Geometry: Brining Circles Together (Illustrations done with Desmos)

Before I get started with today's blog post, I want to share an awesome online mathematical application.  All illustrations on this blog entry were done with the Desmos free on-line graphing calculator.  You can graph functions, parametric equations, table plots, and polar equations – complete with shading.  Many users have fun using Demos to draw interesting and fun graphics.  There is no monetary cost to signing an account.  You can check out Desmos at www.desmos.com.  

The Desmos Screen

The Two Circles

Suppose we have two circles, both having a radius of 1 and are touching at the origin of the Cartesian plane. (The origin is point (0,0).)

An equation that describes the two circles are:

Red Circle (on the left):  y^2 + (x + 1)^2 = 1

Blue Circle (on the right):  y^2 + (x – 1)^2 = 1

The two circles:  each of them are touching at the origin.

Simply put, the area of the two circles is 2*π. 

Moving the Red Circle In

What happens to the total area when I move the red circle to the right?  A portion of the circles will overlap, as illustrated below:

Moving the red circle to the right - note the overlapping ellipse shaded in purple.

 

If we want the total area, we have to take the overlapping ellipse (now shaded in purple) in mind.    One approach to calculating the area of the shape would be:

A = area of the two circles – area of the overlapping ellipse

A = 2 * π  - E

Where E = area of the ellipse

This ensures that there is not a portion of the shape, particularly the purple overlap that otherwise would get counted twice.   Let the variable p represent the distance that the red circle moves.  

The equations become:

Red Circle (on the left):  y^2 + (x + 1 - p)^2 = 1

Blue Circle (on the right):  y^2 + (x – 1)^2 = 1

In general, the area of any ellipse is:  A = π*a*b, where a and b are the length of the semi-major and semi-minor axes.  

The horizontal length of the ellipse is p.  Hence the semi-horizontal (semi-major or semi-minor depending on p) axis of the overlapping ellipse is p/2.  

Finding the semi-vertical axis will require finding the points where the two circles intersect.  Interestingly, the circles intersect when x=p/2.  The center point of the ellipse is (p/2, 0)

Using the equation for the red circle:

y^2 + (x + 1 – p)^2  = 1

y^2 + (p/2 + 1 – p)^2 = 1

y^2 + (1 – p/2)^2 = 1

y^2 = 1 – (1 – p/2)^2

y = ±√(1 – (1 – p/2)^2)

y = ±√(1 – (1 – p + p^2/4))

y = ±√(-p^2/4 + p)

To verify that we get the same answer, let’s substitute x=p/2 for the blue circle:

y^2 + (x – 1)^2 = 1

y^2 + (p/2 – 1)^2 = 1

y^2 = 1 – (p/2 – 1)^2

y = ±√(1 – (p/2 – 1)^2)

y = ±√(1 – (p^4/2 – p + 1))

y = ±√(-p^2/4 + p)

Hence the points where the circles interest are (p/2, ±√(-p^2/4 + p)).  The semi-vertical axis has a length of √(-p^2/4 + p).  The area of the overlapping ellipse, E, is:

E = π*p/2*√(-p^2/4 + p)

Overall, the area of the shape is:

A = 2*π - π*p/2*√(-p^2/4 + p)

Example:  p = 1/2

A = 2*π - π*(1/2)/2*√(-(1/2)^2/4 + 1/2)

A = 2*π – π/4*√(-1/16 + 1/2)

A = 2*π  - π/4 * √7/4 ≈ 5.76370

Eddie

This blog is property of Edward Shore. 2014

Volume of a Bottle (Approximate)

This blog entry will deal with finding two formulas (approximate?) for finding the volume of a bottle. This includes plastic water bottles, beer bottles, and carry bottles.

In order to know how to find the volume, let's loom at the basic parts. This was accomplished by a basic search on Google:

Assume that the neck and body are cylinders. The shape of the shoulder is going to vary. First, let's work with a bottle with a linear shoulder, and one with a parabolic shoulder. The volume of the bottle will be measured in three parts.

V = VN + VS + VB

Where:
VN = volume of the neck
VS = volume of the shoulder
VB = volume of the body

Bottle - Linear Shoulder:

Neck: VN = π * r^2 * a

Body: VB = π * q^2 * c

Shoulder:

Let's use the technique of the Method of Discs.

Top Boundary: y = b

Bottom Boundary: y = 0

Left Boundary: x = 0

Right Boundary: x = (r - q)/b * y + q

The line between the points (r, b) and (q, 0).

Slope:
Δy/Δx = (b - 0)/(r - q) = b/(r - q)

Y Intercept
y = b/(r - q) * x + β

Use point (q, 0) (x = q and y = 0)
0 = b/(r - q) * q + β
β = -b/(r - q) * q

Solving for x:
y = b/(r - q) * x - b/(r - q) * q
(r - q)/b * y = b * x - b * q
x = (r - q)/b * y + q

Volume of the Shoulder:
VS =

b
∫ ((r - q)/b * y + q)^2 dy * π =
0

b
∫ (r - q)/b * ((r - q)/b * y + q)^2 dy * (π * b)/(r - q) =
0

b
[ 1/3 * ((r - q)/b * y + q)^3 ] * (π * b)/(r - q) =
0

(π * b)/(3 * (r - q)) * (r^3 - q^3) =


(π * b)/(3 * (r - q)) * (r - q) * (r^2 + r * q + q^2) =

(π * b)/3 * (r^2 + r * q + q^2)

Total Volume - Bottle: Linear Shoulder:

V = VN + VS + VB =

π * r^2 * a + (π * b)/3 * (r^2 + r * q + q^2) + π * q^2 * c

Bottle: Parabolic Shoulder

Neck: VN = π * r^2 * a

Body: VB = π * q^2 * c

Shoulder:

Let's use the technique of the Method of Discs.

