Problem 13 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 11-16, sketch the region \(R\) bounded by the graphs of the given equations and show a typical horizontal slice. Find the volume of the solid generated by revolving \(R\) about the \(y\)-axis. $$ x=2 \sqrt{y}, y=4, x=0 $$

Step-by-Step Solution

Answer

The volume is \(32\pi\).

Step 1: Understand the Equations and Graph

First, identify and sketch the graphs of the given equations in the xy-plane. The graph of \(x = 2\sqrt{y}\) is a sideways opening parabola. Since \(y = 4\), the line \(y = 4\) is horizontal. The line \(x = 0\) is the y-axis. The bounded region \(R\) is enclosed by these graphs.

Step 2: Identify the Bounded Region

The region \(R\) bounded by these equations is to the right of \(x = 0\), below \(y = 4\), and to the left of \(x = 2\sqrt{y}\). This area extends from \(y = 0\) to \(y = 4\).

Step 3: Determine a Typical Horizontal Slice

A horizontal slice at a height \(y\) within the region will have a width extending from \(x = 0\) to \(x = 2\sqrt{y}\), forming a rectangular strip when drawn parallel to the x-axis.

Step 4: Find the Volume of Revolution Formula

We use the method of washers to find the volume. The volume of a thin washer when rotated about the y-axis is given by the formula \( V = \pi \int_{a}^{b} \left[ R(y)^2 - r(y)^2 \right] \, dy \), where \(R(y)\) is the outer radius and \(r(y)\) is the inner radius.

Step 5: Setup the Integral

For the given problem, the inner radius \(r(y) = 0\) and the outer radius \(R(y) = 2\sqrt{y}\). Therefore, the volume of the solid is given by \( V = \pi \int_{0}^{4} (2\sqrt{y})^2 \, dy = \pi \int_{0}^{4} 4y \, dy \).

Step 6: Evaluate the Integral

Calculate the integral: \( V = \pi \int_{0}^{4} 4y \, dy = 4\pi \left[ \frac{y^2}{2} \right]_{0}^{4} = 4\pi \left( \frac{16}{2} - 0 \right) = 32\pi \).

Step 7: Conclude with the Volume

The volume of the solid generated by revolving the region \(R\) about the y-axis is \(32\pi\).

Key Concepts

Calculus

Calculus is a branch of mathematics that involves studying how things change. This area of mathematics focuses on concepts such as rates of change and accumulations. It is divided mainly into differential calculus and integral calculus.
Differential calculus concerns itself with finding rates of change called derivatives. It helps us understand how a function changes at any given point. On the other hand, integral calculus focuses on accumulating quantities, such as finding areas under curves or volumes of solids.
Both branches of calculus are interconnected. By using these principles, we can solve numerous real-world problems in physics, engineering, biology, and economics, among other fields.

Integration

Integration is one of the core operations in calculus that deals with the accumulation of quantities. Imagine trying to find the total area under a curve on a graph. Integration allows us to determine such areas precisely.

• The process of integration can undo differentiation; this relationship is fundamental to calculus. The tool used to perform integration is called an integral.
• To integrate a function means to sum up infinite small pieces of an area or a volume.

This concept is used not just for curves on a graph but is also extended to finding volumes of three-dimensional objects and identifying quantities like work and energy in physics.

Definite Integral

When we talk about the definite integral, we are specifically looking at how to calculate exact quantities, such as an area or volume, on a bounded interval.
A definite integral has a specific start and end point, defined by the lower limit, denoted by "a," and the upper limit, "b." When calculating a definite integral, we are essentially adding up an infinite number of infinitesimally small quantities between these two bounds. The notation used is \( \int_{a}^{b} f(x) \, dx \).

• Definite integrals yield a numeric value representing total accumulations between these boundaries.
• In the context of our exercise, we use the definite integral to sum up the volumes created by revolving the region around the y-axis.

The outcome of the integration gives the total volume—one of the many practical applications of definite integrals.

Washer Method

The Washer Method is an efficient technique used in calculus to determine the volume of a solid of revolution. This method is used when the solid is generated by revolving a region around a specific axis.

• When revolving around the y-axis, the solid resembles stacked washers (like a series of disks with holes), hence the name.
• Each washer is essentially a disk with a hole in it; the "washers" have an outer radius defined by \( R(y) \) and an inner radius \( r(y) \).
• The formula to find the volume is \( V = \pi \int_{a}^{b} [R(y)^2 - r(y)^2] \, dy \).

In our specific problem, the washer’s outer edge extends to \( 2\sqrt{y} \) and has no inner radius, simplifying the integral to calculate the total volume, which in this case was \( 32\pi \). This method is particularly useful because it allows straightforward computation of volumes formed by rotation without intricate geometry.