Problem 47 - Thomas Calculus

Question

Find the volume of the solid generated by revolving the region bounded by \(y=\sqrt{x}\) and the lines \(y=2\) and \(x=0\) about \begin{equation} \begin{array}{ll}{\text { a. the } x \text { -axis. }} & {\text { b. the } y \text { -axis. }} \\ {\text { c. the line } y=2 .} & {\text { d. the line } x=4}\end{array} \end{equation}

Step-by-Step Solution

Answer

Volumes: a) \(8\pi\), b) \(\frac{64\pi}{5}\), c) \(\frac{16\pi}{3}\), d) \(\frac{32\pi}{5}\).

Step 1: Understand the Region

To find the volume of the solid, first understand the region bounded by the curves. The curve is given by \(y = \sqrt{x}\), and the boundaries are \(y=2\) and \(x=0\). This region lies in the first quadrant.

Step 2: Find the Intersection Points

Determine the intersection of \(y = \sqrt{x}\) and \(y = 2\). Solving \(2 = \sqrt{x}\), we find \(x = 4\). Therefore, the region of interest is from \(x=0\) to \(x=4\).

Step 3a: Solid of Revolution About the x-axis

When revolving around the x-axis, use the disk method. The bounded region is between \(x=0\) to \(x=4\), with disks of radius \(\sqrt{x}\). The volume \(V\) is given by: \[ V = \pi \int_{0}^{4} (\sqrt{x})^2 \, dx = \pi \int_{0}^{4} x \, dx = \pi \left[ \frac{x^2}{2} \right]_0^4 = \frac{16\pi}{2} = 8\pi. \]

Step 3b: Solid of Revolution About the y-axis

For rotation around the y-axis, use the shell method. The height of a shell is \(2 - \sqrt{x}\) and the radius is \(x\). The volume \(V\) is: \[ V = 2\pi \int_{0}^{4} x(2 - \sqrt{x}) \, dx = 2\pi \left[ 2 \int_{0}^{4} x \, dx - \int_{0}^{4} x^{3/2} \, dx \right]. \] Calculating each integral separately, the final volume is \[ V = \frac{64\pi}{5}. \]

Step 3c: Solid of Revolution About the line y = 2

Consider washers with outer radius \(2\) and inner radius \(\sqrt{x}\), using washer method for revolution around \(y=2\). The volume \(V\) is: \[ V = \pi \int_{0}^{4} [(2)^2 - (2 - \sqrt{x})^2] \, dx. \]Simplify and integrate: \[ V = \pi \left[ 4 \int_{0}^{4} \sqrt{x} \, dx \right] = \pi \left[ \frac{16}{3} \right] = \frac{16\pi}{3}. \]

Step 3d: Solid of Revolution About the line x = 4

Here, use the shell method with each shell radius differing \(4 - x\), height \(2 - \sqrt{x}\). The volume is: \[ V = 2\pi \int_{0}^{4} (4-x)(2-\sqrt{x}) \, dx. \]Separate and solve each integral within: \[ V = 2\pi \left[ \frac{16}{5} \right] = \frac{32\pi}{5}.\]

Key Concepts

Disk Method

The Disk Method is a powerful technique to find the volume of a solid generated by rotating a region around an axis. When you have a situation like rotating around the x-axis, you picture the solid being formed through stacks of circular disks.
Each disk has a certain radius, determined by the function or curve you're rotating.For example, consider the curve described by the function \(y = \sqrt{x}\). When you revolve this curve from \(x=0\) to \(x=4\) around the x-axis, the radius of each disk is simply the function itself: \(\sqrt{x}\). When writing your integral, you'll be squaring this radius because the area of a circle (or disk in this case) is \(\pi r^2\).

• Volume using Disk Method is given by the integral \(V = \pi \int_{a}^{b} [f(x)]^2 \, dx\).
• For our example: \(V = \pi \int_{0}^{4} x \, dx = 8\pi\).

The Disk Method is great for when you're revolving around the x-axis because the radii of the disks are parallel to the axis of rotation. It becomes straightforward to set up and solve the volume integral.

Shell Method

The Shell Method provides a different approach for finding the volume of solids of revolution, particularly useful when revolving around the y-axis.
This method envisions the solid as composed of cylindrical "shells." For each shell, the wall thickness can be thought of as a slender rectangle wrapped around the shape.The height of the shell is determined by the distance between the function and the axis of rotation, while the radius is typically \(x\) itself or a similar function of \(x\), depending on the problem.
In our example, for a region bounded by \(y = \sqrt{x}\), revolving around the y-axis involves the height being \(2-\sqrt{x}\) and the radius as \(x\).

• The volume using the Shell Method is \(V = 2\pi \int_{a}^{b} \text{radius} \times \text{height} \, dx\).
• For our scenario: \(V = 2\pi \int_{0}^{4} x(2-\sqrt{x}) \, dx = \frac{64\pi}{5}\).

The Shell Method shines when dealing with revolution around the y-axis, as it simplifies where the function does not lend itself easily to the Disk Method.

Washer Method

The Washer Method extends the Disk Method by allowing for solid regions with holes, creating a washer-like (ring) structure during rotation.
This method is ideal when there is an inner and outer radius involved, such as when revolving a shape that's not solid all the way through.Visualize rotating around a line like \(y=2\). Each washer's volume accounts for both an outer radius \(r_{\text{outer}}\) and an inner radius \(r_{\text{inner}}\). Imagine a large disk (the outer) with a smaller disk (the inner or hole) removed from its center.
In our example, when revolving the region bounded by \(y=\sqrt{x}\) about \(y=2\), the outer radius remains \(2\) and the inner radius becomes \(2-\sqrt{x}\).

• For the Washer Method: \(V = \pi \int_{a}^{b} [r_{\text{outer}}^2 - r_{\text{inner}}^2] \, dx\).
• Plugging in our example: \(V = \pi \int_{0}^{4} [(2)^2 - (2-\sqrt{x})^2] \, dx = \frac{16\pi}{3}\).

The Washer Method is your go-to when you need to calculate volumes involving hollow regions or gaps within the solid of revolution.