Problem 108 - Calculus Volume 1

Question

The base is the region between \(y=x\) and \(y=x^{2}\) . Slices perpendicular to the \(x\) -axis are semicircles.

Step-by-Step Solution

Answer

The volume is \( \frac{\pi}{240} \).

Step 1: Identify the Base Region

The region between the curves \(y = x\) and \(y = x^2\) is the base of the shape. First, find the points of intersection by setting \(x = x^2\). Solve the equation \(x^2 - x = 0\), which factors to \(x(x - 1) = 0\), giving intersection points at \(x = 0\) and \(x = 1\). Thus, the region is bounded between \(x = 0\) and \(x = 1\).

Step 2: Determine the Slice Radius

For a slice perpendicular to the \(x\)-axis between these functions at a given \(x\)-coordinate, the diameter of the semicircle is the vertical distance between \(y = x\) and \(y = x^2\). Thus, the diameter is \(x - x^2\). The radius \(r\) of the semicircle is half of this diameter, so \(r = \frac{x - x^2}{2}\).

Step 3: Calculate the Area of a Semicircular Slice

The area \(A\) of a semicircle is given by \(A = \frac{1}{2} \pi r^2\). Substitute the expression for radius from the previous step to get \(A = \frac{1}{2} \pi \left(\frac{x - x^2}{2}\right)^2\). Simplify this to \(A = \frac{\pi}{8} (x - x^2)^2\).

Step 4: Set up the Integral for Volume

The volume of the solid is the integral of the areas of all the semicircular slices from \(x = 0\) to \(x = 1\). Therefore, the integral expression for the volume is \(V = \int_{0}^{1} \frac{\pi}{8} (x - x^2)^2 \, dx\).

Step 5: Evaluate the Integral

First, simplify \((x - x^2)^2 = x^2 - 2x^3 + x^4\). Substitute back in the integral: \[ V = \frac{\pi}{8} \int_{0}^{1} (x^2 - 2x^3 + x^4) \, dx \]Calculate each term separately: 1. \( \int x^2 \, dx = \frac{x^3}{3} \) evaluated from 0 to 1 gives \(\frac{1}{3} \).2. \( \int -2x^3 \, dx = -\frac{2x^4}{4} = -\frac{x^4}{2} \) evaluated from 0 to 1 gives \(-\frac{1}{2} \).3. \( \int x^4 \, dx = \frac{x^5}{5} \) evaluated from 0 to 1 gives \(\frac{1}{5} \).

Step 6: Combine Results and Find the Volume

Combine results: \(V = \frac{\pi}{8} \left( \frac{1}{3} - \frac{1}{2} + \frac{1}{5} \right)\)Find a common denominator (15):\(V = \frac{\pi}{8} \left( \frac{5}{15} - \frac{7.5}{15} + \frac{3}{15} \right)\)\(= \frac{\pi}{8} \left( \frac{0.5}{15} \right)\)\(= \frac{\pi}{240} \).

Key Concepts

Integral Calculus

Integral calculus is a powerful branch of mathematics that focuses on finding the accumulation of quantities. In simple words, it involves solving problems related to areas, volumes, and other concepts involving continuous sums. One of the main components of integral calculus is the integral itself. An integral can be thought of as a tool to calculate the total amount of something that continuously varies.

When we talk about definite integrals, like the one used in the given problem, it has boundaries or limits of integration. This means we are calculating the total or accumulated value between two points. In this exercise, the purpose of using an integral is to compute the total volume of a solid with a semicircular cross-section by adding up the volumes of an infinite number of infinitesimally small slices.

• Integral calculus helps find areas under curves.
• It is used to calculate the accumulation or total quantity.
• Definite integrals have boundaries or specific limits.

Understanding how to set up and solve these integrals is crucial in many fields such as physics, engineering, and economics, where it helps in solving real-world problems that involve continuously changing variables.

Volume by Integration

Calculating volume using integration leverages the concept of adding up infinitely small elements to measure a larger figure. For shapes with varying cross-sections, integration allows us to sum up all these tiny parts to find the full volume. This is particularly useful for irregularly shaped objects, complex figures with unique cross-sections, such as semicircular slices in this exercise.

To approach these problems, first determine the shape and size of each cross-sectional area, in this case semicircles, at any given point along the base. Then, use integration to accumulate these areas into a total volume. The slice’s area function becomes the integrand, and the integral's bounds are determined by the limits of the base shape, from one end to the other.

• Volume calculation involves summing infinitesimal slices.
• Each slice’s area is integrated over a specific range.
• Integration converts these small areas into total volume.

This method is imperative in engineering and design when creating objects with complex shapes or when precise volumetric data is necessary for material estimation and structural assessment.

Semicircular Cross-Sections

In the context of this exercise, semicircular cross-sections are used to shape the solid whose volume we want to find. When a slice is made perpendicular to the base region defined by the intersection of the functions, we see a semicircle. These cross-sections help determine how the volume changes as you move along the base.

For semicircles, the diameter at any point is given by the vertical distance between the two functions. This diameter is crucial because it allows us to find the radius, which is half of the diameter. Once the radius is identified, the area of the semicircle can be calculated using the formula for the area of a semicircle, which is derived from the circle area formula.

• The diameter is the vertical difference between defining curves.
• The radius is half of the diameter for these semicircular slices.
• Semicircle area aids in finding each slice's contribution to volume.

Understanding the properties of these semicircular sections helps in accurately building up the volume with integration, ensuring that each part of the solid is correctly accounted for in the calculation.