Problem 36 - Thomas Calculus in SI Units

Question

Find the volumes of the solids generated by revolving the regions bounded by the lines and curves about the \(x\) -axis. $$y=2 \sqrt{x}, \quad y=2, \quad x=0$$

Step-by-Step Solution

Answer

The volume of the solid generated is \(2\pi\).

Step 1: Identify the Bounded Region

First, let's identify the region that we are going to revolve around the x-axis. The curves and lines given are \(y = 2\sqrt{x}\), \(y = 2\), and \(x = 0\). The region is bounded underneath \(y = 2\sqrt{x}\), above by \(y = 2\) and to the left by \(x = 0\).

Step 2: Find the Intersection Points

Find the x-values where the curves intersect, specifically where \(y = 2\sqrt{x}\) meets \(y=2\). Set \(2\sqrt{x} = 2\). Solving gives \(\sqrt{x} = 1\), so \(x = 1\). Therefore, the region is bounded between \(x = 0\) and \(x = 1\).

Step 3: Set Up the Volume Integral Using the Disk Method

The disk method formula for volume is \( V = \pi \int_{a}^{b} [f(x)]^2 \, dx \). Here \(f(x) = 2\sqrt{x}\), \(a = 0\), and \(b = 1\).

Step 4: Evaluate the Integral

Compute the integral: \[V = \pi \int_{0}^{1} (2\sqrt{x})^2 \, dx = \pi \int_{0}^{1} 4x \, dx = 4\pi \left[ \frac{x^2}{2} \right]_{0}^{1} = 4\pi \left( \frac{1^2}{2} - \frac{0^2}{2} \right) = 2\pi.\]

Step 5: Interpret the Result

The result of the integral gives the volume of the solid formed by revolving the region under \(y = 2\sqrt{x}\) between \(x = 0\) and \(x = 1\) about the x-axis. It is calculated to be \(2\pi\).

Key Concepts

Disk Method

The disk method is a technique used in integral calculus to find the volume of a solid of revolution. This involves revolving a region around an axis, creating a three-dimensional object. Let's imagine taking a thin slice of the region, perpendicular to the axis of rotation. These slices form disks. To determine the volume, calculate the volume of each disk and sum them up along the interval.

• Each disk's volume is like stacking up many tiny cylinders.
• The formula for the area of a circle is \( \pi r^2 \), and here, \(r = f(x)\) because f(x) is the function that defines the boundary of the region.
• The width of each disk is \(dx\), an infinitesimally small length.

This leads us to the volume formula: \( V = \pi \int_{a}^{b} [f(x)]^2 \, dx \). This integral sums up the volumes of all disks from point \(a\) to \(b\) on the \(x\)-axis. It's an effective way to calculate volumes of solids that are symmetrical around a given axis.

Integral Calculus

Integral calculus is a branch of mathematics focused on finding volumes, areas, and sums, often requiring solving integrals. In the case of the disk method, we use definite integrals to find the volume of the solids formed by revolving a region around an axis.When using integral calculus here:

• We set up an integral based on a function \(f(x)\) which defines the curve's boundary.
• The definite integral helps evaluate the total accumulated change, in this context, resulting in volume.
• The bounds of integration \(a\) and \(b\) mark where the region starts and ends on the \(x\)-axis.

The result of solving this integral provides the exact volume of the solid created. It is one of the powerful applications of integral calculus for solving real-world geometric problems.

Intersection Points

Intersection points are crucial in setup when calculating volumes of solids of revolution as they define the limits of integration. These are the x-values where two curves meet and determine the region that needs to be revolved.In this context:

• We find points where \(y = 2\sqrt{x}\) and \(y = 2\) intersect.
• Setting \(2\sqrt{x} = 2\), solve to find \(x = 1\), where these curves meet.
• This tells us that the region of interest is bounded between \(x = 0\) and \(x = 1\).

Using these intersection points ensures the correct setup of integral bounds, or limits, and accurate volume calculation. This step should not be overlooked as it informs subsequent calculations and eventual results.