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.
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:
- $y = \sqrt {r^2 - x^2}$
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 Right Circular Cylinder.
From the Method of Disks, the volume of this sphere can be found by the definite integral:
- $\ds V = \pi \int_{-r}^r y^2 \rd x$
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:
| \(\ds V\) | \(=\) | \(\ds \pi \int_{-r}^r \paren {\sqrt {r^2 - x^2} }^2 \rd x\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \pi \int_{-r}^r \paren {r^2 - x^2} \rd x\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \intlimits {\pi r^2 x - \pi \frac {x^3} 3} {x \mathop = -r} {x \mathop = r}\) | Linear Combination of Definite Integrals, Integral of Constant, Power Rule | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {\pi r^3 - \pi \frac {r^3} 3} - \paren {\pi \paren {-r^3} + \pi \frac {-r^3} 3}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 2 \pi r^3 - \frac 2 3 \pi r^3\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {4 \pi r^3} 3\) |
$\blacksquare$