Unizor - Geometry3D - Spheres - Problems 2

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Spheres - Problems 2

Just as a reminder, here are a few formulas related to spheres. They are all derived in the previous lectures and are helpful in solving these problems.
Volume of a sphere: 4πR³/3
Surface area of a sphere: 4πR²
Volume of a cap: πH²(3R−H)/3
Surface area of a dome: 2πRH
Volume of a sector: 2πR²H/3

Problem
Given a sphere of some unknown radius.
A cylindrical hole is drilled through it such that the axis of a cylinder goes through a center of a sphere. The height of a hole from edge to edge is h.
What is the volume of remaining part of a sphere?

Answer
V = πh³/6

Solution

The drilled out part of a sphere is a cylinder with two caps on both ends.
To calculate their volumes we need the radius of a sphere R, the radius of a cylinder r, the height of each cap H and the heights of a cylinder h with only the height of a cylinder h given.
Challenging, is not it?

From Pythagorean Theorem:
R² = r² + (h/2)²
The heights of a cylinder and caps is related to the radius of a circle:
R = H+h/2
So, we have only two equations with three unknown - R, r and H.

Let's express r² in terms of known h and unknown H:
r² = (H+h/2)²−(h/2)² = H²+Hh

Let's proceed with calculations of a volume of the remaining after drilling part of a sphere, hoping that the resulting formula would be a function of only cylinder's height h.

The volume of a sphere is
Vsphere = 4πR³/3
The volume of a cylinder is
Vcyl = πr²h
The volume of each cap is
Vcap = πH²(3R−H)/3
The volume of a remaining after drilling part of a sphere is
V = Vsphere − Vcyl − 2Vcap =
=4πR³/3−πr²h−2πH²(3R−H)/3
Let's substitute in this expression
R = H+h/2 and
r² = H²+Hh
The result will be:
V = 4π(H+h/2)³/3 −
− π(H²+Hh)h −
− 2πH²[3(H+h/2)−H]/3 =
= π{4H³+6H²h+3Hh²+h³/2 −
− 3H²h−3Hh² −
− 4H³−3H²h}/3 =
= πh³/6

Unizor - Geometry3D - Spherical Sectors

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Spherical Sector

Consider a spherical cap that is a part of a sphere of radius R with a center O.
Connect each point of a base circle of a cap with center O by a radius of a sphere. All these radiuses form a conical surface with a base circle as a directrix.
An original spherical cap and a cone formed as described above constitute a spherical sector.


A spherical sector is defined by two parameters - radius of a main sphere R, from which it is a part, and the height of a cap on its top H. For "true" spherical sectors, the ones we will be considering here, height H is smaller or equal to radius R of a main sphere.

Since perpendicular from a center of a sphere onto a base circle of a cap falls into a center of this circle, our cone is a right circular one - the only type we considered in this course.

Volume of a Sector

The radius of the base circle (in terms of the radius of a main sphere R and the height of a spherical cap H) equals, by Pythagorean Theorem, to
L = √R²−(R−H)² = √2RH−H²
This is a cone's radius.
The cone's height is, obviously, R−H.
So, the cone's volume is
Vcone = πL²(R−H)/3 =
= π(2RH−H²)(R−H)/3 =
= π(2R²H−3RH²+H³)/3

As we know from the previous lecture about spherical caps, the volume of a cap is
Vcap = = πH²(3R−H)/3

From the two formulas for volumes of a cap and a cone we can determine the volume of a spherical sector:
Vsector = = πH²(3R−H)/3 +
+ π(2R²H−3RH²+H³)/3 =
= 2πR²H/3
A very short formula indeed!

Just for checking, if H=R, our spherical sector occupies exactly half a sphere. The volume of a sphere, as we have determined before, is 4πR³/3.
So, half a sphere has a volume 2πR³/3.
This is exactly a formula we get from the volume of a spherical sector if H=R.

Area of a Dome

Recall the technique we used to calculate the area of a sphere Ssphere, knowing its volume Vsphere.
We considered a sphere's surface as approximated by an infinitely large number of inscribed polyhedrons, whose faces are infinitely small. Then the volume of a sphere would be approximated by a sum of volumes of all the pyramids obtained by connecting each vertex of each face of each polyhedron with a center of a sphere.
Then the volume of a sphere Vsphere would be equal to an area of its surface Ssphere multiplied by one third of its radius R:
Vsphere = Ssphere·R/3.
This consideration allowed us to derive from a formula for the volume of a sphere -
Vsphere=4πR³/3
- its surface area:
Ssphere=4πR².

