# Area of a circle, using $2$ triangular pyramids?

I was thinking about how you can get the area of a circle, by forming a right triangle.

If I have the triangle infinitely increasing to the height of the radius, shouldn't it form a triangular pyramid and if i put $2$ of them together should it form the area of a circle?

Sorry for asking this stupid question, I'm a highschooler.

• How do you get the area of a circle by forming a right triangle? – fleablood May 11 '17 at 19:50
• Please explain what you mean clearly. What do you mean infinitely increasing ti the height of the radius? Do you mean in the third dimension? And what do you mean infinite? The radius is finite. I'm not trying to be discouraging but I don't understand your question. – fleablood May 11 '17 at 19:56
• I just mean, that if you kept stacking right triangles to the radius r then you would get the volume of a half a sphere. Because if you cut a circle in to infinite pieces than it can be rearranged into a triangle. – Ntspamer Desus May 11 '17 at 20:29
• We need to understand whether you're working in two dimensions or three. Circle and triangle are two-dimensional figures. Sphere and pyramid are three-dimensional figures. – John May 11 '17 at 21:07
• Please, you are not being clear. How can stacking two dimensional triangles unto a one dimensional line segment give us a 3 dimensional half sphere. And how can we cut a circle into infinite pieces and a range them into a triangle and how does that help us determine the area. I suspect the you are getting at approximating the area of a circle by infinitely many triangles and are wondering if you can do that with a sphere and slivers of pryamids. You can but you don't get half a sphere you get a third. But please clarify what you are talking about. – fleablood May 11 '17 at 21:31

$\mathbf{Surface \ Area \ of \ Circle:}$

By infinitely partitioning a geometric circle of radius $r$ into symmetrical sectors, and laying each unit against each other in an approximate rectangular shape, one can determine the area of a circle. (Using triangles)

Similarly, by inscribing a polygon of finite side number, bounded within some arbitrary circle of radius $r$, and taking the limit of its area as its side/vertex number approaches infinity, also evaluates to the area of a circle.

Let's consider subdividing a geometric circle of radius $r$ into $n\in \mathbb{N}$ identical sectors. By "trimming" off each arced tail, and placing every triangular unit side by side, we get the following:

If we infinitely subdivide the circle using the same above method, and sum the areas of each triangular unit, we get:

$$Area_{circle}=(\frac 12)2\pi r\cdot r=\pi r^2$$

This however is not the standard approach, as we are assuming the "cut" arced sections of each sector don't cumulatively add to change the final result. (This indeed doesn't happen, but why?)

Rigorously, the circle's area can be evaluateed as follows:

Consider the relation $x^2+y^2=r^2$, which models a circle of radius $r$, bounded by the line $y=\pm r$ on the interval $[-r,r]$.

Then the area of a semicircle evaluates as follows:

$$Area_{semicircle}=\int_{-r}^r(r^2-x^2)^{\frac 12}dx=2\int_0^r(r^2-x^2)^{\frac 12}dx$$

Letting $x=rsin(\theta)\implies dx=rcos(\theta)d\theta$, we have:

$$=2r\int_0^{\frac {\pi}{2}}(r^2-r^2sin^2(\theta))^{\frac 12}cos(\theta)d\theta=2r^2\int_0^{\frac {\pi}{2}}(1-sin(\theta))^{\frac 12}cos(\theta)d\theta$$

$$=2r^2\int_0^{\frac {\pi}{2}}cos^2(\theta)d\theta=2r^2\bigg[\frac {cos(\theta)sin(\theta)+\theta}{2}\bigg]_0^{\frac {\pi}{2}}=2r^2\frac{\pi}{4}=\frac {\pi r^2}{2}$$

To attain the area of a circle, we simply add the area of two identical semicircles, that is:

$$Area_{circle}=2\cdot Area_{semicircle}=2\cdot \frac {\pi r^2}{2}=\pi r^2$$

$\mathbf{Volume \ of \ Sphere:}$

Similarly, by infinitely partitioning a geometric sphere of radius $r$ into symmetrical pyramids of base $B$, and laying each unit against each other in an approximate rectangular prism shape, one can determine the volume of a sphere.

Let's consider infinitely subdividing a geometric sphere of radius $r$ into a series of identical pyramids of arced base. By "trimming" off each curved base, and placing every pyramidal unit side by side, (analogously done with the triangular case above), we can derive a formula for a sphere's volume.

Noting the volume of a square based pyramid is $Volume_{pyramid}=\frac 13Bh$, and surface area of a sphere is $4\pi r^2$, we get the following:

$$Volume_{sphere}=\sum_{k=1}^{\infty}Volume_{pyramid}=\frac r3\sum_{k=1}^{\infty}B_n=\frac r3\cdot 4\pi r^2=\frac 43\pi r^3$$

Where $B_n$ denotes the area of the $n^{th}$ square pyramidal base.

Again, this is not the standard approach, as we must first previously know both the volume of a pyramid, and surface area of a sphere.

Rigorously, the volume of a circle can be evaluated as follows:

Consider the relation $x^2+y^2=r^2$, which models a circle of radius $r$, bounded by the line $y=\pm r$ on the interval $[-r,r]$. Furthermore, by considering the method of washers, and rotating a semicircle's area around the x-axis, we can generate the volume of a sphere, via symmetry.

Thus the volume evaluates as follows:

$$Volume_{sphere}=\int_{-r}^{r}A(x)dx=2\pi \int_0^r (r^2-x^2)dx$$

$$=2\pi \bigg[xr^2-\frac {x^3}{3}\bigg]_0^r=2\pi \bigg[r^3-\frac {r^3}{3}\bigg]=2\pi (\frac {2r^3}{3})=\frac {4\pi r^3}{3}$$

Pictures courtesy of:

http://www.ams.org/samplings/feature-column/fc-2012-02

http://www.k6-geometric-shapes.com/volume-of-a-sphere.html

• I hope this is what you were asking. If you want to know how to determine the circles circumference for the very top approach, just comment and I will add some more to my post! – Mark Pineau May 11 '17 at 22:20
• I think the OP was asking if we can extend the same reasoning to find the volume of a sphere. I was working on an answer but I got stuck on an intutive way without calculus to demonstrate 1) the surface area of a sphere is $4\pi r^2$ and 2) that the volume of a pyramid with base A and height h is $\frac 13Ah$. – fleablood May 11 '17 at 22:25
• Haha yeah, when he says pyramid, it's kind of confusing. If so, it's always good to have more than one solution. – Mark Pineau May 11 '17 at 22:26
• We can do the same thing in 3 dimensions by dividing the unpeeled surface area of a sphere into spikes of height r. But we have to know that the pyramid spikes take up 1/3 (not one half) the volume. (This can be hand-waved by averaging out the three dimensions I guess) and that the surface area of the sphere is $4\pi r^2$ (so $V= \frac 43 \pi r^3$). We can hand wave this as the diameter is 2r and we are unwrapping it in 2 dimensions but.... that's too hand wavy for me to trust. (I'm not even sure it is true.) – fleablood May 11 '17 at 22:38
• Yeah, I can't really see doing this without calculus (volume case), which in fact makes it a lot easier. In my above solution, we already "know" the arc length of an entire circle, yet that requires calculus to attain, which just seems stupid in my opinion. – Mark Pineau May 11 '17 at 22:41