Problem 60 - Calculus: Early Transcendentals

Question

Find the volume of the described solid \( S \). The base of \( S \) is the region enclosed by \( y = 2 - x^2 \) and the x-axis. Cross-sections perpendicular to the y-axis are quarter-circles.

Step-by-Step Solution

Answer

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

Step 1: Find the Intersection Points

To determine the region's limits, find where the parabola intersects the x-axis. Solve for when the equation of the base, \[ y = 2 - x^2 \] becomes zero: \[ 2 - x^2 = 0 \] which simplifies to \[ x^2 = 2 \] thus giving the intersection points as \[ x = \pm\sqrt{2} \]. Therefore, the base region is between \( x = -\sqrt{2} \) and \( x = \sqrt{2} \).

Step 2: Express y in terms of x

For integration with respect to y, express x as a function of y. From \[ y = 2 - x^2 \], solve for x to get \[ x^2 = 2 - y \] and \[ x = \pm\sqrt{2 - y} \]. This tells us how to express the width of the quarter-circle cross-section in terms of y.

Step 3: Determine the Area of a Quarter-Circle Cross Section

A quarter-circle's area, given a radius \( r \), is \[ A = \frac{1}{4}\pi r^2 \]. Here, the radius is the x-distance, given by \[ x = \sqrt{2-y} \]. So the area of each cross-section is \[ A(y) = \frac{1}{4} \pi (\sqrt{2-y})^2 = \frac{1}{4} \pi (2-y) \].

Step 4: Integrate to Find the Volume

Integrate the area of the cross-sections along the y-axis from 0 to 2. The volume \( V \) is given by: \[ V = \int_{0}^{2} \frac{1}{4}\pi (2 - y) \, dy \]. Calculate the integral: \[ V = \frac{1}{4}\pi \int_{0}^{2} (2-y) \, dy = \frac{1}{4}\pi \left[ 2y - \frac{y^2}{2} \right]_0^2 \]. Evaluate the integral: \[ V = \frac{1}{4}\pi ([2(2) - \frac{2^2}{2}] - [2(0) - \frac{0^2}{2}]) = \frac{1}{4}\pi (4 - 2) \]. So, \[ V = \frac{1}{4}\pi \times 2 = \frac{1}{2}\pi \].

Key Concepts

Intersection Points

To solve problems involving the volume of solids, finding intersection points is a crucial first step. Intersection points help identify the limits of integration and the region of interest that needs coverage in a given problem. In this case, you need to determine where the functions meet or, in simpler terms, where the parabola touches or crosses the x-axis.
By solving the base equation of the solid, \( y = 2 - x^2 \), we find the points where this equation equals zero, indicating a crossing at the x-axis. Essentially, you're setting the equation as follows:

• Set the equation to zero: \( 2 - x^2 = 0 \)
• Solve for \( x^2 \) to get \( x^2 = 2 \)
• Take the square root to find \( x = \pm \sqrt{2} \)

Thus, the region of interest spans from \( x = -\sqrt{2} \) to \( x = \sqrt{2} \), which provides the bounds necessary for further calculations. Understanding how to find these points gives you a clearer map to the given problem area and cuts down errors in subsequent steps.

Quarter-Circle Cross Sections

Cross-sectional areas provide essential insights into the solid's structure when calculating volume. Here, each cross-section of the solid perpendicular to the y-axis is a quarter-circle. Understanding how these cross sections work will enable you to grasp how they shape the entire solid.
For a quarter-circle, the area formula is derived from the circle's area and divided by four, given by \( A = \frac{1}{4}\pi r^2 \). In this exercise, the radius \( r \) of each cross-section is the x-distance, allowing you to express it in terms of \( y \). By solving the original quadratic equation for \( x \), we have:

• \( x = \sqrt{2-y} \)

Thus, each quarter-circle's area, \( A(y) \), becomes \( \frac{1}{4} \pi (\sqrt{2-y})^2 = \frac{1}{4} \pi (2-y) \). Such understanding facilitates further steps and ensures that the integration process reflects the actual cross-sectional shape, contributing logically to determining the volume of the solid.

Integral Calculus

Integral calculus serves as the backbone when calculating volumes of complex solids, especially with varying cross-sectional areas. It involves summing up infinitely small parts to find a whole. Here, to find the overall volume, integrating the area function of the cross-sections across the given bounds is necessary.
Calculating the integral is a multi-step process aimed at combining all quarter-circle areas from \( y = 0 \) to \( y = 2 \):

• Set up the integral of the quarter-circle area: \( V = \int_{0}^{2} \frac{1}{4} \pi (2 - y) \, dy \)
• Solve the integral using basic integration rules: \( \int (2 - y) \, dy = [2y - \frac{y^2}{2}] \)
• Calculate the definite integral: \( V = \frac{1}{4}\pi \left[2y - \frac{y^2}{2}\right]_0^2 \)
• Evaluate to find volume: easily completed to \( V = \frac{1}{2}\pi \)

Thus, integral calculus offers a systematic approach to finding the volume by summing all component areas logically and accurately, enabling the understanding and solving of otherwise challenging volumetric problems.