Exactly the same considerations connect the area of a dome with a volume of a spherical sector: Vsector = Sdome·R/3.
Therefore,
2πR²H/3 = Sdome·R/3
Now we can calculate the area of a dome:
Sdome = 2πRH

Unizor - Geometry3D - Spherical Caps

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Spherical Cap

A spherical cap is formed when a plane cuts through a sphere. This cut divides a sphere into two parts, and each can be called, technically, a spherical cap. Usually, however, it's the smaller one that is considered a "true" cap. The part of a cutting plane that is inside a sphere is a base of a spherical cap, and we know from the previous problems that it is a circle. The part of a sphere that belongs to a cap forms its dome.

A spherical cap is defined by two parameters - radius of a main sphere R, from which it is cut, and the height H defined as the length of the longest segment from a dome onto a base along a perpendicular to it. For "true" spherical caps, the ones we will be considering here, height H is smaller or equal to radius R of a main sphere.

Volume of a Cap
Let's start with a volume of a spherical cap of radius R and height H using already familiar process of approximation of this volume as a sum of volumes of cylinders of the same but small altitude and different radiuses stacked on each other.

The radius of the base circle (in terms of the radius of a main sphere R and the height of a spherical cap H) equals, by Pythagorean Theorem, to
L = √R²−(R−H)² = √2RH−H²

Let AB be a segment that lies along the height of our cap. Divide segment AB into N equal parts by points A1, A2,,, AN-1 from the "top" of the dome down (for uniformity, we can designate point A, the center of a base circle, as AN and point B at the "top" of the dome as A0) and draw planes parallel to a base through each such division point.
The intersection of each plane and a sphere is a circle with a center at a corresponding division point Ak because, if we take any two points X and Y on this intersection, lengths of AkX and AkY are equal since right triangles ΔOAkX and ΔOAkY are congruent by a common cathetus OAk and congruent hypotenuses OX=OY=R.

These planes slice our spherical cap into layers. Each layer resembles a cylinder in a way that it is bounded from top and bottom by two parallel planes and is somewhat rounded in shape, but it's not a true cylinder because its side surface is not formed by straight lines parallel to the same generatrix.

The next step is to make a cylinder within each layer preserving it's circular top base and replacing the side surface with a cylindrical surface by dropping perpendiculars from each point on the upper base towards its bottom base.

Now it's time to make a leap of faith and consider a reasonable, intuitively obvious statement that, as N→∞, the total volume of cylinders tends to some limit that we can call the volume of our spherical cap.

Our task is to calculate a sum of volumes of the N cylinders as a formula, that depends on radius of a sphere R, height of the cap H and the number of division points N, and to find its limit as N→∞, which will depend only on radius R and height H.
Let's calculate a volume of a cylinder #k and then summarize it by k∈[1,N].

By Pythagorean theorem, for any point X on a circle with center Ak
AkX² = OX² − OAk²
Point X is on a sphere, therefore OX=R
Point Ak is kth point of division of segment AB into N equal parts, therefore
OAk=R−H·k/N.

Hence, the square of a radius of a base of the kth cylinder equals to AkX² = R²−(R−H·k/N)²
Its altitude is, obviously, H/N.
Therefore, the volume Vk of the kth cylinder is
π[2RHk/N−(Hk/N)²]·H/N =2πRH²k/N² − πH³k²/N³

Now we have to summarize the volumes of all N cylinders and find the limit of this sum as N→∞.
We will use a symbol Σ for summation by k from 1 to N

V = Σ(Vk) = πRH²(N+1)/N − πH³(N+1)(2N+1)/6N²

Note that we have used a known and previously derived expressions for a sum of numbers from 1 to N:
Σk = N(N+1)/2
and for a sum of squares from 1 to N:
Σ(k²)=N(N+1)(2N+1)/6

The limit of (N+1)/N is 1. The limit of (N+1)(2N+1)/6N² is 2/6=1/3.

Therefore, the volume of the spherical cap equals to
Vcap = πRH² − πH³/3 = πH²(3R−H)/3

Unizor - Geometry3D - Spheres - Problems 1

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Spheres - Problems 1

Problem A
Explain that a sphere of radius R divides an entire three-dimensional space into two parts - points outside of a sphere that are at a distance greater than R from its center and points inside a sphere that are at a distance smaller than R from its center.

