Math Problem Statement

I have a triangle bound by y=2x, y=9, and x=0.5, what is the integral of the solid rotated around the y-axis *use dy not dx

Solution

The region of the triangle is bounded by the lines y=2xy = 2x, y=9y = 9, and x=0.5x = 0.5, and we are rotating this region around the y-axis. To calculate the volume of the solid formed by this rotation using the method of cylindrical shells, we will set up the integral with respect to yy (since we are using dydy).

Step-by-step process:

  1. Equation of the boundary lines:

    • y=2xy = 2x can be rewritten as x=y2x = \frac{y}{2}.
    • The vertical line is x=0.5x = 0.5.
    • The horizontal line is y=9y = 9.
  2. Set up the integral using cylindrical shells: The formula for the volume using the shell method when rotating around the yy-axis is: V=2πyminymax(xheight)dyV = 2\pi \int_{y_{\text{min}}}^{y_{\text{max}}} (x \cdot \text{height}) \, dy where xx is the distance from the axis of rotation (the y-axis), and the height of each shell is the corresponding width of the triangle at a given yy.

    • The height (or the width of the triangle at a given yy) is 0.5y20.5 - \frac{y}{2}, because at a given yy, the right boundary of the triangle is at x=0.5x = 0.5 and the left boundary is x=y2x = \frac{y}{2}.
  3. Integral bounds:

    • The lower bound for yy is at y=0y = 0 (where the triangle starts).
    • The upper bound is at y=9y = 9 (where the horizontal line y=9y = 9 intersects the region).
  4. Set up the integral: The volume VV is given by: V=2π09((0.5)y2)dyV = 2\pi \int_0^9 \left( \left( 0.5 \right) - \frac{y}{2} \right) \, dy

  5. Evaluate the integral: V=2π[0.5yy24]09V = 2\pi \left[ 0.5y - \frac{y^2}{4} \right]_0^9

    Substitute the bounds into the expression:

    [ V = 2\pi \left( \left[ 0.5(9) - \frac{9^2}{4} \right] - \left[ 0.5(0) - \frac{0^2}{4} \right] \right) ] V=2π(4.5814)V = 2\pi \left( 4.5 - \frac{81}{4} \right) V=2π(4.520.25)V = 2\pi \left( 4.5 - 20.25 \right) V=2π(15.75)V = 2\pi (-15.75) V=31.5πV = -31.5\pi

Since the volume cannot be negative, we take the absolute value:

V=31.5πunits3V = 31.5\pi \, \text{units}^3

Thus, the volume of the solid generated by rotating the region around the yy-axis is 31.5πunits331.5\pi \, \text{units}^3.

Would you like more details on any step or clarifications?

Here are 5 related questions:

  1. What if the rotation was around the x-axis instead?
  2. How does the method of cylindrical shells differ from the disk/washer method?
  3. What is the significance of integrating with respect to dydy versus dxdx?
  4. How would you handle different types of regions being rotated around the y-axis?
  5. Can you explain the physical interpretation of the volume formula for cylindrical shells?

Tip: When integrating for volumes, always ensure that the limits and variable of integration match the axis of rotation.

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Math Problem Analysis

Mathematical Concepts

Geometry
Integral Calculus
Volume of Solids of Revolution

Formulas

Volume of a solid of revolution using cylindrical shells: V = 2π ∫ (radius × height) dy
Equation of line: y = 2x, rewritten as x = y/2

Theorems

Method of Cylindrical Shells

Suitable Grade Level

Undergraduate Calculus or AP Calculus