Math Problem Statement

Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by y = 0, y = x^2, and y = 10 - 3x about the line x = 4.

Solution

To solve the problem, we'll use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by the curves y=0y = 0, y=x2y = x^2, and y=103xy = 10 - 3x about the line x=4x = 4.

Steps to Solve:

  1. Understand the boundaries:

    • y=0y = 0: This is the x-axis.
    • y=x2y = x^2: This is a parabola opening upwards.
    • y=103xy = 10 - 3x: This is a straight line with a negative slope.

    We need to find the points where these curves intersect to determine the limits of integration.

  2. Find intersection points: Set x2=103xx^2 = 10 - 3x to find the intersection points: x2+3x10=0x^2 + 3x - 10 = 0 Solving this quadratic equation using the quadratic formula: x=3±32+4×1×102×1x = \frac{-3 \pm \sqrt{3^2 + 4 \times 1 \times 10}}{2 \times 1} x=3±492=3±72x = \frac{-3 \pm \sqrt{49}}{2} = \frac{-3 \pm 7}{2} So, x=2x = 2 and x=5x = -5.

    Since the region bounded is positive, we consider the interval x=5x = -5 to x=2x = 2.

  3. Set up the integral using the cylindrical shells method: The formula for the volume using cylindrical shells is: V=2πab(rshell)(hshell)dxV = 2\pi \int_{a}^{b} (r_{\text{shell}})(h_{\text{shell}}) \, dx Here:

    • rshell=4xr_{\text{shell}} = 4 - x (distance from the line x=4x = 4)
    • hshell=(103x)x2h_{\text{shell}} = (10 - 3x) - x^2 (height of the shell, difference between the line and the parabola)

    Therefore, the integral becomes: V=2π52(4x)[(103x)x2]dxV = 2\pi \int_{-5}^{2} (4 - x) \left[(10 - 3x) - x^2\right] dx

  4. Solve the integral: V=2π52[(4x)(103xx2)]dxV = 2\pi \int_{-5}^{2} \left[(4 - x)(10 - 3x - x^2)\right] dx Expanding the integrand: V=2π52[4012x4x210x+3x2+x3]dxV = 2\pi \int_{-5}^{2} \left[40 - 12x - 4x^2 - 10x + 3x^2 + x^3\right] dx Simplify the expression: V=2π52[x3x222x+40]dxV = 2\pi \int_{-5}^{2} \left[x^3 - x^2 - 22x + 40\right] dx Now, integrate term by term: V=2π[x44x3311x2+40x]52V = 2\pi \left[ \frac{x^4}{4} - \frac{x^3}{3} - 11x^2 + 40x \right]_{-5}^{2}

  5. Evaluate the integral: First, substitute x=2x = 2 and then x=5x = -5: V=2π[(24423311(2)2+40(2))((5)44(5)3311(5)2+40(5))]V = 2\pi \left[\left(\frac{2^4}{4} - \frac{2^3}{3} - 11(2)^2 + 40(2)\right) - \left(\frac{(-5)^4}{4} - \frac{(-5)^3}{3} - 11(-5)^2 + 40(-5)\right)\right] Calculate these values and subtract to find the final volume.

Would you like the details for the final calculation or do you have any questions?


Related Questions:

  1. What are the advantages of using the cylindrical shells method over the disk/washer method?
  2. How does rotating about different lines (e.g., y=5y = 5 instead of x=4x = 4) change the volume integral?
  3. How do you determine the limits of integration for problems like this?
  4. Can the volume of a solid be negative, and why?
  5. What happens to the volume if the region is rotated about the yy-axis instead?

Tip:

When solving complex integrals involving rotation, carefully identifying the radius and height functions is crucial for setting up the correct integral.

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Math Problem Analysis

Mathematical Concepts

Calculus
Integration
Volume of Solids of Revolution
Cylindrical Shells Method

Formulas

V = 2π ∫[a,b] (radius)(height) dx

Theorems

Cylindrical Shells Method

Suitable Grade Level

Undergraduate Calculus