Problem 56 - Calculus of a Single Variable

Question

Dividing a Solid In Exercises 55 and 56 , consider the solid formed by revolving the region bounded by \(y=\sqrt{x}, y=0,\) and \(x=4\) about the \(x\) -axis. Find the value of \(x\) in the interval \([0,4]\) that divides the solid into two parts of equal volume. Find the values of \(x\) in the interval \([0,4]\) that divide the solid into three parts of equal volume.

Step-by-Step Solution

Answer

The value of \(x\) that divides the solid into two parts of equal volume is approximately 3.17. The values of \(x\) that divide the solid into three parts of equal volume are approximately 2.3 and 4.

Step 1: Define the volume of solid of revolution

First, understand the concept of solid of revolution. The volume \(V\) of the solid of revolution around the \(x\)-axis is given by the formula \(V = \pi \times \int_a^b f(x)^2 dx\), where \(f(x)\) is the equation of the curve, here \(f(x) = \sqrt{x}\), and \([a,b]\) is the limits of integration, in this case \([0,4]\).

Step 2: Calculate the total volume of the solid

Substitute the given function \(f(x) = \sqrt{x}\) and limits \([a,b] = [0,4]\) into the defined formula, and solve the integral. This results in \(V = \pi \times \int_0^4 x dx = \pi \times [\frac{x^{3/2}}{3/2} ]_0^4 = \frac{32\pi}{3}\).

Step 3: Determine the value that divides the solid into two equal parts

Set up an equation with the integral from 0 to \(x\) equal to half of the total volume we found in Step 2. So, \(\pi \times \int_0^x x dx = \frac{16\pi}{3}\). Solve this equation to determine the \(x\) value. Solving, we obtain \(x = 8^{2/3} \approx 3.17\)

Step 4: Determine the values that divide the solid into three equal parts

Similar to Step 3, now set up two separate equations, each one equal to the one-third of total volume, \(\frac{32\pi}{9}\). The first equation has limits [0, \(x\)] and the second one [\(x\), \(y\)] where \(x \neq y\). Solve these equations to find the \(x\) and \(y\) values which divide the solid into three equal parts. Solving these equations will yield \(x = 24^{2/3}\) and \(y = 4\).

Key Concepts

volume integration

Volume integration is a crucial concept when calculating the volume of a solid of revolution. When a region is revolved around an axis, it forms a three-dimensional shape, and its volume can be found using calculus through a process called integration. This involves integrating the area function over a specific interval, which is determined by the range of rotation.

For this exercise, the formula used is:

• \( V = \pi \times \int_a^b f(x)^2 dx \)

This formula is derived from the idea of accumulating infinitely small disks along the rotation direction.

Substituting in our function, \(f(x) = \sqrt{x}\), and the limits [0,4], the integration finds the total volume:

• \( V = \pi \times \int_0^4 x \,dx = \frac{32\pi}{3} \)

This result provides the step towards finding parts that divide the solid into sections of equal volume.

revolved solid

Creating a revolved solid is a fascinating geometric operation. It involves taking a two-dimensional shape or region and spinning it around an axis to form a solid. In this setup, the region bounded by the curve \(y = \sqrt{x}\), the x-axis, and the vertical line \(x = 4\) around the x-axis creates our solid.

This rotation transforms the flat region into a three-dimensional shape that must be visualized to understand how integration calculates its volume. By revolving these regions, each horizontal section or 'slice' becomes a disk—a flat, circular shape—with a thickness that is infinitesimally small.
This process provides a way to computationally sum these disks' volumes to yield the total volume of the rotated shape.

equal volume division

Understanding equal volume division necessitates splitting the entire solid into distinct sections that each hold the same volume. The primary task is to find specific x-values along the curve that equally divide the total volume obtained.

For instance, when dividing the solid into two equal parts, we set up and solve the equation for the integral from 0 to some point \(x\) so that it equals half the solid's full volume:

• \( \pi \times \int_0^x x \,dx = \frac{16\pi}{3} \)

Solving this gives us \(x \approx 3.17\).

Similarly, to divide the solid into three equal parts, we set the volume from 0 to \(x\) and \(x\) to \(y\) where each integral equals one-third of the total volume:

• \( \pi \times \int_0^x \sqrt{x}^2 \,dx = \frac{32\pi}{9} \)
• \( \pi \times \int_x^y \sqrt{x}^2 \,dx = \frac{32\pi}{9} \)

Solving these equations yields specific values for \(x\) and \(y\) that ensure each section holds an equal share of volume.

integration limits

When dealing with integration, setting the correct integration limits is essential. These limits define the bounds within which the function is integrated, determining the start and end points for calculating the area beneath the curve.

In this problem, the curve is bounded by

• from \(x = 0\) which is the lower limit,
• to \(x = 4\) which serves as the upper limit.

These limits determine the region being revolved to form the solid.

When dividing the solid into smaller volumes, different limits come into play to solve for each segment:

• The first from 0 to \(x\)
• and the second between \(x\) and another point \(y\), adopted from the total interval [0,4].

Such precision in setting these limits is crucial for correctly finding points that divide the solid into equal parts.