Problem 11 - Calculus Early Transcendentals: Pearson New International Edition

Question

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. \(x=y^{2}, x=0, y=3\)

Step-by-Step Solution

Answer

The volume of the solid is \(\frac{81\pi}{2}\).

Step 1: Understand the Equations

We are given the equations \(x = y^2\), \(x = 0\), and \(y = 3\). The equation \(x = y^2\) represents a parabola opening to the right with vertex at the origin. \(x = 0\) is the \(y\)-axis, and \(y = 3\) is a horizontal line. The region \(R\) of interest is bounded by these curves.

Step 2: Sketch the Bounded Region

Sketch the parabola \(x = y^2\), the line \(x = 0\) (the \(y\)-axis), and \(y = 3\). Region \(R\) is the part of the parabola above \(x=0\), bounded vertically from \(y = 0\) to \(y = 3\). This region looks like a `slice` of the parabola cut off by the horizontal line \(y = 3\).

Step 3: Visualize a Horizontal Slice

A typical horizontal slice of the region \(R\) is a thin rectangle at a given \(y\)-value, extending from \(x=0\) to \(x = y^2\). The width of this slice is \(x = y^2\), and the infinitesimally small thickness is \(dy\).

Step 4: Set up the Integral for Volume

We use the method of cylindrical shells to find the volume of the solid formed by revolving the region \(R\) around the \(y\)-axis. The formula for the volume is \[V = 2\pi \int_{a}^{b} (\text{radius})(\text{height}) \, dy\] where \(a=0\) and \(b=3\). Here, radius is \(y\) and height is \(y^2\), therefore \[V = 2\pi \int_{0}^{3} y \cdot y^2 \, dy.\]

Step 5: Solve the Integral

Calculate the integral from the previous step:\[V = 2\pi \int_{0}^{3} y^3 \, dy.\]Evaluate the integral:\[\int y^3 \, dy = \frac{y^4}{4}.\]Then evaluate from \(0\) to \(3\):\[V = 2\pi \left[ \frac{3^4}{4} - \frac{0^4}{4} \right] = 2\pi \left[ \frac{81}{4} \right] = \frac{81\pi}{2}.\]

Step 6: Present Final Answer

After evaluating, the volume of the solid is \(\frac{81\pi}{2}\). This is the volume of the solid generated by revolving the region \(R\) about the \(y\)-axis.

Key Concepts

Cylindrical Shells Method

The cylindrical shells method is a powerful technique for finding the volume of a solid of revolution. When a region in the plane is revolved about an axis of rotation that lies outside the region, the cylindrical shells method is particularly useful. Here's how it works:

• Imagine slicing the region into thin, vertical strips parallel to the axis of rotation. Each strip, when revolved around the axis, forms a cylindrical shell.
• The height of each shell corresponds to the length of the strip, while the radius is the distance from the axis of rotation to the strip.

The volume of one shell is approximated by the formula:\[ V_{ ext{shell}} = 2\pi ( ext{radius}) ( ext{height}) ( ext{thickness}) \]To find the total volume of the solid, integrate the volume of each shell across the interval. This involves setting up a definite integral that sums the volumes of all shells between specified bounds.

Definite Integrals

Definite integrals are fundamental in calculus for finding quantities like area, volume, and accumulation. They represent the accumulated sum of an infinite number of infinitesimally small quantities.

• A definite integral takes a function defined over an interval \([a, b]\). This interval becomes the limits of integration.
• The integral calculates the net "sum" of the function's values over this interval, subtracting negative areas if the function dips below the axis.

In our problem involving the cylindrical shells method, the definite integral is essential. It accumulates the volumes of each infinitesimally small cylindrical shell from the bottom to the top of the region, capturing the entire volume of the 3D solid upon revolving.

Parabolas

A parabola is a symmetric curve defined by a quadratic function, such as the equation \(x = y^2\). In the coordinate plane, parabolas can open vertically or horizontally, depending on their orientation.

• In this exercise, we encounter a parabola that opens to the right, represented by \(x = y^2\). The vertex of this parabola is at the origin (0,0).
• The region of interest is bounded by this parabola upwards to a horizontal line at \(y = 3\), forming a clear boundary for our volume calculation.

Understanding the parabola's shape and boundary conditions helps to set up the integral and accurately portray the region being revolved to create the solid.

Calculus Problems

Calculus problems often involve finding areas, volumes, and understanding the behavior of functions. They require a clear understanding of concepts such as integration, differentiation, and the geometry of curves.

• Problems like the volume of revolution require combining geometric intuition with calculus tools.
• Using methods such as cylindrical shells and understanding the significance of definite integrals helps in solving these problems effectively.

By visualizing the region and applying these mathematical concepts, students can better grasp how to set up and solve real-world calculus problems, turning abstract calculations into tangible understanding.