Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the volume of the solid generated by revolving about the line \(y=2\) the region in the first quadrant bounded by the parabolas \(3 x^{2}-16 y+48=0\) and \(x^{2}-16 y+80=0\) and the \(y\)-axis.

Step-by-Step Solution

Answer

The volume of the solid is calculated using the washer method from \(y=5\) to \(y=12\).

Step 1: Express the Equations in function of y

First, solve each equation for \(x^2\). The equations are given as \(3x^2 - 16y + 48 = 0\) and \(x^2 - 16y + 80 = 0\). Solve the first equation for \(x^2\):\(3x^2 = 16y - 48 \Rightarrow x^2 = \frac{16}{3}y - 16\). Similarly, solve the second equation for \(x^2\): \(x^2 = 16y - 80\).

Step 2: Set Up the Integral for Volume Calculation

Since we're revolving the region around the line \(y = 2\), we use the disk method. The distance from the line \(y = 2\) to \(y = f(x)\) is \((f(y) - 2)\). The radius of the disk will be the difference from \(y = 2\). Calculate the bounds \(y\): Solve for intersections by setting \(\frac{16}{3}y - 16 = 16y - 80\), which simplifies to \(y = 12\). Identify the lower bound as the y-intercepts by setting \(x = 0\), leading to \(y = 5\) and \(y = 5\). The boundaries for integration are from \(y = 5\) to \(y = 12\).

Step 3: Calculate the Volume Integrals

Integrate using the volume formula for the washers, which is \(\pi \int_{a}^{b} [(R(y))^2 - (r(y))^2] dy\). Here, the outer radius \(R(y) = (2 - f_1(y))\) and inner radius \(r(y) = (2 - f_2(y))\). Here, \(f_1(y)\) and \(f_2(y)\) are the respective solutions for \(x^2\) from Step 1. So we have:\[\pi \int_{5}^{12} \left[((16y - 80) - 2)^2 - \left(\left(\frac{16}{3}y - 16\right) - 2\right)^2\right] \,dy\].

Step 4: Evaluate the Integral

Perform the integration \[\pi \int_{5}^{12} \left[ ((16y - 80) - 2)^2 - \left(\frac{16}{3}y - 16 - 2\right)^2 \right] \ dy\].Substitute into the integral and solve. Calculate the integrals separately, then evaluate from y = 5 to y = 12, subtracting the results. This gives the volume of the solid of revolution.

Key Concepts

Volume of Revolution

When we talk about volume of revolution, we are discussing a method to find the volume of a three-dimensional solid. The solid is formed when a two-dimensional region is revolved around a line (known as the axis of revolution). We can think of this rotation as spinning a shape around an axis like a potter spins clay on a wheel to create a shape. Volume of revolution is a key concept in integral calculus and is often used to calculate the volume of objects that are not easily measurable in standard geometric terms. The process involves:

• Identifying the region of interest in the coordinate plane.
• Determining a line around which this region will be revolved.
• Using calculus to integrate and calculate the volume of the formed solid.

In the original exercise, we are revolving the area between two curves around the line \(y=2\). This introduces the concept of the axis not being the x-axis, which is crucial for understanding how the volume will be evaluated.

Disk Method

The disk method is a way to calculate the volume of a solid of revolution when the region is revolved around a horizontal or vertical line. This method is particularly intuitive because it relies on the idea that the solid is made up of many thin disks stacked one on top of each other. Here's how it works:

• The radius of each disk is defined as the distance from the axis of rotation to the curve.
• The thickness of each disk is an infinitesimally small change in \(x\) or \(y\), represented as \(dx\) or \(dy\).
• The volume of each disk is calculated by the formula \( \pi R^2 \times \text{thickness} \).
• We use integration to sum the infinite number of disks that make up the entire solid.

In the scenario given in the original exercise, because the revolution is around \(y=2\), both inner and outer radius functions depend on the value of \(y\). We account for an outer and inner curve which gives us a washer instead of a disk.

Integral Calculus

Integral calculus is a branch of calculus concerned with the concept of integration. Integration can be thought of as finding the accumulation of quantities; in the case of the volumes, it accumulates the infinitesimal pieces (like disks or washers) to find the total volume. In terms of problem-solving, the steps include:

• Setting up the integral, recognizing appropriate bounds for \(x\) or \(y\).
• Determining the function or expressions that will be integrated.
• Executing the integration process, which involves finding the antiderivative of the function.

It's important to know the rules of integration and how to apply them to solve for areas and volumes. In the exercise, we use an integral to sum the volume of washers from \(y = 5\) to \(y = 12\). This is where the magic of calculus shines by simplifying complex geometries into solvable integrals.

Parabolas

A parabola is a symmetrical open curve. Any point on a parabola is equidistant from a fixed point (the focus) and a fixed straight line (the directrix). Parabolas have either a maximum or a minimum point called the vertex, and they can open upward or downward. For the exercise in question, we are given two parabolas represented as quadratic equations in terms of \(y\). The intersection and bounds around these parabolas define the area to be revolved. Understanding parabolas involves:

• Finding the vertex, which is crucial for understanding the shape and orientation.
• Solving for the intersections between parabolas to determine bounds for integration.
• Understanding the impact of coefficients on the parabola’s width and orientation.

In our problem, the parabolas intersect, creating a bounded area that is then rotated around the line \(y=2\). Recognizing where parabolas intersect (like at \(y = 5\) and \(y = 12\)) is essential for setting up integrals, as it determines our integration limits.