34. Volume by Slicing, Disks and Washers

Dated: 30-10-2024

Cylinders

Imagine we have a circle in 2D (the \(xy\) plane¹).
In 3d, it would look something like this

And if we start scaling it into the \(3rd\) axis which is \(z\), we will get a cylinder.

If the circle had a hole inside it, the cylinder would have it too.
To find the volume of this cylinder, we can use the formula

\[V = A \cdot h\]

Where \(A\) is the area of the circle and \(h\) is the unit height in z axis.

This formula can work for any shape which is extended in z axis in similar fashion.

Slicing Method

We can bound this shape on x axis by \(x = a\) and \(x = b\).

Notice how the cross sectional area is uniform across the shape if we look at the surface perpendicular to the x - axis.

Therefore, we can define its volume to be a sum of the areas of these surfaces which come up as a result of slicing the shape into \(n\) chunks.

\[V \approx \sum_{i = 1}^{n} A(x_i) \Delta x_i\]

To get more accuracy, we can make the slices as thin as possible.

\[V = \lim_{max(\Delta x_i \to 0)}\sum_{i = 1}^{n} A(x_i) \Delta x_i = \int_a^b A(x) dx\]

Definition

Let \(S\) be a solid bounded by two parallel planes¹ perpendicular to the x-axis at \(x = a\) and \(x = b\). If , for each \(x\) in \([a,b]\), the cross-sectional area of \(S\) perpendicular to the x-axis is \(A(x)\), then the volume of the solid is

\[V = \int_a^b A(x) dx\]

This definition also works if we replace x axis with y axis.

Volumes of Solids of Revolution by Disks Perpendicular to `x axis`

Imagine we have a continuous function² defined over the interval³ \([a, b]\)

After the rotation, we get a solid which looks like this

Notice how if we look perpendicular to the x axis, we would find the cross sections to be circles.
Therefore,

\[V = \int_a^b A(x) dx = \int_a^b \pi \left(f(x)\right)^2 dx\]

\[\because A = \pi \cdot r^2\]

Example

Find formula for volume of a sphere.

Solution

We can get a sphere by revolving a semi circle around the x axis.

First, let's figure out equation for this semi circle.
We know that a circle consists of 2 semi circles which reflect each other across the x-axis.
If \(r\) is the radius of the circle then.

\[x^2 + y^2 = r^2\]

\[y^2 = r^2 - x^2\]

\[f(x) = y = \pm \sqrt{r^2 - x^2}\]

This gives us equations for both semi circles but we care about only one.

\[f(x) = \sqrt{r^2 - x^2}\]

The interval³ is \([-r, r]\) and therefore,

\[A(x) = \pi \cdot \left(\sqrt{r^2 - x^2}\right)^2\]

\[V = \int_{-r}^{r} \pi \cdot (r^2 - x^2) dx = \frac 4 3 \pi r^3\]