Explanation
(not really a rigorous proof)
Choose any point M outside a sphere and connect it with its center O. The fact that point M is outside a sphere means that segment OM intersects a sphere at some point P lying in between points O and M.
That is, OP⊂OM.
Since the length of segment OP equals exactly to R, the length of segment OM must be greater than R.
Choose any point M inside a sphere and connect it with center O. The fact that point M is inside a sphere means that a continuation of segment OM beyond point M intersects a sphere at some point P, so point M is lying in between points O and P.
That is, OM⊂OP.
Since the length of segment OP equals exactly to R, the length of segment OM must be smaller than R.

Problem B
A plane intersects a sphere, but does not pass through its center.
Prove that a base of a perpendicular from a center of this sphere to this plane is inside the sphere.

Proof
Let the base of a perpendicular from the center of our sphere O to a plane be point M.
Choose any point A on an intersection of a sphere and a plane.
Since OM is perpendicular to an entire plane, it's perpendicular to line MA on it.
Therefore, triangle ΔOMA is a right triangle with catheti OM and MA and hypotenuse OA.
Since hypotenuse is longer than cathetus, the length of OA, equaled to R, is greater than the length of OM.
According to a previous problem, point M must lie inside the sphere.

Problem C
Given a sphere of radius R with center O.
Plane δ intersects this sphere.
Prove that the intersection of this plane with a sphere is a circle of a radius not greater than R and a center being a base of a perpendicular from a sphere's center O onto plane δ.
Determine the radius of this circle, if plane δ is positioned at a distance d (smaller than R) from a center of a sphere O.

Answer:
Drop a perpendicular from center O onto a plane δ. Its base - point M - is inside a sphere, according to a previous problem. The length of segment OM is d.
Choose any two points A and B on intersection of our sphere with plane δ.
We know that OM⊥δ
⇒ OM⊥AM; OM⊥BM
Consider two right triangles ΔAOM and ΔBOM.
They are congruent by common cathetus OM and equal hypotenuses OA and OB, each being a radius of a sphere. Therefore, two catheti AM and BM are equal.
Since A and B are any two points on the intersection of a sphere and plane δ, all points on this intersection are equidistant from point M - a center of a circle and a base of a perpendicular from a sphere's center O onto plane δ.
The radius AM=r can be calculated by Pythagorean theorem as
r² = R² − d²
r = √R² − d²
Technically, this proof is not applicable, if the plane δ passes through a center of a sphere O because in this case we cannot form triangles used above in the proof. However, this situation does not need any special proof since what we have to prove is contained in the definitions of a sphere and a circle. All points lying on the intersection of our sphere and plane δ are lying in one plane and equidistant from one point on this plane - point O - since they belong to a sphere. Therefore, these points form a circle.
The radius of this centrally located circle is exactly R, which is greater than the radius of any other circle formed by intersection of a plane and a sphere represented by a formula above.
By the way, that formula for d=0 (which technically is not applicable) gives the radius of a circle R, which corresponds our intuition that, as a plane moves closer and closer to a center, the radius of a circle it cuts from a sphere becomes greater and greater with a limit R.

Problem D
Consider a sphere of radius R and center O and a plane δ tangential to this sphere, that is the one that has only one point of intersection with this sphere - point P.
Prove that a radius OP connecting a center of a sphere O with a point of tangency of a plane with this sphere P is perpendicular to a plane δ.

Proof
All points of plane δ, except point of tangency to a sphere P, are located outside a sphere. Therefore, their distance from a center of a sphere O is greater than the radius of a sphere R (see Problem A above). Only a point of tangency P is at a distance equal to a radius of a sphere R.
We see now that segment OP is the shortest among all segments connecting center O to points on plane δ. Since we know that a perpendicular to a plane from a point is the shortest connection between them, and any other connection between center O and points on plane δ is greater than R, OP⊥δ.

Unizor - Geometry3D - Cones - Problems 2

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Right Circular Cones - Problems 2

In this course we will be dealing only with right circular cones and will call them simply cones.

Problem A
A doctor has recommended for a sick girl to take some liquid medicine a couple of times a day by 15-20 grams each time.
Mother was going to use a small conical glass that could hold exactly 20 gram of liquid for this. She filled it up, but a girl said that she remembered the taste of this medicine and did not like it.
Then the mother suggested a compromise: "How about you drink only half a glass?"
The girl agreed, drank some medicine and left some in a glass. What she left took exactly half the height of a glass.
How much medicine did a girl drink?

Answer:
17.5 grams - as doctor ordered.

Problem B
A semicircle of a radius R is rolled into a cone.
What is a volume of this cone?

