Problem 156 - Calculus Volume 1

Question

For the following exercises, use technology to graph the region. Determine which method you think would be easiest to use to calculate the volume generated when the function is rotated around the specified axis. Then, use your chosen method to find the volume. \([T] x=\sin \left(\pi y^{2}\right)\) and \(x=\sqrt{2} y\) rotated around the \(x\) -axis.

Step-by-Step Solution

Answer

Use the washer method and integrate from \(y=0\) to \(y=1\) to find the volume.

Step 1: Understanding the Functions

First, identify the functions given: \(x = \sin(\pi y^2)\) and \(x = \sqrt{2} y\). Consider the domain for which these functions intersect. We aim to rotate this bounded region around the \(x\)-axis.

Step 2: Find Points of Intersection

Set the equations \(\sin(\pi y^2) = \sqrt{2} y\) to find points of intersection. Test values within feasible range points. Approximating graphically or numerically shows intersections at points \(y = 0\) and around \(y = 1\).

Step 3: Choose a Method for Volume Calculation

Next, decide whether to use the disk/washer method or the shell method. Since we are rotating around the \(x\)-axis, a disk or washer method is straightforward due to direct function expression in terms of \(y\).

Step 4: Set Up the Integral for Disk Method

For the disk method, volume \(V\) is calculated using the formula \[ V = \pi \int_{a}^{b} [(\text{outer radius})^2 - (\text{inner radius})^2] \, dy \]. Here, outer radius is \(x = \sqrt{2}y\) and inner radius is \(x = \sin(\pi y^2)\).

Step 5: Calculate the Volume Integral

Determine the limits of integration from the intersection points: \(y=0\) to approximately \(y=1\). The integral is then \[ V = \pi \int_{0}^{1} [(\sqrt{2}y)^2 - (\sin(\pi y^2))^2] \, dy \]. Simplify and compute the integral to find the volume.

Step 6: Evaluate the Integral

Perform the integration using a calculator or symbolic software: evaluate \([\pi (2y^2 - \sin^2(\pi y^2))]_{0}^{1}\). Calculate the final numerical value.

Key Concepts

Disk Method

The disk method is a straightforward way to find the volume of a solid of revolution, which is formed by rotating a region around an axis. When using the disk method, imagine taking thin circular disks, like slices of a cylinder, perpendicular to the axis of revolution. The volume of each disk is given by the area of the circle times its thickness. By summing up all these disks, we can approximate the volume. For an exact calculation, we integrate across the interval.If a region is bounded by functions, the formula for the volume is:

• Volume (\(V\)) = \( \pi \int_{a}^{b} [( ext{outer radius})^2 - ( ext{inner radius})^2] \, dy \),

where:

• Outer radius: Distance from the axis to the farthest edge of the slice.
• Inner radius: Distance from the axis to the nearest edge (if present).
• Interval [a,b]: Given by the points of intersection or limits along the axis of revolution.

Use this method when rotating around a horizontal or vertical line, especially the coordinate axes.

Integral Calculus

Integral calculus is a branch of mathematics that deals with the concept of an integral, either as the area under a curve, a surface, or a more complex physical quantity. It allows us to calculate quantities that accumulate, such as areas and volumes. When using the disk method derived from integral calculus:

• We're finding a volume by integrating across a range.
• The integral, \( \int \), is a symbol that represents the sum of infinite small components, which, in this case, are the disks' volumes.

For example, in our problem, the integral \( \pi \int_{0}^{1} [( ext{outer radius})^2 - ( ext{inner radius})^2] \, dy \) calculates the entire volume for the region defined between the intersection points.Calculus makes it possible to determine precise mathematical results, especially when the geometry would be too complicated to work with using basic arithmetic alone. Integrals are essential for continuous evaluation of areas, lengths, and volumes.

Intersection Points

Intersection points play a vital role in determining the limits of integration for volume calculations. These are the points where two curves intersect, defining the bounds of the region that will be revolved around an axis. Identifying these points often requires setting the functions equal to each other and solving for the variable in question.For example, in our exercise, the functions \( x = \sin(\pi y^2) \) and \( x = \sqrt{2} y \) intersect when \( \sin(\pi y^2) = \sqrt{2} y \). To find these intersection points, you might need to:

• Graph the functions to see where they cross visually.
• Use algebraic manipulation and, if needed, approximate these points using technology given that direct solutions might be complex.

Ultimately, these points give us the integration boundaries from \( y=0 \) to approximately \( y=1 \), deciding the slice of the area to be evaluated.

Graphing Technology

Graphing technology bridges the gap between theoretical calculus and practical computation. It allows for visualizing functions and their interactions, making it easier to analyze complex relations and intersections. Here are some ways graphing calculators and software can assist:

• Visualize functions: You can see where curves intersect and understand the region of interest.
• Approximate intersection points: Numeric solvers can provide close estimates for hard-to-solve equations.
• Check work: Ensure that integral limits and assumptions about the functions are accurate.

For students, software like Desmos or a graphing calculator can make these tasks simpler and more accurate, providing real-time insights and reducing computational errors. Combining technology with traditional mathematical techniques improves understanding and accuracy in problem-solving.