Deriving Formulae for Cone, Pyramid and Sphere

In summary, the formulas for the volume of a cone, pyramid, and sphere all involve the number 3 in the denominator because of their geometric shapes. The cone and pyramid can be derived using calculus or by appealing to their geometry, while the sphere can be found using spherical coordinates. Additionally, the concept of steradians is related to the number 4\pi in the formula for the volume of a sphere.
  • #1
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I'm just a little curious with these formulas because I like to understand a formula before I use it. There's no doubt I've been using these since gr3 but I would like a mathmatical derivation of them.

[tex]V_{cone}=\frac{A_bh}{3}[/tex]
[tex]V_{pyramid}=\frac{A_bh}{3}[/tex]
All I would like to know here is why by 3?

And just
[tex]V_{sphere}=\frac{4\pi r^3}{3}[/tex]
With this one I have no clue why it is what it is. There are [itex]4\pi [/itex]radians in a sphere maybe??

Thanks in advance.
 
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  • #2
hm do you mean where does 3 come from?
in integration (calculus), we can find the formula by integration.
for ball, find antiderivative of p f(x)^2 dx
 
  • #3
Well, a cone of height [tex]h[/tex] and with base radius [tex]\alpha h[/tex] can defined by [tex]T = \{ (x,\ y,\ z) \ | \ (\alpha z)^2 \geq x^2 + y^2,\; 0\leq z \leq h \}[/tex]. Clearly it is a solid of revolution (it is symmetric about the z-axis). Its volume is

[tex]V_{\mbox{cone}} = \int \int \int_T dV = \int_0^h \int_{-\alpha z}^{\alpha z} \int_{-\sqrt{(\alpha z)^2 - x^2}}^{\sqrt{(\alpha z)^2 - x^2}} \ dy \ dx\ dz = 2 \int_0^h \int_{-\alpha z}^{\alpha z} \sqrt{(\alpha z)^2 - x^2} \ dx \ dz = 2\int_0^h {(\alpha z)^2 \pi \over 2} \ dz = \alpha^2 \pi \left[{z^3\over 3}\right]_0^h = {\alpha^2 \pi h^3 \over 3}[/tex]

From here all you need to notice is that the base area is [tex] A_b = \alpha^2 h^2 \pi \Longrightarrow V_{\mbox{cone}} = {\alpha^2 \pi h^3 \over 3} = {A_b h \over 3}[/tex] as we wanted. Note that the third-to-last step is valid:

[tex]\int_{-a}^a \sqrt{a^2-x^2} dx = \int_{-{\pi \over 2}}^{\pi \over 2} a^2 \cos^2{t} \ dt = a^2\int_{-{\pi \over 2}}^{\pi \over 2} {1\over 2}\left[1+\cos{2t}\right] \ dt = {a^2 \over 2}\left[t + {\sin{2t} \over 2}\right]_ {-{\pi \over 2}}^{\pi \over 2} = {a^2 \pi \over 2}[/tex]


This is, incidentally, the area of a semicircle of radius [tex]a[/tex], which is to be expected since [tex]0 \leq y \leq \sqrt{a^2-x^2}, -a \leq x \leq a[/tex] defines such a semicircle!

The pyramid can be done in a similar manner, or by appealing to its geometry. You can try it :)

The volume of a sphere is also found in the same way, although this one's easier because you can just use spherical coordinates (who would have guessed!). A sphere of radius [tex]r[/tex] can be defined in polar coordinates to be just [tex] S = \{ ( \rho, \ \phi, \ \theta ) \ | 0 \leq \rho \leq r, \ 0 \leq \phi \leq \pi, \ 0 \leq \theta < 2\pi \} [/tex]. Using the Jacobian transformation we find that since our coordinate transformation is [tex] x = \rho \sin{\phi} \cos{\theta}, \ y = \rho \sin{\phi} \sin{\theta}, z = \rho \cos{\phi}[/tex] we get [tex] dV = \rho^2 \sin{\phi} \ d\rho \ d\phi \ d\theta [/tex] and so our volume is

[tex] V_{\mbox{sphere}} = \int \int \int_S dV = \int_0^{2\pi} \int_0^\pi \int_0^r \rho^2 \sin{\phi} \ d\rho \ d\phi \ d\theta = 2\pi \int_0^\pi \sin{\phi} \ d\phi \int_0^r \rho^2 \ d\rho = 2\pi\left[-\cos{\theta}\right]_0^\pi\left[{\rho^3\over 3}\right]_0^r = (2\pi)(2){r^3 \over 3} = {4 \pi r^3 \over 3}[/tex]

again as we wanted. As to your question about [tex] 4\pi[/tex] radians in a sphere: not quite, but very close. There are [tex]4\pi [/tex] of what we call steradians in a sphere. These are a "unit" (and I use the term loosely, because neither radians nor steradians are really a unit) for a sort of two-dimensional angular arc.
 
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What is the formula for finding the volume of a cone?

The formula for finding the volume of a cone is V = (1/3)πr²h, where V is the volume, π is pi, r is the radius, and h is the height.

How do you derive the formula for the volume of a pyramid?

The formula for the volume of a pyramid can be derived by using the formula for the volume of a cone and considering that a pyramid can be thought of as a cone with a square base. This results in the formula V = (1/3)Bh, where V is the volume, B is the area of the base, and h is the height.

What is the formula for finding the surface area of a sphere?

The formula for finding the surface area of a sphere is A = 4πr², where A is the surface area and r is the radius.

How do you derive the formula for the volume of a sphere?

The formula for the volume of a sphere can be derived by considering a sphere as a stack of infinitely thin disks. The volume of each disk is equal to the area of the base (πr²) multiplied by the height (2r). Integrating this over the entire height of the sphere results in the formula V = (4/3)πr³, where V is the volume and r is the radius.

What is the relationship between the formulas for cone, pyramid, and sphere?

The formulas for cone, pyramid, and sphere are all derived using similar principles, such as using the base area and height to calculate volume. Additionally, the formulas for cone and pyramid can be derived from the formula for sphere by considering different base shapes. However, the formulas are not directly related and must be used separately for each shape.

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