Problem 32 - Thomas Calculus

Question

In Exercises \(31-36,\) find the volumes of the solids generated by revolving the regions about the given axes. If you think it would be better to use washers in any given instance, feel free to do so. The region bounded by \(y=\sqrt{x}, y=2, x=0\) about $$\begin{array}{ll}{\text { a. the } x \text { -axis }} & {\text { b. the } y \text { -axis }} \\ {\text { c. the line } x=4} & {\text { d. the line } y=2}\end{array}$$

Step-by-Step Solution

Answer

(a) 8\pi, (b) 64\pi/5, (c) 128\pi/15, (d) 8\pi/3.

Step 1: Calculate the volume using the washer method around the x-axis

The region is defined by the curves \(y = \sqrt{x}\), \(y = 2\), and \(x = 0\). When revolving around the x-axis, we can use the washer method. The outer radius, \(R(x)\), is 2 (from \(y = 2\)), and the inner radius, \(r(x)\), is \(\sqrt{x}\). The volume \(V\) is calculated using the formula: \[ V = \pi \int_0^{x_1} [R(x)^2 - r(x)^2] \, dx \]where \(x_1\) is found by setting \(\sqrt{x} = 2\), giving \(x = 4\). Thus, \[ V = \pi \int_0^{4} (2^2 - x) \, dx \] Evaluating this, \[ V = \pi \left[ 4x - \frac{x^2}{2} \right]_0^{4} = \pi \left[ 16 - 8 \right] = 8\pi \].

Step 2: Calculate the volume using the shell method around the y-axis

When revolving around the y-axis, use the shell method. The height of the shell is \(y - x = \sqrt{x} - 0\), the radius \(r = x\), and the bounds are \(x = 0\) to \(x = 4\). Therefore, the volume \(V\) is given by:\[ V = 2\pi \int_0^{4} x (\sqrt{x}) \, dx = 2\pi \int_0^{4} x^{3/2} \, dx. \]Evaluating this gives:\[ V = 2\pi \left[ \frac{2}{5}x^{5/2} \right]_0^{4} = 2\pi \cdot \frac{2}{5} \cdot (32) = \frac{64\pi}{5}. \]

Step 3: Calculate the volume around the line x=4 using shell method

Revolving around \(x = 4\), use the shell method again with the radius \(r = 4 - x\) and the same height \(\sqrt{x}\). The bounds remain \(x = 0\) to \(x = 4\):\[ V = 2\pi \int_0^{4} (4 - x)(\sqrt{x}) \, dx = 2\pi \int_0^{4} (4x^{1/2} - x^{3/2}) \, dx. \]Breaking this into parts, \[ V = 2\pi \left[ \left( \frac{8}{3}x^{3/2} - \frac{2}{5}x^{5/2} \right) \right]_0^{4} \]Evaluates to:\[ V = 2\pi \left( \frac{8}{3} \cdot 8 - \frac{2}{5} \cdot 32 \right) = \frac{128\pi}{15}. \]

Step 4: Calculate the volume around the line y=2 using washer method

Revolving the region around \(y=2\), switch to washer method. The outer radius \(R = 2\), and the inner radius \(r = 2 - \sqrt{x}\) within the integral limits \(x=0\) to \(x=4\):\[ V = \pi \int_0^{4} [2^2 - (2 - \sqrt{x})^2] \, dx \].Simplifying,\[ V = \pi \int_0^{4} (4 - (4 - 4\sqrt{x} + x)) \, dx = \pi \int_0^{4} (4\sqrt{x} - x) \, dx. \]Evaluating the integral, \[ V = \pi \left[ \frac{8}{3}x^{3/2} - \frac{x^2}{2} \right]_0^{4} = \frac{32\pi}{3} - 8\pi = \frac{8\pi}{3}. \]

Key Concepts

washer method

The washer method is a powerful technique used in calculus to find the volume of a solid of revolution. It involves revolving a region around an axis, and the resulting solid is composed of nested disks akin to washers, hence the name. Each washer has a hole in the center, characterized by an inner and outer radius.

To visualize this method, imagine stacking multiple washers along the axis of rotation. The outer radius, denoted as \(R(x)\), corresponds to the distance from the axis to the outer edge of the region being revolved. The inner radius, \(r(x)\), is the distance from the axis to the inner edge or hole.

• Volume is calculated by subtracting the volume of the hole from the larger solid disk.
• The integral formula used is: \[ V = \pi \int_a^b \left( R(x)^2 - r(x)^2 \right) \, dx\]
• Perfect for scenarios where the solid isn't touching the axis directly, creating a natural "hole" (like the region between \(y = \sqrt{x}\) and 2).

Using the washer method ensures a systematic approach to such problems, allowing for precise calculation of the volume of complex shapes.

shell method

The shell method is another essential technique used to compute the volume of solids of revolution, especially when revolving around the y-axis. This method visualizes the solid as being made up of cylindrical shells. The approach is beneficial when the given region is a function of \(x\), revolving around a vertical axis.

Here’s how the shell method works:

• The radius of each cylindrical shell is the distance from the line of revolution, involving expressions like \(x\) or adjustments like \(4-x\) when revolving around an off-center line.
• The height of each shell is the value of the function being revolved, such as \(\sqrt{x}\).
• The formula used is: \[ V = 2\pi \int_a^b x \cdot f(x) \, dx\]This formula captures the essence of adding up the cylindrical shell slices from \(x=a\) to \(x=b\).

The shell method simplifies problems where the region is being revolved around a vertical axis, aiding in the straightforward calculation of volumes by considering lateral surface areas of cylindrical shells.

calculus integration

Calculus integration is an overarching tool that underpins both the washer and shell methods. It systematically accumulates quantities, crucial for calculating areas, volumes, and more in mathematics. Integration in calculus helps in summing infinitesimally small pieces to find the total volume of an irregular shape.

In volume problems:

• Integration offers precision by considering infinite slabs or slices, representing the thickness of washers or shells.
• It utilizes definite integrals with set bounds, \([a, b]\), to calculate a total area or volume based on the given region.
• The fundamental theorem of calculus connects the function's antiderivative with the area under a curve, a key concept in volume calculations.

By coupling integration with geometric formulas, like those from the washer and shell methods, calculus allows for a deeper understanding of how volumes are constructed and computed. Each integral takes into account the shape’s particular dimensions, offering a methodical way to solve complex geometry problems.