Math Problem Statement

Set up, but do not evaluate, an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. x = y^2, y = 0, x = 4; about x = 9

Solution

To set up the integral for the volume of the solid obtained by rotating the region bounded by the curves x=y2x = y^2, y=0y = 0, and x=4x = 4 about the line x=9x = 9, we will use the shell method because the axis of rotation, x=9x = 9, is parallel to the region but not on the boundary.

Step-by-step setup using the shell method:

  1. Visualize the region:

    • The curve x=y2x = y^2 is a parabola opening to the right.
    • The region is bounded by the line x=4x = 4 (a vertical line) and the horizontal line y=0y = 0.
    • This region will be rotated about the vertical line x=9x = 9.
  2. Shell radius: The distance from the line of rotation x=9x = 9 to a point on the curve x=y2x = y^2 is: Radius of shell=9x=9y2\text{Radius of shell} = 9 - x = 9 - y^2 where x=y2x = y^2.

  3. Shell height: The height of each shell is the horizontal distance between the line x=y2x = y^2 and the vertical boundary at x=4x = 4. So, the height is: Height of shell=4y2\text{Height of shell} = 4 - y^2

  4. Volume element (dV): The volume element using the shell method is: dV=2π×(radius of shell)×(height of shell)×dydV = 2\pi \times (\text{radius of shell}) \times (\text{height of shell}) \times dy Substituting in the expressions for the radius and height: dV=2π(9y2)(4y2)dydV = 2\pi (9 - y^2)(4 - y^2) \, dy

  5. Limits of integration: The limits for yy come from the boundaries of the region. From the curve x=y2x = y^2, we know: y=0toy=2y = 0 \quad \text{to} \quad y = 2 (since at x=4x = 4, y=4=2y = \sqrt{4} = 2).

  6. Integral setup: The integral for the volume is: V=022π(9y2)(4y2)dyV = \int_{0}^{2} 2\pi (9 - y^2)(4 - y^2) \, dy

This integral represents the volume of the solid obtained by rotating the region about x=9x = 9.

Would you like more details or have any questions about the setup?


Here are 5 related questions that build on this topic:

  1. How would the integral change if the rotation axis were x=0x = 0?
  2. Can you use the disk/washer method for this problem instead? If so, how would you set it up?
  3. What if the boundary was extended to x=9x = 9? How would that affect the integral?
  4. How does changing the rotation axis to a different line, like x=5x = 5, affect the volume?
  5. How would the setup change if the region were rotated around the y-axis instead?

Tip: The shell method is especially useful when rotating around a vertical line that is not the boundary of the region, as in this case!

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Math Problem Analysis

Mathematical Concepts

Calculus
Integral Calculus
Volume of Solids of Revolution
Shell Method

Formulas

Shell radius: r(y) = 9 - y^2
Shell height: h(y) = 4 - y^2
Volume element: dV = 2π(9 - y^2)(4 - y^2) dy
Integral setup: V = ∫[0, 2] 2π(9 - y^2)(4 - y^2) dy

Theorems

Method of Shells for Volumes of Revolution

Suitable Grade Level

Grades 11-12 (Advanced High School Calculus) or Early College Level