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.
 A: $\mathbf{Surface \ Area \ of \ Circle:}$
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)
Please see this link:  http://www.ams.org/samplings/feature-column/fc-2012-02
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
