Problem 45 - Thomas Calculus

Question

In Exercises 45 and \(46,\) find the volume of the solid generated by revolving each region about the given axis. \begin{equation} \begin{array}{l}{\text { The region in the first quadrant bounded above by the curve }} \\ {y=x^{2}, \text { below by the } x \text { -axis, and on the right by the line } x=1 \text { , }} \\ {\text { about the line } x=-1}\end{array} \end{equation}

Step-by-Step Solution

Answer

The volume is \( \frac{7\pi}{6} \) cubic units.

Step 1: Identify Region and Axis of Revolution

The region is in the first quadrant, bounded by the curve \( y = x^2 \), the x-axis, and the line \( x = 1 \). We will revolve this region about the vertical line \( x = -1 \).

Step 2: Understand the Method of Cylindrical Shells

Since the region is revolved around a vertical line \( x = -1 \), we use the method of cylindrical shells. The formula for the volume using cylindrical shells is: \[ V = 2\pi \int_{a}^{b} (radius\,\text{of shell})(height\,\text{of shell})\,dx \]

Step 3: Determine the Elements for Shell Method

For this problem, the radius of the shell is the horizontal distance from \( x \) to \( x = -1 \), which is \((x - (-1)) = (x + 1)\). The height is \( y = x^2 \). The limits of integration for \( x \) are from 0 to 1.

Step 4: Set Up The Integral

Input the known terms into the cylindrical shell formula:\[ V = 2\pi \int_{0}^{1} (x+1)(x^2) \: dx \]This simplifies to: \[ V = 2\pi \int_{0}^{1} (x^3 + x^2) \: dx \]

Step 5: Calculate the Integral

Perform the integration:\[ \int_{0}^{1} (x^3 + x^2) \: dx = \int_{0}^{1} x^3 \: dx + \int_{0}^{1} x^2 \: dx \]\[ = \left[ \frac{x^4}{4} \right]_{0}^{1} + \left[ \frac{x^3}{3} \right]_{0}^{1} \]\[ = \left( \frac{1^4}{4} - \frac{0^4}{4} \right) + \left( \frac{1^3}{3} - \frac{0^3}{3} \right) \]\[ = \frac{1}{4} + \frac{1}{3} \]

Step 6: Simplify and Solve

Combine the results:\[ V = 2\pi \left( \frac{1}{4} + \frac{1}{3} \right) \]\[ = 2\pi \left( \frac{3}{12} + \frac{4}{12} \right) \]\[ = 2\pi \left( \frac{7}{12} \right) \]\[ = \frac{7\pi}{6} \] cubic units.

Key Concepts

Volume of Solid of Revolution

The concept of the volume of a solid of revolution is pivotal in calculus as it helps us find the volume of solids that are formed by rotating a shape around an axis. Imagine you have a thin, flat region. If you spin this region around a line (the axis of revolution), it sweeps out a three-dimensional shape. The resulting solid's volume can be calculated using integral calculus.

In our problem, the region is a part of the plane that is bounded by the parabola \( y = x^2 \), the \( x \)-axis, and the line \( x = 1 \). By revolving it around the line \( x = -1 \), we create a hollow shell-like solid.

• The orientation of the axis of revolution (vertical in this case) determines the choice of method.
• The method of cylindrical shells is particularly useful when dealing with revolution around vertical lines if the region is better expressed with horizontal slices.
• This approach focuses on summing up the volumes of numerous thin cylindrical shells that make up the solid.

Integration

Integration plays a crucial role in calculating the volume of solids of revolution. It works by accumulating infinitely many small elements, each contributing to the total volume.

With the cylindrical shell method, the integration process is applied to each shell's attributes. The integration formula includes:

• The radius of the shell, measured from the axis of revolution to the shell.
• The height of the shell, which corresponds to the function defining the boundary of the region.

• For our problem, the radius is \( x + 1 \), given that the axis is \( x = -1 \), hence the shells have an increased radius by 1 unit.
• The height is defined by \( y = x^2 \).
• The limits of integration go from \( x = 0 \) to \( x = 1 \), covering the region from the origin to the left boundary.

Conducting the integration \( \int_{0}^{1} (x+1)x^2 \, dx \) involves employing the basic rules of integration, including the power rule for polynomials. This results in a simplification followed by definite substantiation, giving us the beautiful result: \( V = \frac{7\pi}{6} \).

Axis of Revolution

The axis of revolution is fundamental to the configuration of the solid of revolution. It is the line around which a region is rotated, influencing the size and shape of the resulting 3D object.

In this exercise, identifying the axis correctly is critical. We revolve around the vertical line \( x = -1 \), not around a typical axis like \( x = 0 \) or \( y = 0 \).

• Revolving around \( x = -1 \) shifts every point 1 unit further than if it were \( x = 0 \) – account for this extra distance in radius calculation.
• It's essential to accurately measure from the region to this axis to ensure the correct volume is captured in shells.

Misidentifying the axis could lead to calculating a vastly different volumetric shape. Hence, always ensure to carefully distinguish your axis of revolution from the very start.