Volume of Cone

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1. Volume of Cylinder: Archimedes 2. Volume of Cylinder: Cavaliries 3. Cone and Pyramid 4. 5. 6.

Volume of Cylinder: Archimedes

A cylinder can be seen as a collection of multiple congruent disks stacked one above the other. To calculate volume, we calculate the space occupied by each disk and then add them up.

Although a cylinder is technically not a prism, it shares like properties of a prism, so volume is found by multiplying the base area by its height.

Therefore, volume of cylinder is
Volume =Base Area x Height
orVolume \(= \pi r^2 \times h\)
orVolume \(= \pi r^2 h\)

According to the Archimedes, if a conical object is placed inside a cylinder in which the radius of the cone is equal to the radius of the cylinder, and height of the cone is equal to the height of the cylinder. Also if a spherical object is placed inside a cylinder such that radius of the sphere is equal to the radius of the cylinder, and diameter of the sphere is equal to the height of the cylinder. Then the volume of the cylinder is equal to the sum of the volume of sphere and volume of cone.
The volume of three objects are in the ratio of 1:2:3

	+		=
Cone in the Cylinder	+	Sphere in the Cylinder	=	Cylinder

Therefore, volume of cone is
Volume =\(\frac{1}{3}\) volume of cylinder
orVolume \(= \frac{1}{3} \pi r^2 \times h\)
orVolume \(= \frac{1}{3} \pi r^2h\)

Therefore, volume of sphere is
Volume =\(\frac{2}{3}\) volume of cylinder
orVolume \(= \frac{2}{3} \pi r^2 \times h\)
orVolume \(= \frac{2}{3} \pi r^2 \times (2r)\)
orVolume \(= \frac{4}{3} \pi r^3 \)

Therefore, volume of sphere is
Volume =\(2 \times\) volume of cone
orVolume \(= 2 \times \frac{1}{3} \pi r^2 \times h\)
orVolume \(= \frac{2}{3} \pi r^2 \times (2r)\)
orVolume \(= \frac{4}{3} \pi r^3 \)

Volume of Cylinder: Cavaliries

Let us take a cylinder of radius \(r\), height \(r\). From the top of the cylinder, immerse a cone of radius \(r\)¸ height \(r\), so that the vertex of the cone coincide at the bottom center of the cylinder. Consider this composition of cylinder and cone as object \(1\).
Also consider a semi-sphere of radius \(r\). Consider this as object \(2\).
Now, we take section of both objects by a plane parallel to the base of the objects and is at height \(h\) from the base.

In this section, the cross section of object \(1\) is a ring. The area of this cross section (ring) is
Area of cross section (ring) = Area of outer circle-Area of inner circle
= \(\pi \times r^2-\pi \times h^2 \)
or =\( \pi \times (r^2-h^2)\)(1)
In this section, cross section of object \(2\) is a circle. The area of this cross section (circle) is
Area of cross section (circle) = \(\pi \times \text{ Radius}^2\)
= \(\pi \times \left( \sqrt{r^2-h^2} \right) ^2\)
or= \( \pi \times (r^2-h^2)\)(2)
Here, the area of cross sections in both objects are equal. Therefore, by Cavaliries principle [SMSG, 22], volume of both objects are equal, thus,
Volume of hemisphere = Volume of cylinder – Volume of cone
= \(\pi r^2h-\frac{1}{3}\pi r^2h\)
or= \( \pi r^2 \times r-\frac{1}{3}\pi r^2\times r\)
or=\( \frac{2}{3}\pi r^3\)
Now,
Volume of sphere =Volume of hemisphere x 2
=\( \frac{2}{3}\pi r^3 \times 2\)
or= \(\frac{4}{3}\pi r^3 \)

Pyramid is a solid formed by polygon base and all other triangles faces with a common vertex. Whereas, Cone is a solid formed by circular base and rays emanating from a fixed point (vertex) and passing through the base.
In the limit, as the number of faces in pyramid approaches to infinity, the shape is a cone. There are four types of cones: circular, elliptical, right, and oblique. A cone is also a pyramid, and it is the fifth type as n = ∞.
In the figure left, keep sliding the number of sides and see that as the number of triangular faces increases, the pyramid begins to look more and more like a cone.