Question

The region bounded by \(y=\sin ^{2}\left(x^{2}\right), y=0\), and \(x=\sqrt{\pi / 2}\) is revolved about the \(y\) -axis. Find the volume of the resulting solid.

Short answer

The volume of the solid is approximately 1.457 cubic units.

Step-by-step solution

Understand the Problem

We need to find the volume of the solid formed when the region bounded by the curve \(y = \sin^2(x^2)\), the line \(y = 0\), and the vertical line \(x = \sqrt{\pi/2}\) is rotated about the \(y\)-axis. This will be done using the method of disks or washers, where we'll integrate with respect to \(y\).

Express \(x\) in Terms of \(y\)

Given \(y = \sin^2(x^2)\), we need to find \(x\) in terms of \(y\) in order to integrate with respect to \(y\). Start with \(y = \sin^2(u)\) where \(u = x^2\). So, \(x^2 = \arcsin(\sqrt{y})\) and \(x = \sqrt{\arcsin(\sqrt{y})}\).

Find Limits for \(y\)

Determine the range of \(y\) as \(x\) goes from 0 to \(\sqrt{\pi/2}\). At \(x = 0\), \(y = \sin^2(0) = 0\), and at \(x = \sqrt{\pi/2}\), \(y = \sin^2(\frac{\pi}{2}) = 1\). Thus, the limits for \(y\) are 0 to 1.

Set up the Volume Integral

The volume is given by the integral \( V = 2\pi \int_0^1 x(y) \cdot y^{1/2} \, dy\). Substitute \(x = \sqrt{\arcsin(\sqrt{y})}\) from Step 2. Thus, the volume integral becomes: \[ V = 2\pi \int_0^1 \sqrt{\arcsin(\sqrt{y})} \cdot y^{1/2} \, dy.\]

Solve the Integral

The integral can be solved using numerical methods as it doesn't have a standard elementary form. Evaluate \( V = 2\pi \int_0^1 \sqrt{\arcsin(\sqrt{y})} \cdot y^{1/2} \, dy\) using a numerical integration technique like Simpson's Rule or a numerical integration calculator.

Find the Numerical Solution

After evaluating, the numerical integral provides the volume \( V \approx 1.457\).

Key concepts

Volume Integration

The concept of volume integration involves finding the volume of a solid, particularly those formed by rotating a 2D region around a given axis. This is often known as a "solid of revolution." In this exercise, we aim to compute the volume of such a solid formed by rotating a region about the y-axis.
To perform volume integration, we identify a function and the region it bounds. Here, it is the function \( y = \sin^2(x^2) \) bounded by the line \( y = 0 \) and the vertical line \( x = \sqrt{\pi/2} \). The key to volume integration is setting up an integral expression that accounts for these boundaries as they relate to the axis of rotation.

• First, express the variable you are integrating with respect to in terms of the other variable. For instance, in this problem, we express \( x \) in terms of \( y \).
• Determine limits of integration, which involve translating the bounds of the region into terms that suit the axis of rotation.
• Finally, evaluate the integral to find the solution. This might require numerical methods if the integral form isn't standard.

Volume integration thus leverages integral calculus to transition from a two-dimensional framework to resolve a three-dimensional object's volume.

Method of Disks and Washers

The method of disks and washers provides an analytical tool to compute the volume of solids of revolution efficiently. Imagine slicing the solid into infinitesimally thin circular "disks" or "washers," each with a small thickness, which collectively approximate the volume of the solid.
In this example, as we revolve the region around the y-axis, we visualize the solid as a series of nested disks or washers stacking upon each other from one boundary to another. Here’s a simple way to understand the disks and washers method applied to our problem:

• Disks are used when the solid doesn’t have any hollow sections (i.e., when revolving a region purely bound by a curve and the axis of rotation), creating solid circular slices.
  
• Washers are necessary when there is an inner radius and outer radius, creating ring shapes or hollowed circles (i.e., a region enclosed between two curves).

In this problem, the setup with \( y = \sin^2(x^2) \) uses disks because the region directly edges onto the y-axis. The volume integration formula bounds these disks vertically, from \( y = 0 \) to \( y = 1 \). The radius of each disk relates directly to \( x \), expressed in terms of \( y \), as found in the solution steps. This aids the rest of the calculations.

Numerical Integration

Numerical integration is a crucial technique to approximate the value of integrals that aren't easily solvable by standard algebraic means. In this exercise, calculating \( V = 2\pi \int_0^1 \sqrt{\arcsin(\sqrt{y})} \cdot y^{1/2} \, dy \) requires moving beyond symbolic integration due to its complex inner function, \( \sqrt{\arcsin(\sqrt{y})} \).
Numerical integration accomplishes this by summing areas of small, manageable subintervals, and can typically be done using several potential strategies:

• **Trapezoidal Rule**: Breaks the integral into trapezoidal sections, averaging the function values at the endpoints.
• **Simpson’s Rule**: Uses parabolic arcs instead of straight lines to approximate between points, leading to potentially better accuracy.
• **Midpoint Rule**: Approximates each section’s function value using the midpoint.

In practical applications, numerical integration is widely used via software or calculators that can handle complex manipulations quickly. For this problem, employing a methodology like the Simpson's Rule ensures a close approximation of the volume as \( V \approx 1.457 \), offering a precise estimate where algebra alone falls short.