Problem 41 - Essential Calculus

Question

Use the disk method to verify that the volume of a sphere is \(\frac{4}{3} \pi r^{3}\).

Step-by-Step Solution

Answer

Therefore, using the disk method, it's verified that the volume of a sphere is \(\frac{4}{3} \pi r^{3}\)

Step 1: Visualize the Sphere as a Rotation of Disks

Picture the sphere as a set of infinitely thin disks stacked together. The sphere can be obtained by rotating a semi-circle on the xy-plane around the x-axis. The equation for such a semi-circle is \( y = \sqrt{r^2 - x^2} \). Here, the radius, \( r \), is a constant, while \( x \) varies from \( -r \) to \( r \). The volume element is a thin disk with thickness \( dx \), at a distance \( x \) from the origin, and with radius \( \sqrt{r^2 - x^2} \). Therefore, the volume of this disk is \( \pi [r^2 - x^2] dx \).

Step 2: Integrate to Find the Volume

The total volume of the sphere, \( V \), is the integral of the volume of these infinitely thin disks as \( x \) varies from \( -r \) to \( r \). That is \( V = \int_{-r}^{r} \pi (r^2 - x^2)dx \). This integral can be split into two parts: \( V = \int_{-r}^{r} \pi r^2 dx - \int_{-r}^{r} \pi x^2 dx \). The first integral gives \( \pi r^2 x \) which evaluates to \( 2 \pi r^3 \) when the limits of \( -r \) and \( r \) are substituted. The second integral is \( \frac{1}{3} \pi x^3 \) when evaluated from \( -r \) to \( r \) gives \( \frac{2}{3} \pi r^3 \). Therefore, the total volume is \( 2 \pi r^3 - \frac{2}{3} \pi r^3 \).

Step 3: Simplify the Expression

Simplify the expression to get the final volume, \( V = \frac{4}{3} \pi r^{3} \)

Key Concepts

Disk Method

When it comes to understanding the geometry of three-dimensional shapes, the disk method is a valuable tool that can simplify complex volume calculations. It involves slicing the shape into a series of very thin disks or washers and then adding up the volumes of these disks to find the total volume. Imagine slicing a loaf of bread; each slice represents a disk.

For a sphere, which is perfectly symmetrical, the radius of each disk can be expressed as a function of its position along the x-axis. If you visualize a semi-circle with radius \( r \), the radius of the disks changes as you move from the bottom to the top of the sphere. Specifically, the radius of a disk at a distance \( x \) from the center is given by \( y = \[\sqrt{r^2 - x^2}\] \). To calculate the volume of a single disk, we multiply its area (\(\pi y^2 \)) by its infinitesimally small thickness (\( dx \)). Summing up all these minuscule volumes from one side of the sphere to the other offers the sphere's total volume. This is where integral calculus steps in to streamline the process.

Integral Calculus

Integral calculus is all about the accumulation of quantities and the determination of total size or value, such as areas under curves and, as we've seen, volumes of solids. By definition, an integral represents the area under a curve on a graph, which translates to summing an infinite number of infinitesimally small data points across a particular interval.

In the context of finding the volume of a sphere using the disk method, integral calculus allows us to calculate the volume of infinitely many disks by integrating their areas across the limits that define the sphere. In technical terms, we're implementing a definite integral, where we calculate the integral of the function representing the disk area from \( x = -r \) to \( x = r \), the limits spanning the entire diameter of the sphere. This process provides the exact volume of the sphere with just a few steps of calculation, illustrating the power of integral calculus in transforming complex summation problems into manageable ones.

Volume Integration

Volume integration is a specific application of integral calculus focused on finding the volume of solid objects with clearly defined bounds. To obtain the volume of a 3D shape like a sphere using the disk method, volume integration takes into account the changing dimensions of each disk. Instead of working with the full 3D object, volume integration decomposes it into 2D cross-sections that are easy to calculate.

The process requires setting up an integral that represents the sum of the volumes of all the individual disks. For the sphere, we integrate the area of a circle (which is the base of each disk) across the interval that represents the sphere's diameter. This calculation is not just pure mathematics but a powerful tool that demonstrates the practicality of calculus in solving real-world problems by decomposing them into simpler, more understandable parts. Without volume integration, calculating the volume of complex shapes would often be much more challenging, if not practically impossible.