Problem 28 - Thomas Calculus

Question

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 a. the \(x\) -axis b. the \(y\) -axis c. the line \(x=4\) d. the line \(y=2\)

Step-by-Step Solution

Answer

a) 8\pi; b) 16\pi/3; c) 32\pi/15; d) 8\pi/3.

Step 1: Understand the Region

First, let's visualize the region: it's bounded by the curve \(y=\sqrt{x}\), the horizontal line \(y=2\), and the vertical line \(x=0\). This region lies in the first quadrant.

Step 2: Setup for the x-axis Revolution

To revolve around the \(x\)-axis for part (a), we use the disk method. The outer radius is the curve \(y=\sqrt{x}\), and the inner radius is 0 since we're revolving around the \(x\)-axis.

Step 3: Integration Bounds and Expression for x-axis

The integration bounds for part (a) are from \(x = 0\) to \(x = 4\) because \(y = 2\) corresponds to \(x = 4\) when \(2 = \sqrt{x}\). The volume is \[V = \pi \int_{0}^{4} (\sqrt{x})^2 \, dx.\]

Step 4: Integration to find Volume for x-axis

Calculate the integral: \[V = \pi \int_{0}^{4} x \, dx = \pi \left[ \frac{x^2}{2} \right]_{0}^{4} = \pi \left( \frac{16}{2} - 0 \right) = 8\pi.\]

Step 5: Setup for the y-axis Revolution

For revolving around the \(y\)-axis, the method of cylindrical shells is suitable. The height of each shell is \(2 - y\), and the radius is \(x = y^2\).

Step 6: Integration Bounds and Expression for y-axis

For part (b), the bounds are \(y=0\) and \(y=2\). The volume is \[V = 2\pi \int_{0}^{2} y^2(2-y) \, dy.\]

Step 7: Integration to find Volume for y-axis

Calculate the integral: \[V = 2\pi \int_{0}^{2} (2y^2 - y^3) \, dy = 2\pi \left[ \frac{2y^3}{3} - \frac{y^4}{4} \right]_{0}^{2} = 16\pi/3.\]

Step 8: Setup for x=4 Revolution

Revolve around \(x = 4\) using the washer method. The outer radius is \(4\) and the inner radius is \(4 - y^2\).

Step 9: Integration Bounds and Expression for x=4

The integration bounds remain \(y=0\) to \(y=2\), thus: \[V = \pi \int_{0}^{2} ((4^2) - (4-y^2)^2) \, dy.\]

Step 10: Integration to find Volume for x=4

Calculate the integral: \[V = \pi \int_{0}^{2} (16 - (16 - 8y^2 + y^4)) \, dy = \pi \int_{0}^{2} (8y^2 - y^4) \, dy\] which evaluates to \(32\pi/15\).

Step 11: Setup for y=2 Revolution

Revolve around \(y=2\) using the disk method. Outer radius: \(2 - 0 = 2\) and inner radius: \(2 - \sqrt{x}\).

Step 12: Integration Bounds and Expression for y=2

Bounds are \(x=0\) to \(x=4\), so the expression is \[V = \pi \int_{0}^{4} ((2^2) - (2 - \sqrt{x})^2) \, dx.\]

Step 13: Integration to find Volume for y=2

Calculate the integral: \[V = \pi \int_{0}^{4} (4 - (4 - 4\sqrt{x} + x)) \, dx = \pi \int_{0}^{4} (4\sqrt{x} - x) \, dx\], which evaluates to \(8\pi/3\).

Key Concepts

Disk Method

The disk method is a powerful technique for finding the volume of a solid of revolution. Imagine slicing the solid perpendicular to the axis of rotation. Each slice looks like a thin disk or circle. To compute the volume of the entire solid, we sum the volumes of these tiny disks.

Key points for the disk method include:

• The axis of rotation plays a pivotal role. When revolving around the x-axis or y-axis, the slices are aligned perpendicular to this axis.
• The radius of each disk is crucial. For rotations about the x-axis, this radius is the value of the function, f(x).
• The formula for volume when using the disk method is: \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \] where \(a\) and \(b\) are the limits of integration that define the segment of the curve being revolved.

With this method, the volume becomes a matter of evaluating the integral, offering precision and clarity in understanding three-dimensional shapes.

Cylindrical Shells

Cylindrical shells offer an elegant way to find the volume of solids of revolution when the disk method proves unwieldy. Imagine wrapping the function's graph into cylinders rather than stacking disks, particularly when rotating around an axis not parallel to the typical axes.

For the method of cylindrical shells, remember:

• The height of the shell is the difference in function values, typically the upper curve minus the lower curve. In our problem, this is often \( 2 - y \).
• The radius of the shell directly relates to the axis of rotation. For revolution around the y-axis, the radius is the x-value, often \( y^2 \) when calculated in context.
• The volume formula using cylindrical shells is: \[ V = 2\pi \int_{c}^{d} (\text{radius})(\text{height}) \, dy \]

This method can simplify calculations significantly, especially with boundaries defined in terms of one variable, revealing the structure of the solid more intuitively.

Washer Method

The washer method extends the disk method, particularly useful when there is a gap around the axis of rotation. Visualize washers, or disks with holes, stacked to form the solid.

With the washer method, consider:

• An outer and an inner radius. The outer radius is taken from the farther curve from the axis of rotation, and the inner radius from the closer curve.
• Revolution creates both outer and inner edges. Their difference defines the washer's volume.
• The formula reads: \[ V = \pi \int_{a}^{b} \left[ R^2 - r^2 \right] \, dy \] where \(R\) and \(r\) are the outer and inner radii, respectively.

This method is indispensable when the region bounded has separate curves defining its boundaries, allowing us to calculate volume accurately by removing the inner void from the outer solid.