# Sphere Volume:$V=\frac 4 3 \pi r^3$ Surface area: $A=4\pi r^2$ ## Derivation of the Formula for the Volume of a Sphere If we slice a sphere each slice would look like a disk. The volume of a sphere can be thought of as the sum of the volumes of all disks. If we position the sphere such that its center is at the origin of a coordinate system, and we cut our slices along the $y$ axis, each disk would have a radius of $r=\sqrt {R^2-x^2}$ where $R$ is the radius of the sphere. This follows from the definition of a circle as $x^2+y^2 = R^2$. In our case the circle we're referring to now is the sphere as we look at it head on (looking at the $xy$ plane. From the definition of a volume of a disk, each slice would have a volume of $ \begin{align} V_{disk}&=r^2\pi h \\ V_{disk}&=(R^2-x^2)\pi h \end{align} $ As the sum of the sphere is the sum of volumes of all disks we can claim that $ \begin{align} V_{sphere} &= \int \limits_{-R}^R \left(R^2-x^2\right) \pi \,dx\\ &=\int \limits_{-R}^R R^2\pi - x^2\pi \, dx \\ &= \pi \bigg[ R^2 x -\frac 1 3 x^3 \bigg]_{-R}^R \\ &= \pi \bigg[ \bigg( R^3-\frac 1 3R^3 \bigg)-\bigg({-R^3} + \frac 1 3 R^3 \bigg) \bigg] \\ &= \pi R^3 \bigg ( \frac 2 3 + \frac 2 3 \bigg) = \frac 4 3 \pi R^3 \end{align} $ ## Derivation of the Formula for the Surface Area of a Sphere If we slice a sphere vertically along the $y$ axis each slice would look like a disk whose cross-section would be a circle with a radius $r$. ![[Screenshot 2025-06-15 at 13.16.48.png|200]] We can define an angle $\theta$ of each disk that is formed with the line connecting the center of the sphere and the center of the disk, and the line that connects the center of the sphere and the edge of the disk. For a slice defined by $\theta$, its radius would be: $r = R\sin \theta$ where $R$ is the radius of the sphere. A disk can be completely defined via the difference of angles between two neighboring slices ($\Delta \theta$), and the angle of the slice. For smaller and smaller $\Delta \theta$, the surface area of each disk approximates the surface area of a rectangle whose width is the circumference of cross-section circle (with radius $r$), and whose height is $R \,\Delta\theta$ (the length of the angle arc). As the thickness of each slice (and correspondingly $\Delta \theta$) approximate zero, the surface area of the whole sphere can be defined as: $ \begin{align} S &= \int \limits_0^\pi 2\pi r \cdot R \, d\theta \\ &= \int \limits_0^\pi 2\pi R \sin \theta \cdot R \, d\theta \\ &= \int \limits_0^\pi 2\pi R^2 \sin \theta \, d\theta \\ &= -2\pi R^2 \cos \pi - (-2\pi R^2 \cos 0) \\ &= 2\pi R^2 + 2\pi R^2 \\ &=4\pi R^2 \end{align} $