Problem 30 - Thomas Calculus

Question

Find the volumes of the solids generated by revolving the regions bounded by the lines and curves about the \(y\)-axis. The region enclosed by \(x=y^{3 / 2}, \quad x=0, \quad y=2\)

Step-by-Step Solution

Answer

The volume of the solid is \(4\pi\).

Step 1: Understand the problem

We need to find the volume of the solid formed by revolving the region bounded by the curve \(x = y^{3/2}\), the line \(x = 0\), and the horizontal line \(y = 2\) about the \(y\)-axis.

Step 2: Sketch the region

Draw the graph of \(x = y^{3/2}\) from \(y = 0\) to \(y = 2\) and the line \(x = 0\) which is the y-axis. The region is enclosed by these two lines up to \(y = 2\).

Step 3: Setup the integral using the disk method

Since we are revolving around the \(y\)-axis, and we have the boundaries in terms of \(y\), we use the disk method. The formula for the volume \(V\) will be \(V = \pi \int_{c}^{d} [f(y)]^2 \,dy\), where \(f(y)\) represents \(x = y^{3/2}\).

Step 4: Determine the limits of integration and the function

The limits of integration are from \(y = 0\) to \(y = 2\), and the function is \(f(y) = y^{3/2}\). Thus, the integral becomes \(V = \pi \int_0^2 (y^{3/2})^2 \,dy\).

Step 5: Simplify the integrand

Simplify \((y^{3/2})^2\) to \(y^3\), so the integral is now \(V = \pi \int_0^2 y^3 \,dy\).

Step 6: Integrate the function

Find the antiderivative of \(y^3\), which is \(\frac{y^4}{4}\). Evaluate this from \(y = 0\) to \(y = 2\).

Step 7: Evaluate the definite integral

Compute \(\left. \frac{y^4}{4} \right|_0^2 = \frac{2^4}{4} - \frac{0^4}{4} = \frac{16}{4} - 0 = 4\).

Step 8: Calculate the volume

Multiply the result from Step 7 by \(\pi\) to get the volume of the solid of revolution: \(V = 4\pi\).

Key Concepts

Disk Method

The disk method is a way to find the volume of a solid of revolution. It's particularly useful when the axis of rotation is along the boundary of the region. When using the disk method, you visualize slicing the solid into thin disks or circles. These disks have a small thickness and each disk's volume can be calculated.

• The radius of each disk is given by the function value at a specific point.
• In the exercise, the function was given in terms of y, because we revolved around the y-axis.
• The formula to calculate each disk's volume is elements: radius squared, times element_THICKNESS.

For a solid formed by rotating around the y-axis, the volume is found by integrating the function representing the x-values, squared. The method sets up as an integral of the form \( V = \pi \int [f(y)]^2 \, dy \). This is perfect for cases where the function is expressed in terms of y, just like our problem.

Integral Calculus

Integral calculus handles the concept of summing infinitely small parts to get a whole. It's indispensable for finding areas under curves, volumes of solids, and more. In our exercise, we used integral calculus to find the solid's volume by integration. - The problem involved finding the area of an infinitely small disk and then summing the areas to get the entire volume.To do this:

• Consider each infinitesimally small piece, represented as a mathematical function.
• Find the antiderivative, which is the function that describes accumulating these areas.

In the volume example, Integral calculus was used to perform the operation \( \int_0^2 (y^3) \,dy \).By solving the integral \( \int_0^2 y^3 \,dy = \frac{y^4}{4} \), we determine the accumulated volume under the curve fromy=0 toy=2. This provides the slice's mass, contributing to the total volume.

Solid of Revolution

A solid of revolution is created when a two-dimensional area is rotated around an axis. Imagine taking a flat shape and spinning it around a line to create a 3D object. That's a solid of revolution!

• In our problem, the solid was generated by rotating a region about the y-axis.
• The boundaries defined were the curve x = y^3/2, and the vertical line x = 0.

Visualizing this can help you understand the shape formed: - Imagine holding a shape along a line and spinning it. - As it turns, it "sweeps" out a volume. Using the disk method, we calculated how the slices of this revolved area contribute to the solid's full volume. This is crucial because it switches the problem from 2D areas to 3D objects.

Definite Integral

A definite integral is a key concept in calculus for computing overall quantities from change rates. Think of it like a magic tool that sums up continuous data.

• In our problem, it provided the total volume of the solid of revolution.
• The bounds of integration established the scale, or area to evaluate, between y=0 and y=2.

The process of evaluating a definite integral includes:- Finding the antiderivative.- Applying the limits of integration.- Calculating the difference.In simpler terms:Evaluate \( \int_0^2 y^3 \,dy \),which becomes \( \left. \frac{y^4}{4} \right|_0^2 = 4 \).This result is then multiplied by \( \pi \)to represent both the area summation across the axis and the initial volume computed by the integral.Thus, Definite Integrals enable precise calculation of the solid's full volume.