Problem 16 - 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. \(x=\sqrt{4-y^{2}}, x=0\)

Step-by-Step Solution

Answer

The volume is \( \frac{64\pi}{3} \).

Step 1: Analyze the Equations

The given equations are of the form \( x = \sqrt{4 - y^2} \) and \( x = 0 \). The equation \( x = \sqrt{4 - y^2} \) is half of a circle with a radius of 2, centered at the origin, but only for non-negative values of \( x \). The equation \( x = 0 \) is the y-axis. Thus, the region \( R \) is the right half of a circle, bounded by \( -2 \le y \le 2 \).

Step 2: Sketch the Region R

Draw the region \( R \) which is bounded by \( x = 0 \) (the y-axis) and \( x = \sqrt{4 - y^2} \), which represents a semicircle extending from \( y = -2 \) to \( y = 2 \). Note that the region lies entirely in the right half of the plane.

Step 3: Recognize the Horizontal Slice

A typical slice of the region \( R \) horizontal to the y-axis will be a thin rectangle. The length of this rectangle at any \( y \) is given by the difference in the x-values: \( \Delta x = \sqrt{4 - y^2} - 0 \). The height \( \Delta y \) is infinitesimally small.

Step 4: Set Up the Integral for Volume

To find the volume of the solid formed by revolving the region around the y-axis, use the method of shell integration. The formula for the volume \( V \) is:\[V = \int_{-2}^{2} 2\pi x \cdot \text{height} \cdot \Delta y\]Where \( x = \sqrt{4 - y^2} \) and the height of the shell at a slice is \( \sqrt{4 - y^2} \).

Step 5: Simplify and Solve the Integral

Refine the integral expression:\[V = \int_{-2}^{2} 2\pi (\sqrt{4 - y^2})^2 \, dy\]This simplifies to:\[V = 2\pi \int_{-2}^{2} (4 - y^2) \, dy\]Evaluate this integral:\[= 2\pi \left[ 4y - \frac{y^3}{3} \right]_{-2}^{2} = 2\pi \left[ (8 - \frac{8}{3}) - (-8 + \frac{8}{3}) \right]\]Solve the arithmetic:\[= 2\pi \left[ \frac{16}{3} + \frac{16}{3} \right] = \frac{64\pi}{3}\]

Step 6: Final Result

The volume of the solid formed by rotating the region \( R \) around the y-axis is \( \frac{64\pi}{3} \).

Key Concepts

Shell Integration

Shell integration is a useful technique for finding volumes of solids of revolution, especially when revolving shapes around an axis other than the one they are typically aligned with. In this method, we visualize our solid as a series of cylindrical shells. Each shell is generated by rotating a thin vertical or horizontal slice of the region around the axis of rotation.

• The shell method is particularly effective when the axis of rotation is parallel to the slices of the region.
• The volume of each cylindrical shell can be thought of as the product of its circumference, its height, and its thickness.
• The integral of these volumes over the entire region yields the total volume of the solid.

This process involves setting up an integral that accounts for the "shells" created by each horizontal or vertical slice of the region as it rotates. For our example, since the region is rotated around the y-axis, we consider the horizontal slices to form cylindrical shells.

Horizontal Slicing

Horizontal slicing is a straightforward method of breaking down the region of interest into manageable pieces that are easier to compute. Imagining the shape as being sliced horizontally helps in understanding how these slices contribute to the overall volume.

• Each horizontal slice is a thin rectangle stretching from the y-axis to the edge of the semicircle.
• The length of this rectangle is given by our function, in this case, \(x = \sqrt{4 - y^2}\).
• The thickness is an infinitesimally small change in the y-direction, \(\Delta y\).

By revolving this slice around the y-axis, we visualize forming a thin cylindrical shell. The combination of all such shells, calculated across the bounds of the region, allows us to approximate and compute the volume.

Integral Calculus

Integral calculus is the mathematical tool used here to accumulate quantities like area and volume over a given interval. It forms the backbone of the method used to derive the volume of a solid of revolution. Understanding how to set up and compute these integrals is crucial.

• An integral takes the sum of infinitely many infinitesimally small quantities, providing exact values for areas and volumes.
• In this example, the integral computes the sum of the volume of cylindrical shells.
• The bounds of the integral, from \(-2\) to \(2\), correspond to the full extent of the region along the y-axis.

By analyzing these integral bounds and functions, students can effectively set up and solve integrals to find exact volumes for varied shapes when revolved around an axis.

Volume Calculation

Volume calculation using integral calculus allows us to determine the total space occupied by a solid. Revolving a defined region around an axis converts a 2D area into a 3D volume.

• For a solid formed around the y-axis, the circumference of each shell is \(2\pi x\), where \(x\) is the radius.
• The height of each shell corresponds to the function description, \(\sqrt{4 - y^2}\) in this case.
• Multiplying this by 2 times \(\pi\) and integrating along the y-axis gives us the volume.

For our semi-circle example, the final integral is simplified and computed as:
\[V = 2\pi \int_{-2}^{2} (4 - y^2) \, dy\]
Finally, solving provides the solid volume as \(\frac{64\pi}{3}\), showcasing how calculus helps solve complex geometric problems smoothly.