The formula for finding a volume of a solid formed by revolving a smooth and continuous function $f(x)$ where $a \leq x \leq b$ , 360° around the $x$ axis is

$\pi \int^b_a \big( f(x) \big)^2 dx$

Let’s find out why. Here’s the outline of the discussion:

Examine the simple case, $f(x)=3$ .
Agree that the function $V(x)$ exists.
Agree that if $f(x)$ is smooth and continuous, then $V(x)$ is smooth and continuous.
Conclude that $V'(x)=\pi\big(f(x)\big)^2$ which leads to $V(x)=\pi \int^b_a\big(f(x)\big)^2dx$

Examine a simple case

Let’s take $f(x)=3$ , from $x=0$ to $x=b$ .

By rotating the straight, horizontal line $y=3$ around the $x$ axis, we generate a cylinder. We already know that the volume of a cylinder can be found using the formula:

$V=\pi r^2 h$

Where $r$ is the radius of the cross sectional circle, and $h$ is the length or height of the cylinder.

We also remember that a function maps a value from one set (domain) to another set (range). Let’s create a function $V$ that maps the length of the cylinder to the volume of the cylinder. Our domain is the length of the cylinder, which in our case is the measurement on the $x$ axis. If we keep one end of the cylinder parked at $0$ , then the length of the cylinder is just $b$ .

The cylinder length $=b$ . The cylinder radius $=3$ . Using the volume of cylinder formula given above, our volume function is:

$V(b)=\pi \times 3^2 b$

We see that this agrees with $\pi \int^b_0 \big( f(x) \big)^2 dx$ when $f(x)=3$ as follows:

$\pi \int^{b}_0 \big( 3 \big)^2 dx=\pi\big[9x\big]^{b}_0= \pi \times 9b$

Let’s check out the volume of the cylinder when $a=0$ and $b=10$ , using our function $V(b)$ :

$V(10)=\pi \times 3^2 \times 10=90\pi = 283\text{ (3 s.f.)}$

Now let’s check out the volume of the cylinder when $a=0$ and $b=4$ .

$V(4)=\pi \times 3^2 \times 4=36\pi = 113\text{ (3 s.f.)}$

Lastly, let’s check out the volume of the cylinder when $a=4$ and $b=10$ . We can calculate this by taking the longer cylinder (from zero to ten) and subtracting the shorter cylinder (from zero to 4).

$V(10)-V(4)=\pi \times 3^2 \times 10-\pi \times 3^2 \times 4=90 \pi - 36 \pi = 54 \pi$

The calculation above is exactly the calculation

$\pi \int^{10}_4 \big( 3 \big)^2 dx=\pi\big[9x\big]^{10}_4=\pi\big[90-36\big]=54\pi$

Let’s move on to understand why this calculation works for any smooth, continuous function $f(x)$ , not just the simple straight line that generates the familiar cylinder.

Next step: Agree that the function $V(x)$ exists for other smooth continuous functions. (under construction)

Volume of Revolution Menu

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Volume of Revolution Discussion i

Examine a simple case