane at a distance x from the center.

The volume of one segment is twice the volume of the other.

The Answer of the equation 3x3−9x+2=0, 0<x<1 is x.

Calculation:

Find the volume of the sphere as shown below.

V=43πr3 (3)

Substitute 1 for r in Equation (3).

V=43π(1)3=43π

Consider the smaller segment has height h=1−x.

Refer to part (a).

Volume of the segment of a sphere as shown below:

V=13πh2(3r−h) (4)

Substitute 1 for r and 1−x for h in Equation (4).

V=13π(1−x)2(3×1−(1−x))=13π(1−x)2(x+2)

This volume must be 13 of the total volume of the sphere.

Substitute 43π for V in the above Equation.

13(43π)=13π(1−x)2(x+2)43=(1+x2−2x)(x+2)x+x3−2x2+2+2x2−4x=43x3−3x+2=43

3x3−9x+6=43x3−9x+2=0

Apply Newton’s method as shown below.

xn+1=xn−f(xn)f′(xn) (5)

Consider f(x)=3x3−9x+2

Differentiate both sides of the Equation.

f′(x)=9x2−9

Substitute 3x3−9x+2 for f(x) and 9x2−9 for f′(x) in Equation (5).

xn+1=xn−xn3−9xn+23xn2−9 (6)

Consider x1=0.

Substitute 1 for n in Equation (6).

x2=x1−x13−9x1+23x12−9=0−0+20−9=0.2222

Substitute 2 for n in Equation (6).

x3=x2−x23−9x2+23x22−9=0.2222−0.22223−9×0.2222+23×0.22222−9=0.2222+0.0013≈0.2235

Substitute 2 for n in Equation (6).

x4=x3−x33−9x3+23x32−9=0.2235−0.22353−9×0.2235+23×0.22352−9=0.2222+0.0013≈0.2235

x3≈0.2235≈x4

Therefore, the value of x by using Newton’s method is 0.2235.

Given information:

Wooden sphere of radius as 0.5 m.

Specific gravity of the wooden sphere is 0.75.

The depth x to which a floating sphere of radius r sinks in water is a root of the equation as follows:

x3−3rx2+4r3s=0

Calculation:

Show the root of the equation as shown below.

x3−3rx2+4r3s=0 (7)

Substitute 0.5 m for r and 0.75 for s in Equation (7).

x3−3(0.5)x2+4(0.5)3×0.75=0x3−32x2+4×18×34=08x3−12x2+3=0

Consider f(x)=8x3−12x2+3

Differentiate both sides of the Equation.

f′(x)=24x2−24x

Apply Newton’s method.

Substitute 8x3−12x2+3 for f(x) and 24x2−24x for f′(x) in Equation (5).

xn+1=xn−8xn3−12xn2+324xn2−24 (8)

Consider x1=0.5.

Substitute 1 for n in Equation (8).

x2=x1−8x13−12x12+324x12−24=0.5−8×0.53−12×0.52+324×0.52−24=0.5+0.0556=0.5556

Substitute 2 for n in Equation (8).

x3=x2−8x23−12x22+324x22−24=0.5556−8×0.55563−12×0.55562+324×0.55562−24=0.5556+0.0402=0.5958

Substitute 3 for n in Equation (8).

x4=x3−8x33−12x32+324x32−24=0.5958−8×0.59583−12×0.59582+324×0.59582−24=0.5958+0.0279=0.6237

Approximately the depth should be x=0.6736.

Therefore, the sinking depth of the sphere is 0.6736 m.

Given information:

The shape of the bowl is hemisphere.

The radius of hemispherical bowl is 5 inches.

Water is running into the bowl at the rate of 0.2 in3/s.

Depth of water is 3 inches.

Calculation:

Show the volume of the segment of a sphere as follows:

V=13πh2(3r−h)=πrh2−13πh3

Differentiate both sides of the Equation with respect to time t.

dVdt=2πrhdhdt−πh2dhdt (9)

Find the speed of the water level in the bowl rising at the instant as shown below.

Substitute 5 inches for r, 0.2 in3/s for dVdt, and 3 inches for h in Equation (9).

0.2=2π×5×3dhdt−π×32dhdt0.2=(94.248−28.274)dhdtdhdt=0.003 in/s

Therefore, the speed of rising the water level in the bowl from instant of water level at 3 inches deep is 0.003 in/s_

Given information:

The radius of hemispherical bowl is 5 inches.

Water is running into the bowl at the rate of 0.2 in3/s.

Depth of water is 4 inches.

Calculation:

Find the volume of the water required to fill the bowl from the instant that the water is 4 in. depth as follows:

V=12×43πr3−13πh2(3r−h)=23πr3−πrh2+13πh3

Substitute 5 inches for r and 4 inches for h in the Equation.

V=23π×53−π×5×42+13π×43=23×125π−80π+643π=π(250−240+64)3=743π

Find the time required to fill the bowl as shown below.

T=VdVdt=743π0.2=387.5 sec×1 min60 sec=6.5 min

Therefore, The time taken to fill the bowl from 4 inch level of water is 6.5 min_.