Math Problem Statement

Use the shell method to find the volume of the solid generated by revolving the shaded region about the​ x-axis. Set up the integral that gives the volume of the solid. upper = y=sqrt3, lower = x=y^2

Solution

To find the volume of the solid generated by revolving the shaded region about the x-axis using the shell method, we need to set up the integral for the volume. The shell method involves integrating with respect to the variable perpendicular to the axis of revolution, which in this case is the yy-axis, since we are revolving around the xx-axis.

Step-by-Step Solution:

  1. Identify the bounds and functions:

    • Upper boundary: y=3y = \sqrt{3}
    • Lower boundary: x=y2x = y^2
    • We are rotating about the xx-axis, so the radius of each shell will be yy.
  2. Determine the radius and height of the shell:

    • Radius of the shell: yy (distance from the yy-axis to the xx-axis).
    • Height of the shell: y2y^2 (given by x=y2x = y^2).
  3. Set up the integral:

    • The formula for the volume of a solid using the shell method is: V=2πab(radius)×(height)dyV = 2\pi \int_a^b (\text{radius}) \times (\text{height}) \, dy
    • Substituting in the radius and height: V=2π03y(y2)dyV = 2\pi \int_0^{\sqrt{3}} y \cdot (y^2) \, dy
  4. Simplify the integral:

    • Combine the terms inside the integral: V=2π03y3dyV = 2\pi \int_0^{\sqrt{3}} y^3 \, dy
  5. Evaluate the integral:

    • Integrate y3y^3: V=2π[y44]03V = 2\pi \left[\frac{y^4}{4}\right]_0^{\sqrt{3}}
    • Substitute the limits: V=2π((3)44044)V = 2\pi \left(\frac{(\sqrt{3})^4}{4} - \frac{0^4}{4}\right)
    • Simplify: V=2π(94)=9π2V = 2\pi \left(\frac{9}{4}\right) = \frac{9\pi}{2}

Final Integral Setup:

V=2π03y3dyV = 2\pi \int_0^{\sqrt{3}} y^3 \, dy

The evaluated volume is: V=9π2V = \frac{9\pi}{2}

Would you like further details on any of these steps or have any questions?

Here are some related questions for further practice:

  1. How would you set up the shell method integral if the region was rotated around the y-axis?
  2. What changes in the setup if the upper function was x=y+1x = y + 1 instead?
  3. How would the volume change if the boundaries were between y=1y = 1 and y=3y = \sqrt{3}?
  4. Can you set up the volume integral using the disk/washer method instead?
  5. How do you determine the limits of integration for the shell method?

Tip: When using the shell method, always ensure your radius and height expressions are in terms of the integration variable for accuracy.

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Math Problem Analysis

Mathematical Concepts

Calculus
Volume of Solids of Revolution
Shell Method
Integration

Formulas

Shell Method Formula: V = 2π ∫ (radius) × (height) dy
Integral of y^3: ∫ y^3 dy = (y^4 / 4)

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Grades 11-12, College Level