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

Solution

The Shell Method is useful when rotating around an axis that is parallel to the axis of integration, in this case, the x-axis. For the problem given, we have the equation y=1xy = 1 - x, and we are revolving the region around the x-axis.

Step 1: Identify the region

The equation y=1xy = 1 - x is a straight line with y-intercept 1 and slope -1. The region of interest lies between y=1xy = 1 - x, the x-axis (i.e., y=0y = 0), and the limits where the curve intersects the axes.

The line intersects the x-axis when y=0y = 0, so: 1x=0x=11 - x = 0 \quad \Rightarrow \quad x = 1 Thus, the region is bounded between x=0x = 0 and x=1x = 1.

Step 2: Set up the shell method formula

The Shell Method formula for rotating around the x-axis is: V=2πab(radius)×(height)dxV = 2\pi \int_{a}^{b} \text{(radius)} \times \text{(height)} \, dx

  • Radius: The distance from a point xx to the axis of rotation, which is simply yy (since we're rotating around the x-axis).
  • Height: The value of xx at each point, which is xx.

Thus, for this problem, the radius is y=1xy = 1 - x and the height is xx.

Step 3: Set up the integral

The bounds of integration are from x=0x = 0 to x=1x = 1, and the integral becomes: V=2π01(1x)xdxV = 2\pi \int_{0}^{1} (1 - x) x \, dx

Step 4: Evaluate the integral

Now, expand the integrand: V=2π01(xx2)dxV = 2\pi \int_{0}^{1} (x - x^2) \, dx

Next, compute the integral: V=2π[x22x33]01V = 2\pi \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1}

Evaluate the expression at the limits: V=2π((122133)(022033))V = 2\pi \left( \left( \frac{1^2}{2} - \frac{1^3}{3} \right) - \left( \frac{0^2}{2} - \frac{0^3}{3} \right) \right) V=2π(1213)V = 2\pi \left( \frac{1}{2} - \frac{1}{3} \right)

Simplify: V=2π(3626)=2π×16=π3V = 2\pi \left( \frac{3}{6} - \frac{2}{6} \right) = 2\pi \times \frac{1}{6} = \frac{\pi}{3}

Final Answer:

The volume of the solid generated by revolving the region around the x-axis is: V=π3V = \frac{\pi}{3}


Would you like more details on any step? Here are 5 related questions for further exploration:

  1. How would the volume change if the region was revolved around the y-axis instead?
  2. Can the Disk/Washer Method be applied to solve this same problem?
  3. How does the Shell Method differ from the Disk Method in terms of applications?
  4. What is the general formula for the Shell Method, and when is it best applied?
  5. How can you extend this method to more complex curves?

Tip: When using the Shell Method, always double-check the orientation of the axis of rotation to set up the radius and height correctly!

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Math Problem Analysis

Mathematical Concepts

Calculus
Volume of Solids of Revolution
Integration

Formulas

Shell Method: V = 2π ∫(radius) × (height) dx

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Grades 11-12, College