Problem 38 - Thomas Calculus

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 is \(2\pi\).

Step 1: Understand the Problem

We need to find the volume of a solid that is created by rotating the area between three curves around the x-axis. These curves are the function \(y = 2\sqrt{x}\), the line \(y = 2\), and the line \(x = 0\).

Step 2: Determine the Boundaries

The curve \(y = 2\sqrt{x}\) reaches \(y = 2\) when \(x = 1\). This gives the boundaries for integration as \(x = 0\) to \(x = 1\).

Step 3: Set Up the Integral for the Volume

Since we are rotating around the x-axis, we use the disk method. The formula for the volume \(V\) is \[ V = \pi \int_{a}^{b} f(x)^2 \, dx \] where \(f(x) = 2\sqrt{x}\) and our limits \(a = 0\), \(b = 1\).

Step 4: Substitute the Function into the Integral

Substituting \(f(x) = 2\sqrt{x}\) into the formula gives us:\[ V = \pi \int_{0}^{1} (2\sqrt{x})^2 \, dx \] Simplifying inside the integral, we have:\[ (2\sqrt{x})^2 = 4x \]

Step 5: Integrate the Function

Now compute the integral:\[ V = \pi \int_{0}^{1} 4x \, dx \] \[ = \pi [2x^2]_{0}^{1} \] Evaluating the definite integral:\[ = \pi (2(1)^2 - 2(0)^2) \] \[ = \pi \times 2 \] \[ = 2\pi \]

Step 6: Conclusion

The volume of the solid obtained by rotating the region bounded by \(y = 2\sqrt{x}\), \(y = 2\), and \(x = 0\) around the x-axis is \(2\pi\).

Key Concepts

Volume of Solids

In calculus, calculating the volume of solids is a key application. When a region on a plane is spun around a line, such as an axis, it sweeps out a 3-dimensional shape.
Understanding the geometry of these shapes helps in determining their volumes using integrals.
This particular exercise involves finding the volume of a solid formed by revolving a plane region around the x-axis. The boundary lines and curves shape the region, and as it revolves, a solid is formed.
To solve this, one must calculate, how much space this 3-dimensional object occupies.

Disk Method

The disk method is a straightforward technique for finding volumes of solids of revolution. It is especially useful when the region is rotated around the x-axis or y-axis.
The idea is to slice the solid into many thin disks, or circles, perpendicular to the axis of rotation. Each disk has a small width, called \(dx\), and a radius given by the function that describes the curve.
By summing up the volumes of these disks, using integration, the total volume of the solid is found.

• The radius of each disk is determined by the distance from the axis to the curve.
• The volume of each disk is \(\pi f(x)^2 dx\).
• The integral sums up all these small volumes from one boundary to the other.

Integral Calculation

Integrating is crucial in calculus as it allows calculation of accumulated areas, volumes, and other quantities. In this example, the integral calculates the sum of the slices’ volumes to find the total volume of the solid.
When you see an integral in the solution, it's essentially adding up many tiny, infinitesimal pieces.
Integral calculus involves solving these calculations to find precise values. Here, our task is to find the integral of the square of the function from the starting point to the endpoint, within established boundaries.

Definite Integral

A definite integral, unlike an indefinite integral, computes a number representing the accumulation over a specific interval. It gives us a precise quantity, such as area under a curve or volume.
In the context of this exercise, the definite integral tells us the volume of the solid formed by the rotation.

• The limits of integration define the section of the curve being turned into a solid.
• For this problem, the boundaries are \(x = 0\) and \(x = 1\).
• Evaluating the definite integral involves substituting these limits into the antiderivative.

This result offers the exact volume of the solid.

Curve Rotation

Rotating a curve creates symmetrical 3D objects around an axis. This concept is central to problems in calculus involving solid volumes.
By revolving functions around an axis, you modulate the 2D shape into a complete solid form.
Here, the curve given by \(y = 2\sqrt{x}\) is rotated around the x-axis, forming a solid that resembles a paraboloid.

• This rotation stretches the curve into a surface that defines the solid’s shape.
• The entire strip of area from the curve to the axis contributes to this formation.
• The simplicity of calculations often increases when using symmetry in rotational solids.

Understanding how curves turn into geometric solids through rotation is crucial for solving such problems.