Volume Calculation Method of Cross-Sections
The cross-section method calculates the volume of a solid by integrating the area of its cross-sectional slices. These slices, perpendicular to an axis, vary in shape and size along the solid. By summing their areas, this method effectively builds up the total volume, useful for irregular shapes or when the solid lacks symmetry.
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The Cross-Section Method calculates a solid’s volume by integrating the area of its cross-sections. These sections, perpendicular to an axis, vary in shape across the solid. By summing these area slices, the method accumulates the total volume, adept for solids with non-uniform cross-sections, providing a precise measure of space occupied by complex geometrical figures.
Utilizing the Cross-Section Method for volume involves defining a function \(A(x)\) that represents the area of a cross-section at any point (x) along a chosen axis. The integral \(\int_{a}^{b} A(x) dx\) spans the limits where the solid extends. This process mathematically aggregates the product of each area and its infinitesimal thickness, yielding the solid’s entire volume. Essential for irregular shapes, this method bridges geometry and calculus, offering a robust tool for volumetric analysis in multidisciplinary applications.
Let’s calculate the volume of a solid whose cross-sectional area perpendicular to the x-axis is a square, with the side of the square at any point (x) given by \( s(x) = 4x \). We’ll find the volume of this solid between \( x = 1 \) and \( x = 3 \).
Step-by-Step Solution:
- Define the Area of a Cross-Section:
- The area of a square is \( A = s^2 \), so the area of each cross-section is \( A(x) = (4x)^2 = 16x^2 \).
- Set Up the Integral for Volume:
- The volume \( V \) of the solid is found by integrating the area function \( A(x) = 16x^2 \) from \( x = 1 \) to \( x = 3 \):
\( V = \int_{1}^{3} 16x^2 \, dx \)
- Calculate the Integral:
- The antiderivative of \( 16x^2 \) is \( \frac{16x^3}{3} \), so:
\( V = \left[ \frac{16x^3}{3} \right]_{1}^{3} \)
- Evaluate the Integral:
- Substitute the limits into the antiderivative:
\( V = \left( \frac{16(3)^3}{3} \right) – \left( \frac{16(1)^3}{3} \right) \)
\( V = \left( \frac{16 \cdot 27}{3} \right) – \left( \frac{16}{3} \right) \)
\( V = 144 – \frac{16}{3} \)
The volume of the solid, with cross-sectional areas as squares where the side length at any point \(x\) is \(4x\), calculated between \(x = 1\) and \(x = 3\), is approximately \(138.67\) cubic units.
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