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8.7 Volumes with Cross Sections: Squares and Rectangles

Syllabus
2020
Topic
8.7
Level
—

Learning objectives

—CHA-5.B—Calculate volumes of solids with known cross sections using definite integrals• CHA-5.B Calculate volumes of solids with known cross sections using definite integrals.• CHA-5.B.1 Volumes of solids with square and rectangular cross sections can be found using definite integrals and the area formulas for these shapes.• Enduring understanding CHA-5: Definite integrals allow us to solve problems involving the accumulation of change in area or volume over an interval.

Build Volume from Square or Rectangular Slices

For a solid whose cross sections perpendicular to the xx-axis are known, a thin slice of thickness dxdx has volume approximately A(x) dxA(x)\,dx. Adding all slices gives the exact volume. The base region supplies a length such as s(x)=top−bottoms(x)=\text{top}-\text{bottom}; the named cross-sectional shape determines how that length becomes area.

V=\int_a^b A(x),dx,\qquad A_{\text{square}}(x)=[s(x)]^2,\qquad A_{\text{rectangle}}(x)=\ell(x)w(x)

  1. Identify the slicing direction and interval.
  2. Express the required side length or lengths from the base region.
  3. Use the square or rectangle area formula to create A(x)A(x).
  4. Integrate A(x)A(x) across the interval.
  5. Report a nonnegative result in cubic units.

Example: the base is between y=xy=x and y=x2y=x^2 on 0≤x≤10\le x\le1, and perpendicular cross sections are squares. Here s(x)=x−x2s(x)=x-x^2, so A(x)=(x−x2)2A(x)=(x-x^2)^2. Therefore V=∫01(x2−2x3+x4) dx=[x33−x42+x55]01=130V=\int_0^1(x^2-2x^3+x^4)\,dx=\left[\frac{x^3}{3}-\frac{x^4}{2}+\frac{x^5}{5}\right]_0^1=\frac{1}{30} cubic units.

Do not integrate the slice length itself. A distance such as s(x)s(x) has linear units; the integrand must be the cross-sectional area A(x)A(x), which has square units. Multiplying by slice thickness through integration then produces cubic units.