Math Problem Statement

A region is formed by the functions y = x^3 + x^2 + 4 and y = 0 and bounded by the line x = 1. Calculate the volume of the object formed by rotating the region about the x-axis.

Solution

The problem involves calculating the volume of the solid of revolution obtained by rotating the given region about the xx-axis. The region is defined by the curve:

y=x3+x2+4,y = x^3 + x^2 + 4,

and bounded by y=0y = 0 and the vertical line x=1x = 1.

Steps to solve:

  1. Volume formula for revolution about the xx-axis: The formula for the volume of a solid of revolution about the xx-axis is: V=πab[f(x)]2dx,V = \pi \int_a^b [f(x)]^2 \, dx, where f(x)f(x) is the function describing the curve, and aa and bb are the bounds of xx.

  2. Set up the integral:

    • The bounds are from x=0x = 0 to x=1x = 1.
    • The function is f(x)=x3+x2+4f(x) = x^3 + x^2 + 4.
    • The square of the function is: [f(x)]2=(x3+x2+4)2.[f(x)]^2 = \left(x^3 + x^2 + 4\right)^2.

    Substituting this into the volume formula: V=π01(x3+x2+4)2dx.V = \pi \int_0^1 \left(x^3 + x^2 + 4\right)^2 \, dx.

  3. Expand the square: Expand (x3+x2+4)2\left(x^3 + x^2 + 4\right)^2: (x3+x2+4)2=x6+2x5+9x4+16x3+8x2+16.\left(x^3 + x^2 + 4\right)^2 = x^6 + 2x^5 + 9x^4 + 16x^3 + 8x^2 + 16.

    So the integral becomes: V=π01(x6+2x5+9x4+16x3+8x2+16)dx.V = \pi \int_0^1 \left(x^6 + 2x^5 + 9x^4 + 16x^3 + 8x^2 + 16\right) \, dx.

  4. Integrate term by term: Compute each term separately: 01x6dx=x7701=17,\int_0^1 x^6 \, dx = \frac{x^7}{7} \bigg|_0^1 = \frac{1}{7}, 012x5dx=2x6601=13,\int_0^1 2x^5 \, dx = \frac{2x^6}{6} \bigg|_0^1 = \frac{1}{3}, 019x4dx=9x5501=95,\int_0^1 9x^4 \, dx = \frac{9x^5}{5} \bigg|_0^1 = \frac{9}{5}, 0116x3dx=16x4401=4,\int_0^1 16x^3 \, dx = \frac{16x^4}{4} \bigg|_0^1 = 4, 018x2dx=8x3301=83,\int_0^1 8x^2 \, dx = \frac{8x^3}{3} \bigg|_0^1 = \frac{8}{3}, 0116dx=16x01=16.\int_0^1 16 \, dx = 16x \bigg|_0^1 = 16.

  5. Add the results: Combine all terms: V=π(17+13+95+4+83+16).V = \pi \left(\frac{1}{7} + \frac{1}{3} + \frac{9}{5} + 4 + \frac{8}{3} + 16\right).

    First, simplify the fractions: 17+13+95+4+83+16=105105+35105+189105+420105+280105+1680105.\frac{1}{7} + \frac{1}{3} + \frac{9}{5} + 4 + \frac{8}{3} + 16 = \frac{105}{105} + \frac{35}{105} + \frac{189}{105} + \frac{420}{105} + \frac{280}{105} + \frac{1680}{105}.

    Combine: 105+35+189+420+280+1680105=2709105.\frac{105 + 35 + 189 + 420 + 280 + 1680}{105} = \frac{2709}{105}.

  6. Simplify: V=π2709105=2709π105.V = \pi \cdot \frac{2709}{105} = \frac{2709\pi}{105}.

    Reduce the fraction: V=903π35.V = \frac{903\pi}{35}.

Final Answer:

V=903π35units3.V = \frac{903\pi}{35} \, \text{units}^3.

Would you like detailed steps or have any questions?

Related Questions:

  1. How do you handle the volume of revolution when the region is bounded by two curves?
  2. How does the disk method differ from the shell method?
  3. What if the region is revolved around a vertical axis instead?
  4. Can this problem be solved using numerical integration instead?
  5. How does symmetry simplify volume of revolution problems?

Tip:

Always double-check your expanded polynomial terms to avoid calculation errors!

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Math Problem Analysis

Mathematical Concepts

Volumes of Revolution
Definite Integrals
Polynomial Expansion

Formulas

Volume of revolution about x-axis: V = π ∫[a to b] [f(x)]^2 dx
Polynomial expansion: (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

Theorems

Disk Method for Volumes of Revolution

Suitable Grade Level

Grade 11-12 or introductory college-level calculus