Answer:
V = πR³√3/24

Problem C
A right circular cylinder of a radius r is inscribed into a cone of a radius R (greater than r) and an altitude H such that the lower base of a cylinder lies in the same plane as the base of a cone with the centers of these bases coinciding, the cylinder height ends at an intersection of a side of a cylinder with a side of a cone.
What is a volume of the cylinder?

Answer:
V = πr²(R−r)H/R

Problem D
Given a cone with a radius of a base R and height H.
Determine a central angle of a circular sector obtained if we cut a side of a cone along a generatrix and roll it out on a plane (in radians).

Answer:
φ = 2πR/√R²+H²

Problem E
Consider a right circular cylinder. Two congruent cones are inscribed into it. One cone has its base coinciding with the bottom base of a cylinder and has its apex at the center of an upper base of a cylinder. Another cone has its base coinciding with the upper base of a cylinder and has its apex at the center of a bottom base of a cylinder.
Find the ratio of a volume of the common part of these two cones (part of their intersection) to a volume of one of them.

Answer:
1/4

Unizor - Geometry3D - Cones - Problems 1

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Right Circular Cones - Problems 1

In this course we will be dealing only with right circular cones and will call them simply cones.

Problem A
Find a volume of a truncated cone obtained by cutting-off its top by a plane parallel to its base in terms of two radiuses of two bases R1 and R2 and its height H (the distance between bases).

Answer:
πH(R1²+R1R2+R2²)/3

Problem B
The side surface of a cone, rolled out on a plane, is a circular sector with a central angle 120o and area S.
Determine the volume of this cone.

Answer:
2S√6πS /27π

Problem C
Given an area S of a side surface of a cone and a distance d from a center of its base to any generatrix, find a volume of a cone.

Answer:
S·d /3

Problem D
Side surface area of a cone is twice as big as an area of its base.
A section of a cone obtained by cutting it by a plane passing through its main axis (that is, a plane that goes through an apex and a center of a base) has an area S.
Find a volume of a cone.

Answer:
(27^1/4)πS√S/9

Problem E
An altitude of a cone of a volume V is divided into N equal parts by points Ai, where i∈[1,N−1]. For convenience, let A0 be an apex of a cone and AN - a center of its base.
Then we draw a plane through each division point parallel to a base of a cone. These planes divide the cone into N parts, all of them except the top are truncated cones.
Find the volume of a truncated cone #k between planes going through points Ak−1 and Ak.

Answer
V[k³−(k−1)³]/N³

Unizor - Geometry3D - Sphere - Volume and Surface Area

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Volume and Surface Area of a Sphere

Volume

Let's start with a volume of a sphere of a radius R and a center at point O using already familiar process of approximation of this volume as a sum of volumes of other geometric objects with known formulas for volume. We will use cylinders of the same but small altitude and different radiuses stacked on each other as an approximation of a sphere.

Imagine a plane that cuts our sphere of a radius R in two equal halves by going through its center. We will calculate the volume of a half a sphere and then double it.

From a center of a base O draw a perpendicular to this plane "upward" to intersection with our half a sphere at point P on its "top". The length of this perpendicular OP is, of course, R, since this is a radius of an original sphere.

Divide this perpendicular into N equal parts by points A1, A2,,, AN-1 (for uniformity, we can designate point P at the end of this perpendicular as AN) and draw planes parallel to a base through each such division point. The intersection of each plane and a sphere is a circle with a center at a corresponding division point Ak.

The next step is to make a cylinder within each layer preserving it's circular top base and replacing the side surface with a cylindrical surface by dropping perpendiculars from each point on the upper base towards its bottom base.

By Pythagorean theorem, for any point X on a circle with center Ak
AkX² = OX² − OAk²
Point X is on a sphere, therefore OX=R
Point Ak is kth point of division of a radius OP into N equal parts, therefore OAk=R·k/N.
Hence, the square of a radius of a base of the kth cylinder equals to
AkX² = R²−(R·k/N)²
The above is a radius of the kth cylinder.
Its altitude is, obviously, R/N.
Therefore, the volume of the kth cylinder is
Vk = π[R²−(R·k/N)²]·R/N =
= πR³/N − πR³·k²/N³
Now we have to summarize the volumes of all N cylinders and find the limit of this sum as N→∞. Doubling this (since we were dealing with half a sphere) will give us the volume of a sphere.
The result, the volume of the whole sphere of radius R will then be
Vsphere = 4πR³/3

Surface Area

Let's inscribe a convex polyhedron into a sphere by choosing a sufficiently large number of points on its surface (they will be vertices of a polyhedron) and connecting each point with its closest neighbors (these connections will be edges of the polyhedron and each triangle formed by these edges will be its face).
Connecting all the vertices with a center of a sphere, we divide a polyhedron into pyramids with different faces and, generally speaking, different altitudes.

