Problem 59 - Calculus Early Transcendentals

Question

Find the volume of the following solids using the method of your choice. The solid whose base is the region bounded by \(y=x^{2}\) and the line \(y=1,\) and whose cross sections perpendicular to the base and parallel to the \(x\) -axis are semicircles

Step-by-Step Solution

Answer

Question: Determine the volume of the solid formed by the base region bounded by the curve \(y=x^2\) and the line \(y=1\), with cross sections perpendicular to the base and parallel to the x-axis being semicircles. Answer: The volume of the given solid is \(\frac{2\pi}{15}\).

Step 1: Find the range of x values for the base region

To know the range of values where the solid is formed, we must find the x values where \(y=x^2\) and the line \(y=1\) intersect. Hence, we will solve the following equation: $$x^2 = 1$$ By solving the equation we get \(x=-1\) and \(x=1\). So, the base region is defined between \(x=-1\) and \(x=1\).

Step 2: Find the radius of semicircle as a function of x

The radius of the semicircle at a given x-value will be the difference between the y-values formed by the line \(y = 1\), and the curve of the parabola \((y = x^2)\). Thus, the radius can be defined as: $$r(x) = 1 - x^2$$

Step 3: Find the area of the semicircle as a function of x

Now, we will find the area of the semicircle using the function \(r(x)\). The area of a full circle is given by \(\pi r^2\), hence, the area of a semicircle will be: $$A(x) = \frac{1}{2}\pi (1-x^2)^2$$

Step 4: Integrate the area function over the range of x values

Finally, we integrate the area function \(A(x)\) over the range of x values, from \(x=-1\) to \(x=1\), to find the volume of the solid: $$V = \int_{-1}^1 \frac{1}{2}\pi (1-x^2)^2 dx$$

Step 5: Evaluate the integral and find the volume

To evaluate the integral, we can use u-substitution or other integration techniques. The result of the evaluation is: $$V = \int_{-1}^1 \frac{1}{2}\pi (1-x^2)^2 dx = \frac{2\pi}{15}$$ Hence, the volume of the given solid is \(\frac{2\pi}{15}\).

Key Concepts

Integration

Integration is a fundamental concept in calculus used to calculate the total size, length, area, or volume of a given entity. In this context, integration helps us determine the volume of complicated shapes by breaking them into smaller and more manageable parts. The method involves summing infinitesimal volume slices, which form the solid's overall volume when accumulated.
When dealing with problems involving solids, integration takes advantage of the area function across a specific range of values. It involves:

• Identifying the area function that describes the cross-section.
• Integrating this function across the bounds established by the intersections or limits of the region.

By conducting these steps, one can find the total volume of the solid. In our exercise, integration is used to integrate the area of the semicircle cross-sections from the predetermined range of -1 to 1. This integration yields the volume of the solid from these individual slices.

Cross Sections

Cross sections provide a method to visualize and compute the volume of a solid by analyzing a slice perpendicular to a certain axis. Imagine slicing through a loaf of bread. Each slice represents a cross-section of the loaf, just as a slice through a solid provides a cross-sectional area.
For solids where cross-sections are regularly-shaped, like semicircles in our problem, this regularity allows for calculating volume via integration, which sums up all these slices.
To solve the problem:

• We consider slices perpendicular to the x-axis.
• These slices produce semicircular cross-sections, each described by a specific radius function involving the curve's position.

Understanding this allows us to more accurately define the shape and size of each slice and ultimately calculate its area and integrate these to determine the solid's total volume.

Semicircles

In the problem we are dealing with, the cross sections of the solid are semicircles. A semicircle is half of a circle, and its key feature in calculating the area or volume of related solids is its radius.
For our exercise, the radius of each semicircle is defined as the difference between the horizontal line and the parabolic curve, mathematically expressed as \( r(x) = 1 - x^2 \).
The formula for the area of a full circle is \( \pi r^2 \), so for a semicircle, it is half of that: \( A = \frac{1}{2}\pi r^2 \). When plugged into our problem, this results in the area function: \( A(x) = \frac{1}{2}\pi (1-x^2)^2 \).
After calculating the area of a cross-sectional semicircle, integration over the required range provides the solid's total volume. In this instance, the semicircles play a critical role in visualizing and solving for the volume.