Problem 44 - Single Variable Calculus

Question

In Exercises \(43-46\), sketch a plane region, and indicate the axis about which it is revolved so that the resulting solid of revolution has the volume given by the integral. (The answer is not unique.) $$ \pi \int_{0}^{1} y^{2 / 3} d y $$

Step-by-Step Solution

Answer

The plane region can be represented by the function \(f(y) = y^{1/3}\) for the interval \(0 \leq y \leq 1\). When this region is revolved around the \(y\)-axis, the resulting solid of revolution has a volume given by the integral \(\pi \int_{0}^{1} (y^{1 / 3})^{2} \, dy\).

Step 1: Find the Plane Region

First, we need to find a function whose square will represent the area of the region when multiplied by \(\pi\). Since the integral includes a power function, \(y^{2/3}\), we can assume that a possible function is \(f(y) = y^{1/3}\). Thus, we can rewrite the integral as: \[ \pi \int_{0}^{1} (y^{1 / 3})^{2} \, dy \] Now, let's sketch the plane region for \(f(y) = y^{1/3}\) for the interval \(0 \leq y \leq 1\).

Step 2: Sketch the Plane Region and the Axis of Revolution

The graph of the function \(f(y) = y^{1/3}\) has an upward curve that intersects the origin and the point \((1,1)\) on the \(x\)-axis. We will revolve the region enclosed by the function, the \(y\)-axis, and the interval \(0 \leq y \leq 1\), around the \(y\)-axis. By revolving the region around the \(y\)-axis, we create a solid of the revolution. The volume of this solid is given by the provided integral.

Step 3: Confirm the Solid's Volume

By using the disk method, we can find the volume of the solid as: \[ V = \pi \int_{0}^{1} R(y)^2 \, dy \] In this case, the radius of the solid, \(R(y)\), is given by the function \(f(y) = y^{1/3}\), and the volume can be calculated as: \[ V = \pi \int_{0}^{1} (y^{1 / 3})^{2} \, dy \] This is the same integral we started with, so our choice of plane region and the axis of revolution yields a solid with the desired volume.

Key Concepts

Volume Calculations

Understanding how to calculate the volume of a solid of revolution is a fundamental concept in calculus. In simple terms, we are interested in finding the amount of space inside a 3-dimensional shape created by revolving a 2-dimensional area around an axis.
One common method to compute such a volume is to slice the solid into smaller, simpler shapes—like disks, washers, or shells—and then sum their volumes.

For instance, when dealing with solids formed by revolving a function around an axis, we can utilize the integral:

• \[ V = \int_{a}^{b} A(x) \, dx \]
• Here, \(A(x)\) represents the cross-sectional area, which often is a circle (for the disk method), so \(A(x) = \pi R(x)^2\)

The integration bounds \(a\) and \(b\) specify the interval over which we are revolving the region.
By understanding this calculation method, students can efficiently find volumes for different solids, as it ties in concepts of areas, functions, and integration.

Disk Method

The disk method is a popular technique for finding the volume of a solid of revolution, particularly useful when the solid is symmetrical around an axis. This method involves slicing the solid perpendicular to the axis of revolution, resulting in many disk-shaped cross-sections.

If you imagine these disks stacked along an axis, the volume of the entire solid can be visualized as the sum of their individual volumes, each contributing a small, cylindrical sliver to the total volume.

To apply this method mathematically, consider the integral:

• \[ V = \int_{a}^{b} \pi [R(y)]^2 \, dy \]
• The term \(\pi [R(y)]^2\) represents the area of each disk's circular face

The limits \(a\) and \(b\) indicate the part of the graph being considered for rotation.
This approach is particularly effective when dealing with functions that can be easily expressed in terms of one variable along the axis of rotation.

Plane Region

A key step in calculating the volume of a solid of revolution is understanding and correctly identifying the plane region to revolve. In many exercises, the function provided will dictate this region.
For instance, consider the function \(f(y) = y^{1/3}\), which, when revolved around the \(y\)-axis from \(y = 0\) to \(y = 1\), forms a specific solid of revolution.

The plane region is the part of the graph that, when rotated, forms the intended 3D shape. To visualize this, plot the function within the specified bounds and identify the enclosed area.

It's vital to properly map out these regions, as revolving incorrect or partial sections may produce undesired shapes or volumes. Thus, correctly sketching and identifying the plane region ensures that the integral used in calculation reflects the actual solid being analyzed.
Understanding this concept simplifies tasks across calculus and helps in accurately solving problems related to volumes of solids of revolution.