As we add new points on a sphere, evenly distributing them in the empty spots on a sphere's surface, we will increase the number of faces of these polyhedrons, increasing the number of pyramids these polyhedrons are divided. What's interesting is that altitudes of all the pyramids will be closer and closer to a radius of a sphere and the surface area of a polyhedron becomes closer and closer to a surface area of sphere.

In the limit, as all faces are getting smaller and smaller, and the distance between a center of a sphere and all these faces becomes closer and closer to a radius of a sphere, we can say that the volume of a polyhedron (that approximates the volume of a sphere) approximately equals to one third of its surface area (that approximates the surface area of a sphere) multiplied by the radius of a sphere (that approximates the altitudes of all pyramids). The approximation is better as the largest face area of a polyhedron tends to zero.

The above logic is an intuitive base for the following statement about the relationship between the volume of a sphere Vsphere of a radius R and its surface area Ssphere:
Vsphere = Ssphere·R/3

Knowing the formula for the volume of a sphere, we can derive from this the formula for the surface area:
4πR³/3 = Ssphere·R/3
form which immediately follows:
Ssphere = 4πR²

Unizor - Geometry3D - Cones - Area and Volume

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Area and Volume of a Cone

Let's define two important parameters that fully characterize a cone.
They are:
(a) radius of a base circle, which we will refer to as radius of a cone and denote as R,
(b) altitude or height of a cone (the distance from an apex to a bottom circular base), which we denote as H.
Note that the distance from an apex to any point on a circular base it is connected with by a straight line on a side conical surface is constant and, according to Pythagorean Theorem, is equal to
L = √R²+H²

Surface Area

There are different approaches to defining an area of a cone. More rigorous approach involves full force of the theory of limits, but we would suggest here a different approach.

First of all, consider the side surface of a cone.
Since this side surface is formed by a straight lines connecting points of a circular directrix with an apex, it is intuitively obvious that, if we cut the side surface of a cone along one of these straight lines, we will be able to "flatten" it on a plane without stretching or squeezing, that is without any change to its area.

As a result of this transformation, we will obtain a circular sector with radius equal to a distance between an apex and each point on a circular base of a cone that we have calculated above as
L = √R²+H².

Another characteristic of this sector, that we will use to determine its area, is the length of an arc of this sector. Obviously, it is equal to a circumference of a base of a cone, that is
C = 2πR.

Therefore, the area of a side surface of a cone is equal to the area of a circular sector with a radius
L = √R²+H²
and an arc length
C = 2πR.

The sector's area is a part of an area of a circle of the same radius L. The ratio between the sector's area and the area of a circle, which this sector is a part of, equals to the ratio between their arcs.

We know the arc of a sector, it is equal to 2πR.
We also know the arc (that is, circumference) of a circle, which our sector is a part of, it is equal to 2πL.
And we know this circle's area, it is equal to πL².
We can determine the side area Sside of a cone from the following ratio:
Sside /πL² = 2πR /2πL

The solution to this is
Sside = πR·L = πR√R²+H²

To determine a full area of a cone, we have to add an area of its base Sbase=πR².
The final result for a full area of a cone Sfull:
Sfull = Sside + Sbase =
= πR√R²+H² + πR² =
= πR(R+√R²+H²)

Volume

The situation with volume of a cone is similar to that of a volume of a cylinder, and we will not be able to escape considerations based on the limit theory.

Let's inscribe into a circular base of a cone a regular N-sided polygon. Then construct a pyramid with this polygon being a base and the same apex as that of a cone. We obtain a pyramid inscribed into a cone.

Without rigorous proof, it is intuitively obvious that, as we increase the number of vertices N, the regular polygon inscribed into a circular base of a cone becomes closer and closer to a circle itself, and the pyramid, based on this polygon inscribed into a circular base of a cone, becomes closer and closer to a cone. So, the volume of a cone is a limit of the volumes of inscribed in this manner pyramids as N→∞.

Since a volume of a pyramid is one third of a product of an area of its base by height and, as N→∞, the area of the N-sided polygon inscribed into a circle of a radius R tends to the area of a circle itself, that is πR², while the height H remains constant, we conclude that the volume of a pyramid tends to
V = πR²·H /3