Question

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

Short answer

The volume of the solid is \(\pi (\sqrt{2} - 1)\).

Step-by-step solution

Set Up the Volume Integral

To find the volume of the solid formed by revolving a region about the y-axis, we use the method of cylindrical shells. The volume of a small shell with thickness \(dx\), radius \(x\), and height \(h(x)\) is \(2\pi x h(x) dx\). Here, the height \(h(x)\) is the absolute difference between \(\sin(x^2)\) and \(\cos(x^2)\), thus \(|\cos(x^2) - \sin(x^2)|\) because \(\cos(x^2)\) is above \(\sin(x^2)\) within the region where they intersect.

Find Points of Intersection

Identify the range of integration (points of intersection) by solving the equation \(\sin(x^2) = \cos(x^2)\). In the first quadrant, this occurs when \(x^2 = \frac{\pi}{4} \), so \(x = \frac{\sqrt{\pi}}{2}\). Hence, our integration range is from \(x = 0\) to \(x = \frac{\sqrt{\pi}}{2}\).

Write the Integral Expression for Volume

Using the range from 0 to \(\frac{\sqrt{\pi}}{2}\), the volume \(V\) is expressed as: \[ V = \int_{0}^{\frac{\sqrt{\pi}}{2}} 2\pi x (\cos(x^2) - \sin(x^2)) \, dx \] because the cosine function is above the sine function within the range.

Simplify the Integral Expression

Factor out \(2\pi\) for clarity:\[ V = 2\pi \int_{0}^{\frac{\sqrt{\pi}}{2}} x (\cos(x^2) - \sin(x^2)) \, dx \] Use substitution for simplification. Let \(u = x^2\), which gives \(\frac{du}{dx} = 2x\) or \(x\,dx = \frac{1}{2}\,du\). Adjust the bounds accordingly: when \(x=0, u=0\) and when \(x=\frac{\sqrt{\pi}}{2}, u=\frac{\pi}{4}\). The integral becomes: \[ V = \pi \int_{0}^{\frac{\pi}{4}} (\cos u - \sin u) \, du \]

Evaluate the Integral

Compute the integral: \[ \int (\cos u - \sin u) \, du = \sin u + \cos u \]Evaluate this expression from 0 to \(\frac{\pi}{4}\): \[ \left[ \sin u + \cos u \right]_{0}^{\frac{\pi}{4}} = (\sin(\frac{\pi}{4}) + \cos(\frac{\pi}{4})) - (\sin 0 + \cos 0) \] \[ = \left( \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \right) - (0 + 1) \] \[ = \sqrt{2} - 1 \] Thus, the volume \(V\) becomes:\[ V = \pi (\sqrt{2} - 1) \]

Key concepts

Volume of Solids

Calculating the volume of solids formed by revolving a region around an axis is a fundamental concept in calculus. This is commonly applied in problems where a two-dimensional area is rotated around a line to create a three-dimensional object, useful in engineering and physics applications. In our case, a region bound by trigonometric functions is revolved about the y-axis.

To find the volume of this solid, we can use a specific technique that helps bridge the two-dimensional shape and the resulting solid form. Two common methods include the disk method and the method of cylindrical shells. For our exercise, the latter is more applicable due to the configuration and axis of revolution of the given region. The idea is to break the solid into a series of infinitesimally thin cylindrical shells.

Understanding the initial setup of the problem is crucial. Establishing the boundaries and intersection points of the curves helps define the limits of integration, crucial for accurate volume calculation.

Cylindrical Shells

The method of cylindrical shells is a fantastic integration technique used to find the volume of a solid of revolution, particularly effective when revolving around the y-axis or an axis parallel to it. Compared to other methods, this approach can be especially convenient when the solid involves vertical segments.

In this approach, the idea is to think about slicing the solid as a series of hollow cylindrical 'shells.' Imagine peeling these shells off as you would layers from an onion. Each of these layers has a certain radius, height, and thickness. The volume of each shell is computed and then summed (integrated) over the chosen limits.

• **Radius:** Given by the distance from the axis of rotation; in this problem, the radius is the x-coordinate.
• **Height:** Derived from the difference between the higher and lower function values. Here it is \(|\cos(x^2) - \sin(x^2)|\).
• **Thickness:** Simply the infinitesimally small dx.

This method requires setting up an integral that sums the volumes of these shells to find the total volume.

Integration Techniques

Integration is at the heart of finding the volumes of solids through calculus. It allows the computation of the continuous summation of infinitesimal elements. Here, the integration process starts by expressing the problem appropriately for integral setup, looking especially at bounds and simplifying where necessary.

The task at hand involves not just the basic process of integrating a simple function but doing so over a bounded interval determined by the intersection of trigonometric functions. Also, we leverage substitution to simplify and evaluate the integral effectively.

In this exercise, the substitution technique is used—choosing a part of the function as a new variable \(u = x^2\) simplifies the integrand since it involves a trigonometric component. This method reduces complexity and makes the integration feasible over a transformed interval. Additionally, integrating trigonometric functions starts from understanding and manipulating standard identities and integrals like sine and cosine.

Trigonometric Functions

Trigonometric functions such as sine and cosine often appear in calculus, particularly in problems involving intervals and areas under curves. These functions have distinct properties and relationships pivotal for setting up and solving integrals.

In this exercise, we deal with sine and cosine functions squared in their arguments, integrating these through products, and correctly applying trigonometric identities. The key point is understanding how these functions interact over their period and how to handle them within the integration.

• The equation \(\sin(x^2) = \cos(x^2)\) sets the points of intersection, crucial for determining integration bounds.
• We consider \(x^2 = \frac{\pi}{4}\) to derive limits and ensure that we rotate and encompass the correct area within valid conditions.

Trigonometric identities, like the fundamental Pythagorean identity, can assist in simplifying problems and ensuring solutions are reached correctly.