Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-12, find the volume of the solid generated when the region \(R\) bounded by the given curves is revolved about the indicated axis. Do this by performing the following steps. (a) Sketch the region \(R\). (b) Show a typical rectangular slice properly labeled. (c) Write a formula for the approximate volume of the shell generated by this slice. (d) Set up the corresponding integral. (e) Evaluate this integral. \(x=\sqrt{2 y}+1, y=2, x=0, y=0 ;\) about the line \(y=3\)

Step-by-Step Solution

Answer

The volume is \(8\pi\).

Step 1: Sketch the Region

Consider the curves: \(x = \sqrt{2y} + 1\), \(y = 2\), \(x = 0\), and \(y = 0\). Plot these curves on the coordinate plane. The region \(R\) is bounded by these curves. Specifically, \(x = \sqrt{2y} + 1\) is a sideways parabola opening to the right, bounded below by \(y = 0\) (the x-axis), on the left by \(x = 0\) (the y-axis), and above by \(y = 2\).

Step 2: Label a Typical Slice

Consider a vertical slice at a typical \(y\) value within the region, which acts as a shell when revolved about the line \(y = 3\). The thickness of the slice is \(\Delta y\), and its height is given by the horizontal length from \(x = 0\) to \(x = \sqrt{2y} + 1\).

Step 3: Approximate Volume of the Shell

The shell method gives the approximate volume of the shell as: \( \Delta V = 2\pi (3 - y) (\sqrt{2y} + 1) \Delta y \), where \((3 - y)\) is the distance from the axis of rotation to the shell.

Step 4: Set Up the Integral

Integrate the expression for \(\Delta V\) over the range of \(y\) values from \(0\) to \(2\) to find the total volume. The integral setup is: \[ V = \int_{0}^{2} 2\pi (3 - y) (\sqrt{2y} + 1) \, dy \]

Step 5: Evaluate the Integral

Simplify the integral: \[ V = 2\pi \int_{0}^{2} \left((3 - y)(\sqrt{2y} + 1)\right) \, dy \] Distribute the terms: \[ V = 2\pi \int_{0}^{2} \left(3\sqrt{2y} + 3 - y\sqrt{2y} - y\right) \, dy \] Separate the integrals: \[ V = 2\pi \left( \int_{0}^{2} 3\sqrt{2y} \, dy + \int_{0}^{2} 3 \, dy - \int_{0}^{2} y\sqrt{2y} \, dy - \int_{0}^{2} y \, dy \right) \] Solve each integral: \( \int_{0}^{2} 3\sqrt{2y} \, dy = 2\sqrt{2} \times \frac{6}{3} \) \( \int_{0}^{2} 3 \, dy = 6 \) \( \int_{0}^{2} y\sqrt{2y} \, dy = 4 \) \( \int_{0}^{2} y \, dy = 2 \)So, the volume \( V = 2\pi \times (4 + 6 - 4 - 2) = 8\pi \).

Step 6: Conclusion

The volume of the solid generated by revolving the region around the line \(y = 3\) is \(8\pi\).

Key Concepts

Shell Method

The Shell Method is a way to calculate the volume of a solid of revolution. It's especially useful when the solid is generated by rotating a region about an axis that is parallel to the axis of the cylinder (or shell) formed. The Shell Method involves the following:

• Visualizing the solid as a series of cylindrical shells.
• Calculating the volume of one representative cylindrical shell.
• Integrating these cylindrical slices over the entire region.

To set up the Shell Method, identify a typical slice parallel to the axis of rotation. The volume of this shell is approximated as \[ \Delta V = 2\pi r h \Delta y \]where:

• r is the average radius of the shell.
• h is the height of the cylindrical segment.
• \Delta y is the thickness of the shell.

In our example, revolving the region about the line \(y = 3\), the radius \(r\) is \(3 - y\), and the height \(h\) is \(\sqrt{2y} + 1\). These definitions allow for effectively applying the Shell Method to find the exact volume.

Integral Calculus

Integral Calculus is a branch of mathematics that helps us find volumes, areas, and other quantities accumulated over a range. It's one of the two main branches of calculus, alongside Differential Calculus. Fundamentally, Integral Calculus revolves around

• Definite Integrals – which calculate accumulated quantities.
• Indefinite Integrals – which capture an entire range of functions or antiderivatives.

Here, integral calculus steps in to compute the volume by integrating the expression of the shell volumes over a defined interval.
For our problem, we have the integral setup as:\[ V = \int_{0}^{2} 2\pi (3 - y)(\sqrt{2y} + 1) \, dy \]This integral accumulates the volume of all infinitesimally thin shells across the region from \(y = 0\) to \(y = 2\). Thus, we are summing up all these volumes to find the total volume of the solid.

Definite Integrals

Definite Integrals are crucial in volume problems because they allow us to calculate the total accumulation of volume between specific bounds. In calculus, a definite integral has a start limit (lower bound) and an end limit (upper bound), making it essential for precision.To find the volume of a solid of revolution:

• Define the bounds corresponding to the region.
• Express the function that represents the volume element, like the shell in this instance.
• Evaluate it over the specified bounds.

In the exercise, the definite integral \[ V = \int_{0}^{2} 2\pi \left(3\sqrt{2y} + 3 - y\sqrt{2y} - y\right) \, dy \] was broken down into separate, manageable integrals. Each integral represented a different part of the shell's volume. After evaluating each component, the results were summed to give the total volume of \(8\pi\).
This process makes definite integrals an indispensable tool in solving calculus volume problems accurately.