#### To determine

(a)

**To show:**

The volume of a segment of height h of a sphere of radius r is

V=13πh2(3r-h)

#### Answer

The volume of a segment of height h of a sphere of radius r is

V=13πh2(3r-h)

#### Explanation

**1) Given:**

Radius of sphere is r, and segment of height is h.

**2) Calculations:**

The equation of a circle is x2+y2=r2

x2=(r2-y2)

x=r2-y2

From the figure, limits of integral are from r-hto r.

So the volume of the segment of height h of sphere is

V=π∫r-hrr2-y2dy

By integrating,

V=πr2y-y33rr-h

By applying the Fundamental Theorem Calculus,

V=πr2·r-r33-r2r-h-r-h33

After simplification,

V=13π2r3-r-h3r2-r2-2rh+h2

=13πh2(3r-h)

**Conclusion:**

The volume of a segment of height h of a sphere of radius r is

V=13πh2(3r-h)

#### To determine

**(b)**

**To show:**

If a sphere of radius 1 is sliced by a plane at a distance *x * from the center in such a way that the volume of one segment is twice the volume of the other, then x is a solution of the equation 3x3-9x+3=0 where 0<x<1.

Also find x accurate to four decimal places.

#### Answer

If a sphere of radius 1 is sliced by a plane at a distance *x * from the center in such a way that the volume of one segment is twice the volume of the other, then x is a solution of the equation 3x3-9x+3=0 where 0<x<1.

And x≈0.2261.

#### Explanation

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

Therefore, the smaller segment has height h=1-x.

Let V1 be the volume of smaller segment.

Therefore, V2=2V1 is the volume of the larger segment.

V2=Vsphere-V1

From part (a), volume of the segment with height h, V=13πh2(3r-h)

Vsphere= 43πr3

Therefore,

23πh2(3r-h)=43πr3-13πh2(3r-h)

Multiply by 3π , and

2h23r-h=4r3-h2(3r-h)

Note that r=1, so

3h23-h=4

Since the height of the smaller part is h=1-x

31-x23-1-x=4

31-2x+x22+x=4

32+x-4x-2x2+2x2+x3=4

After simplifying,

6-9x+3x3=4

After rearranging the terms,

3x3-9x+2=0

Therefore, x is a solution of the equation 3x3-9x+3=0 where 0<x<1.

Now,

By Newton’s Method,

xn+1=xn-f(xn)f'(xn)

f=3x3-9x+2

f'=9x2-9

Therefore,

xn+1=xn-3xn3-9xn+29xn2-9

For x1=0,

x2=0--29≈0.2222

x2≈0.2222

x3= x2-3x23-9x2+29x22-9

=0.2222-30.22223-90.2222+290.22222-9

Simplifying,

x3≈0.2261

x4= x3-3x33-9x3+29x32-9

=0.2261-30.22613-90.2261+290.22612-9

x4≈0.2261

So, x accurate to four decimal places is x≈0.2261.

**Conclusion:**

If a sphere of radius 1 is sliced by a plane at a distance *x * from the center in such a way that the volume of one segment is twice the volume of the other, then x is a solution of the equation 3x3-9x+3=0 where 0<x<1.

x accurate to four decimal places is x≈0.2261.

#### To determine

**(c)**

**To calculate:**

Depth to which the sphere will sink if the radius of the wooden sphere is 0.5m and specific gravity is 0.75.

#### Answer

0.6736m

#### Explanation

**1) Given:**

Radius of the wooden sphere is 0.5m and specific gravity is 0.75.

**2) Calculations:**

Substitute r=0.5 and s=0.75 in the given equation x3-3rx2+4r3s=0.

Then x3-30.5x2+40.530.75=0

After simplifying,

x3-1.5x2+0.375=0

Now, use Newton’s Method with

fx=x3-1.5x2+0.375

f'x=3x2-3x

So, by Newton’s Formula,

xn+1=xn-xn3-1.5xn2+0.3753xn2-3xn

Take x1=0.5,

x2=0.5-0.53-1.50.52+0.37530.52-30.5

Simplifying,

x2≈0.6667

Take x2=0.6667

x3=0.6667-0.66673-1.50.66672+0.37530.66672-30.6667

Simplifying,

x3≈0.6736

Take x3=0.6736

x4=0.6736-0.67363-1.50.67362+0.37530.67362-30.6736

x4≈0.6736

x3≈0.6736≈x4

So, up to four decimal places, depth is 0.6736.

**Conclusion:**

The depth x to which a floating sphere of radius r sinks in water is a root of the equation x3-3rx2+4r3s=0, and the depth to which the sphere will sink if radius of wooden sphere is 0.5m and specific gravity is 0.75 is 0.6736m.

#### To determine

**(d)**

**To calculate:**

i) How fast is the water level in the bowl rising at the instant the water is 3 inches deep?

ii) At a certain instant, the water is 4 inches deep. How long will it take to fill the bowl?

#### Answer

i) The water level in the bowl rising at the instant the water is 3 inches deep is 0.003 in/s.

ii) It will take 6.5min to fill the bowl when the water is 4 inches deep.

#### Explanation

**1) Given:**

Radius of the bowl is 5 inches, and rate of flow is 0.2 in3/s.

**2) Calculations:**

i) From part (a), the volume of water in the bowl when height is h is

V=13πh23r-h

Substitute r=5 in above equation.

V=13πh235-h=13πh215-h=5πh2-13πh3

Rate of flow of water is 0.2 in3/s

That is,

dVdt=0.2

To find dhdt when h=3,

Differentiate V with respect to t.

dVdt= 10πhdhdt-πh2dhdt=0.2

Solving for dhdt

dhdt=0.2π(10h-h2)

When h=3,

dhdt= 0.2π(30-9)≈0.003

Therefore, the water level in the bowl is rising at the instant the water is 3 inches deep at 0.003 in/s.

ii) From part (a), the volume of water required to ﬁll the bowl from the instant that the water is 4 in. deep is

V=12·43π53-13π4235-4

=23·125π-113·16π=743π

To find the time required to fill the bowl, divide this volume by rate

Time =743π0.2

=370π3

≈387s

≈6.5 min

Therefore, it will take 6.5min to fill the bowl when the water is 4 inches deep.

**Conclusion:**

i) The water level in the bowl is rising at the instant the water is 3 inches deep at 0.003 in/s.

ii) It will take 6.5min to fill the bowl when the water is 4 inches deep.