Problem 2 - Thomas Calculus in SI Units

Question

Find the volumes of the solids. The solid lies between planes perpendicular to the \(x\) -axis at \(x=-1\) and \(x=1 .\) The cross-sections perpendicular to the \(x\) -axis are circular disks whose diameters run from the parabola \(y=x^{2}\) to the parabola \(y=2-x^{2}\).

Step-by-Step Solution

Answer

The volume of the solid is \(\frac{16 \pi}{15}\).

Step 1: Identify the Cross-Sectional Radius

To find the volume, we must first understand the dimensions of each cross-sectional area. The cross-section is a circular disk with a diameter running from the first parabola, \(y = x^2\), to the second parabola, \(y = 2 - x^2\). The distance from one parabola to the other gives the diameter of the disk: \( ext{Diameter} = (2 - x^2) - x^2 = 2 - 2x^2 \). The radius \( r \) of the disk is half this diameter, so \( r = \frac{2 - 2x^2}{2} = 1 - x^2 \).

Step 2: Write the Formula for Volume

The volume of the solid is found by integrating the area of the cross-sectional circular disks along the \(x\)-axis, from \(x = -1\) to \(x = 1\). The area \(A\) of a circle with radius \(r\) is given by \(A = \pi r^2\). Substituting in the radius from Step 1, the area becomes \(A(x) = \pi(1 - x^2)^2\). The integral to find the volume \(V\) is given by \( V = \int_{-1}^{1}A(x)\,dx = \int_{-1}^{1}\pi(1 - x^2)^2\,dx \).

Step 3: Simplify the Integrand

Expand the expression \((1 - x^2)^2\) before integrating: \((1 - x^2)^2 = 1 - 2x^2 + x^4\). Therefore, the integrand becomes \( \pi (1 - 2x^2 + x^4) \).

Step 4: Integrate the Function

Set up the integral as \( V = \pi \int_{-1}^{1} (1 - 2x^2 + x^4) \, dx \). Integrate each term separately: \(1 \rightarrow x , \quad -2x^2 \rightarrow -\frac{2}{3} x^3 , \quad x^4 \rightarrow \frac{1}{5} x^5 \). Thus, the integral becomes \[V = \pi \left[ x - \frac{2}{3}x^3 + \frac{1}{5}x^5 \right]_{-1}^{1} \].

Step 5: Evaluate the Definite Integral

Calculate the integral at the bounds: \([1 - \frac{2}{3}(1)^3 + \frac{1}{5}(1)^5] - [-1 - \frac{2}{3}(-1)^3 + \frac{1}{5}(-1)^5]\). Simplify this: \(1 - \frac{2}{3} + \frac{1}{5} = \frac{30}{30} - \frac{20}{30} + \frac{6}{30} = \frac{16}{30} = \frac{8}{15}\) and for the negative evaluates the same \([1 - (-\frac{2}{3}) + (-\frac{1}{5})]\) results the same \(= \frac{8}{15}\). Thus, \(V = \pi \left( \frac{8}{15} + \frac{8}{15}\right) = \frac{16 \pi}{15}\).

Key Concepts

Cross-sectional Area

Understanding the concept of cross-sectional area is crucial in finding the volume of a solid of revolution. A cross-sectional area in this context refers to the area of the shape that forms when a solid is sliced through perpendicular to an axis of interest. In this exercise, the cross-sections are circular disks formed when the parabolas intersect within the given boundaries, specifically at any cut from the parabola line down to the axis of interest.

For our given problem, the parabolas are defined by the equations \(y = x^2\) and \(y = 2 - x^2\). The cross-section at any point \(x\) between \(-1\) and \(1\) is a circular disk. The formula to determine the cross-sectional area (here the area of the circle of the disk) is derived from the radius of the disk, which in turn originates from the distance between the two parabolas, acting as the diameter of our circle.

Circular Disks

In problems involving the volume of a solid of revolution, the term circular disks usually refer to the solid slices that are perpendicular to the axis of revolution. They are circular because of the symmetry around the axis.

• Each disk's diameter is determined by examining the shape from which it is carved. In this problem, the diameter of each disk is the vertical distance between the parabolas \(y = 2 - x^2\) and \(y = x^2\).
• To find the radius of the disks, one calculates the diameter and divides by two. Accordingly, here the radius is \(r = \frac{2 - 2x^2}{2} = 1 - x^2\).
• Every disk is then represented as a circle with area \(A = \pi r^2\). This area depends on the length of the radius at a specific point \(x\).

Thus, the concept of circular disks is paramount in understanding the makeup of the cross-sections, leading us to calculate the entire volume by integration.

Definite Integral

The definite integral is a fundamental concept that allows us to calculate the exact volume of a solid through integration. It involves summing up an infinite number of infinitesimally small areas to find a total quantity. For the problem at hand, we use the definite integral to sum up the areas of the circular cross-section disks across all slices from one point to another along the \(x\)-axis.

Here, the integration bounds are set between \(-1\) and \(1\), which are the points where planes cut the solid. The integral expression is \( V = \int_{-1}^{1}\pi(1 - x^2)^2 \,dx \). This formula calculates the volume by integrating the function \(\pi(1 - x^2)^2\), the area of a cross-sectional disk at any given \(x\), over the interval of \(-1\) to \(1\).

By evaluating this definite integral, you effectively add up the volumes of all those tiny disks to find the total volume of the solid of revolution. This method is due to the application of calculus, greatly simplifying what would otherwise be an incredibly complex task by providing a single formulaic solution.