Problem 124 - Thomas Calculus in SI Units

Question

Find the volumes of the solids. The solid lies between planes perpendicular to the \(x\) -axis at \(x=-\sqrt{2} / 2\) and \(x=\sqrt{2} / 2 .\) The cross-sections perpendicular to the \(x\) -axis are a. circles whose diameters stretch from the \(x\) -axis to the curve \(y=2 / \sqrt[4]{1-x^{2}}\). b. squares whose diagonals stretch from the \(x\) -axis to the curve \(y=2 / \sqrt[4]{1-x^{2}}\).

Step-by-Step Solution

Answer

a. \(\frac{\pi^2}{2}\); b. \(\pi\).

Step 1: Understanding the Region for Integration

The problem indicates that the solid lies between planes at \(x = -\frac{\sqrt{2}}{2}\) and \(x = \frac{\sqrt{2}}{2}\) with cross-sections perpendicular to the \(x\)-axis being defined by the curve \(y = \frac{2}{\sqrt[4]{1-x^2}}\). We need to integrate from \(x = -\frac{\sqrt{2}}{2}\) to \(x = \frac{\sqrt{2}}{2}\).

Step 2: Volume Calculation for Circular Cross-Sections

For part (a), the cross-sections are circles with a diameter from the curve to the \(x\)-axis. The diameter is the \(y\)-value, \(D = \frac{2}{\sqrt[4]{1-x^2}}\), so the radius \(r\) is \(\frac{1}{\sqrt[4]{1-x^2}}\). The area \(A\) of a circle is given by \(A = \pi r^2\). Therefore, \(A = \pi \left(\frac{1}{\sqrt[4]{1-x^2}}\right)^2 = \frac{\pi}{\sqrt{1-x^2}}\). Integrating from \(x = -\frac{\sqrt{2}}{2}\) to \(x = \frac{\sqrt{2}}{2}\) gives the volume: \[ V = \int_{-\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}} \frac{\pi}{\sqrt{1-x^2}} \, dx \].

Step 3: Integration of Circular Cross-Section

To solve \(\int \frac{1}{\sqrt{1-x^2}} \, dx\), recognize it as the derivative of \(\sin^{-1}(x)\). Thus, \[ V = \pi \left[ \sin^{-1}(x) \right]_{-\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}} \]. The definite integral evaluates to \(\pi \times \left( \sin^{-1}\left(\frac{\sqrt{2}}{2}\right) - \sin^{-1}\left(-\frac{\sqrt{2}}{2}\right) \right)\). Since \(\sin^{-1}\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}\), the volume is \(\pi \times \frac{\pi}{2} = \frac{\pi^2}{2}\).

Step 4: Volume Calculation for Square Cross-Sections

For part (b), each cross-section is a square with diagonal equal to \(\frac{2}{\sqrt[4]{1-x^2}}\). If \(d\) is the diagonal, then each side \(s\) of the square is given by \(s = \frac{d}{\sqrt{2}}\). Thus, \(s = \frac{2}{\sqrt{2}\cdot \sqrt[4]{1-x^2}} = \sqrt{2} \cdot \frac{1}{\sqrt[4]{1-x^2}}\). Hence, the area of the square is \(A = s^2 = 2\cdot \frac{1}{\sqrt{1-x^2}}\). Intgrate to find the volume: \[ V = \int_{-\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}} 2 \cdot \frac{1}{\sqrt{1-x^2}} \, dx \].

Step 5: Integration of Square Cross-Section

This integral is similar to the previous one. We know \[ \int \frac{1}{\sqrt{1-x^2}} \, dx = \sin^{-1}(x) \]. Thus, \[ V = 2 \cdot \left[ \sin^{-1}(x) \right]_{-\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}} \].We already computed \(\left[ \sin^{-1}(x) \right]_{-\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}} = \frac{\pi}{2}\). So the volume of the solid is \(2 \cdot \frac{\pi}{2} = \pi\).

Key Concepts

Integral Calculus

Integral Calculus helps us find the volume of complex shapes, like the solid formed by revolving a region around an axis. It extends our understanding of summation into continuous domains, allowing for the precise calculation of areas, volumes, and other quantities. In contexts like ours, we derive the formula for the volume of a shape from an infinite sum of cross-sectional areas.

When you are dealing with solids of revolution, the integral represents this infinite sum along the specified axis with the specified limits. This means, by integrating the area of infinitesimally thin regions (or "slices"), we can calculate the entire volume between those boundaries. Calculating these volumes is often done through the use of definite integrals, which gives us the ability to evaluate using real numerical values.

To work out these calculations, particularly for volumes, we rely on the technique where the cross-sectional area, expressed as a function of x, is integrated with respect to x. Each tiny slice is summed from the start to the end of the solid's defined region, completing a full representation of the volume.

Cross-sectional Area

The cross-sectional area is crucial in determining the volume of a solid. Imagine slicing through a loaf of bread: each slice represents a cross-section. The shape of this cross-section affects how we calculate its area, and by extension, the volume of the solid.

In the exercise, the cross-sections of the solid are both circles and squares, each depending on the function given. For circles, the area is derived from the diameter given by the function, and for squares, the length of the diagonal as given by the function represents the cross-section in relation to the curve.

Understanding how to find and integrate these cross-sectional areas is vital.

• For circles, use the formula for the area of a circle, \(A = \pi r^2\), where the radius is linked to the diameter derived from the function.
• For squares, adjust the formula to account for diagonal lengths, involving basic geometry.

The integral of these areas gives the exact volume of the solid as the slices are stacked up over the interval.

Definite Integral

The use of definite integrals is key in finding the exact values of measures like area and volume. A definite integral \(\int_a^b f(x) \, dx\) boils down to calculating the "net area" under the curve between points \(a\) and \(b\).

This process gives us a finite number that represents accumulated values across these bounds.

• It is an extension of sums to an infinitesimally small scale, allowing precise calculations impossible with more straightforward arithmetic calculations.
• Definite integrals accumulate signed values, giving depth to calculations involving areas and volumes in calculus and physics.

In the exercise, the definite integral forms connect to trigonometric functions since this type of integration requires special techniques.

With a thorough grasp of definite integrals, you can determine how these complicated forms translate to defined, tangible measures like precise volumes.

Inverse Trigonometric Functions

Inverse trigonometric functions are used when you're finding integrals involving expressions like \(\frac{1}{\sqrt{1-x^2}}\). They inverse standard trigonometric functions, which means they "undo" what these functions do.

One such function is \(\arcsin(x)\) or \(\sin^{-1}(x)\), which is the inverse of the \(\sin(x)\) function. Here, it helps us resolve the integral for the volume of the solid. Understanding inverse trig functions is crucial because they often appear in calculus involving arcs or sections of circles.

• The integral \(\int \frac{1}{\sqrt{1-x^2}} \, dx\) utilizes \(\sin^{-1}(x)\) to convert the integral into a solvable problem.
• The outcome showcases the power and flexibility these functions offer in simplifying and solving real-world mathematical issues efficiently.

Getting familiar with these functions, their graphs, and properties helps to tackle integrations that initially seem complex at first glance.