Top Boundary: y = b

Bottom Boundary: y = 0

Left Boundary: x = 0

Right Boundary:

Parabolic Equation with roots x = -q and x = q and the curve concave downward, an equation to describe this curve can be:

y = -x^2 + q^2
y + x^2 = q^2
x^2 = q^2 - y (note we have x^2)

Volume of the Shoulder:
VS =

b
∫ q^2 - y dy * π =
0

b
[ q^2 * y - y^3/3 ] * π =
0

π * (b * q^2 - b^3/3)

Total Volume - Parabolic Shoulder:

V = VB + VS + VN =

π * q^2 * c + π * (b * q^2 - b^3/3) + π * r^2 * a


Example:
a = 1 in, b = 1 in, c = 4 in, r = 0.9 in, q = 1.5 in
V = π * 1.5^2 * 4 + π * (1 * 1.5^2 - 1^3/3) + π * 0.9^2 * 1 ≈ 36.84041 in^3


These are two ways to approximate the volume of the bottles.

Eddie

This blog is property of Edward Shore. 2014

Area: Interesting Results

Area 1 (see figure 1 above)

This is a simple rectangle.

A = r * h


Easy, right? Suppose that two of the sides are not straight lines but sinusoidal curves, like in Figure 2 below.

yTOP = sin(2 * π * x/r) + h
yBTM = sin(2 * π * x/r)

Left boundary: x = 0
Right boundary: x = r

Area:

r
∫ yTOP - yBTM dx
0

r
∫ sin(2 * π * x/r) + h - sin(2 * π * x/r) dx
0

r
∫ h dx
0

r * h


Hmmm. Let's try something with curves shaped parabolically, like in Figure 3.

yTOP = -x^2 + r * x + h
yBTM = -x^2 + r * x

Left boundary: x = 0
Right boundary: x = r

Area:

r
∫ yTOP - yBTM dx
0

r
∫ -x^2 + r * x + h - (-x^2 + r * h) dx
0

r
∫ h dx
0

r * h

Interesting that the same result is obtained in all three cases, A = r * h.

Keep in mind that these are specific shapes. Who said math isn't fun? :)

Eddie


This blog is property of Edward Shore. 2014

Spherical Hourglass

Let the above picture represent a spherical hourglass, where the bulbs are sections of spheres. Assume that the two bulbs have equal size. In order to calculate the volume of a spherical hourglass, double the volume of a single spherical bulb.

A cross section of a sphere can be describe by the equation x^2 + y^2 = r^2. Surprised? A sphere is a three-dimensional circular object.

By the diagram above, we calculate the volume of a spherical bulb giving radius r and height h. Using the Method of Discs with the discs rotating around the y-axis (x=0):

Top constraint: y = h
Bottom constraint: y = 0
Left constraint: x = 0
Right constraint: x = √(r^2 - y^2)

r(y) = √(r^2 - y^2)

And the volume of one of the spherical bulbs is:

h
∫ (r(y))^2 dy * π
0

h
∫ r^2 - y^2 dy * π
0

h
[ r^2 * y - y^3/3 ] * π
0

(r^2 * h - h^3/3) * π

---

. To get the volume of the spherical hourglass, double the volume of a spherical bulb:

V = 2 * π * (r^2 * h - h^3/3)

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Note that if h = r, the bulbs are two half-spheres and that hourglass' volume:

V = 2 * π * (r^2 * r - r^3/3) = 2 * π * 2/3 * r^3 = 4/3 * π * r^3

Turns out to be the volume of a sphere.


Interesting how the mathematics checks out. With that I wish you a great day/night!

Eddie


This blog is property of Edward Shore. 2014

Parabolic Hourglass

Let the above picture represent a "parabolic hourglass", where the two bulbs are outlined by a parabolic curve. Assume that each of the bulb is equal in size.

To calculate the volume of the parabolic hourglass, first calculate the volume of one of the two bulbs. The total volume of the parabolic hourglass is twice the volume of a bulb.

Examining one of the bulbs, suppose the edge of the bulb can be described by the equation y = x^2 - b. See the diagram below, where we impose a cross section of the bulb on the Cartesian plane.

Note that:

(1) The small art of the bulb is placed on the x-axis, and

(2) The origin (point (0,0)) is placed in the center of the base.

To calculate the volume, we are going to use the Disc Integration Method with the discs rotating around the y-axis. The general formula for this method with y-axis
(x = 0) as the axis of rotation is:

d
∫ (r(y))^2 dy * π
c

We have the following constraints:

Upper: y = h
Lower: y = 0
Left: x = 0
Right: x = √(y + b)

And r(y) = √(y + b) - 0 = √(y + b)

With c = 0 and d = h, the volume of one bulb is:

h
∫ (√(y + b))^2 dy * π
0

h
∫ y + b dy * π
0

h
[ y^2/2 + b*y ] * π
0

π * (h^2/2 + b*h) (I)

If we want to determine the volume of the bulb in terms of a (outer radius), use the equation y = x^2 + b and the point (a, h) to determine that:

h = a^2 - b

b = a^2 - h (II)

Substitute equation (II) into (I) to get:

π * (h^2/2 + a^2 * h - h^2)

π * (a^2 * h - h^2/2) (III)

Remember that this the volume of one bulb. The parabolic hourglass consists of two equally sized bulbs.

---

Therefor the volume of the parabolic hourglass is:

V = 2 * π * (h^2/2 + b * h) = 2 * π * (a^2 * h - h^2/2)

---

Eddie


This blog is property of Edward Shore. 2014