Math Problem Statement

Use the shell method to write and evaluate the definite integral that represents the volume of the solid generated by revolving the plane region about the x-axis. y = 1 x The x y-coordinate plane is given. A curve, a region, and two representative rectangles are graphed. The curve starts at (1, 1), goes down and right becoming less steep, and ends at (4, 1⁄4). The region is below the curve, above the x-axis, to the right of x = 1, and to the left of x = 4. The region is divided into two parts by a dashed horizontal line at y = 1⁄4. The first horizontal representative rectangle goes from (1, 0.67) to (1.5, 0.67). The second horizontal representative rectangle goes from (1, 0.13) to (4, 0.13).

Solution

To find the volume of the solid generated by revolving the region under the curve y=1xy = \frac{1}{x} about the x-axis using the shell method, we follow these steps:

1. Set up the Shell Method Formula:

The general formula for the volume of a solid of revolution using the shell method, when revolving around the x-axis, is:

V=2πy1y2(radius)(height)dyV = 2\pi \int_{y_1}^{y_2} (\text{radius}) (\text{height}) \, dy

2. Identify Variables:

In this case, we revolve the region about the x-axis, and the height of the shells will be given in terms of yy, while the radius will simply be yy because we're revolving around the x-axis. From the curve y=1xy = \frac{1}{x}, we solve for xx in terms of yy:

y=1x    x=1yy = \frac{1}{x} \implies x = \frac{1}{y}

Thus, the radius is yy, and the height of the shell is the distance between x=1x = 1 and x=1yx = \frac{1}{y} at each yy, which is given by 1y1\frac{1}{y} - 1.

3. Determine the Limits of Integration:

From the given information, the region lies between x=1x = 1 and x=4x = 4, which corresponds to the values of y=1xy = \frac{1}{x}, or y=1y = 1 to y=14y = \frac{1}{4}. Therefore, the limits of integration for yy are:

y1=14,y2=1y_1 = \frac{1}{4}, \quad y_2 = 1

4. Write the Integral:

Now, we substitute everything into the formula for the volume:

V=2π141y(1y1)dyV = 2\pi \int_{\frac{1}{4}}^{1} y \left( \frac{1}{y} - 1 \right) \, dy

5. Simplify the Integral:

Distribute the yy term inside the parentheses:

V=2π141(1y)dyV = 2\pi \int_{\frac{1}{4}}^{1} (1 - y) \, dy

6. Evaluate the Integral:

First, integrate 1y1 - y:

(1y)dy=yy22\int (1 - y) \, dy = y - \frac{y^2}{2}

Now, evaluate this from y=14y = \frac{1}{4} to y=1y = 1:

V=2π[(1122)(14(14)22)]V = 2\pi \left[ \left( 1 - \frac{1^2}{2} \right) - \left( \frac{1}{4} - \frac{\left(\frac{1}{4}\right)^2}{2} \right) \right]

Simplifying the expressions:

V=2π[(112)(14132)]V = 2\pi \left[ \left( 1 - \frac{1}{2} \right) - \left( \frac{1}{4} - \frac{1}{32} \right) \right] V=2π[12(832132)]V = 2\pi \left[ \frac{1}{2} - \left( \frac{8}{32} - \frac{1}{32} \right) \right] V=2π[12732]V = 2\pi \left[ \frac{1}{2} - \frac{7}{32} \right] V=2π[1632732]V = 2\pi \left[ \frac{16}{32} - \frac{7}{32} \right] V=2π932V = 2\pi \cdot \frac{9}{32}

7. Final Answer:

V=9π16V = \frac{9\pi}{16}

Thus, the volume of the solid generated by revolving the region under the curve y=1xy = \frac{1}{x} from x=1x = 1 to x=4x = 4 about the x-axis is 9π16\frac{9\pi}{16}.


Would you like further details or have any questions?

Here are 5 related questions to consider:

  1. How would the solution change if we revolved around the y-axis instead?
  2. Can you derive the volume using the disk/washer method instead of the shell method?
  3. What if the region was bounded between x=2x = 2 and x=5x = 5?
  4. How do you set up the shell method for different types of curves?
  5. What is the geometric interpretation of the shell height and radius in this example?

Tip: The shell method is often easier for regions where the function is best described in terms of yy.

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Math Problem Analysis

Mathematical Concepts

Calculus
Volume of Solids of Revolution
Integration

Formulas

V = 2π ∫(radius)(height) dy
y = 1/x
x = 1/y

Theorems

Shell Method

Suitable Grade Level

Grades 11-12