Volume of Sphere/Proof by Method of Disks

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Theorem

The volume $V$ of a sphere of radius $r$ is given by:

$V = \dfrac {4 \pi r^3} 3$


Proof

Construction

Describe a circle on the $x y$-plane.

Let its center be the origin.

By Equation of Circle, this circle is the locus of:

$x^2 + y^2 = r^2$

where $r$ is a constant radius.

Solving for $y$:

$y = \pm \sqrt {r^2 - x^2}$

Considering only the upper half of the circle:

$y = \sqrt {r^2 - x^2}$
Semicircle.png

This plane region is a semicircle whose radius is $r$ and whose extremes are at $x = -r$ and $x = r$.

By Euclid's definition of a sphere, the solid of revolution of this plane region about the $x$-axis is a sphere whose radius is $r$.


Proof

Note that this proof uses the Method of Disks and thus is dependent on Volume of Cylinder.

From the Method of Disks, the volume of this sphere can be found by the definite integral:

$\displaystyle V = \pi \int_{-r}^r y^2 \rd x$

where $y$ is the function of $x$ describing the curve which is to be rotated about the $x$-axis in order to create the required solid of revolution.

By construction, $y = \sqrt {r^2 - x^2}$.

The volume, then, is given by:

\(\displaystyle V\) \(=\) \(\displaystyle \pi \int_{-r}^r \paren {\sqrt {r^2 - x^2} }^2 \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \pi \int_{-r}^r \paren {r^2 - x^2} \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \intlimits {\pi \ r^2 x - \pi \frac {x^3} 3} {x = -r} {x = r}\) Linear Combination of Integrals, Integral of Constant, Power Rule
\(\displaystyle \) \(=\) \(\displaystyle \paren {\pi r^3 - \pi \frac {r^3} 3} - \paren {\pi \paren {-r^3} + \pi \frac {-r^3} 3}\)
\(\displaystyle \) \(=\) \(\displaystyle 2 \pi r^3 - \frac 2 3 \pi r^3\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {4 \pi r^3} 3\)

$\blacksquare$


Sources