Fix the mistake: Applications of Integration (Volumes and Lengths of Curves)
Volume of a Solid (Integration of Cross-Sectional Areas)
If the cross-sectional area of a solid at x = t is S(t), its volume is
V = ∫_a^b S(x) dx
Example ① The volume of a cone with a circular base of radius r and height h.
Solution: Let the vertex be the origin and the axis be the x-axis. The radius at position x is (r/h)x, so the cross-sectional area is S(x) = π(r/h)²x².
V = ∫_0^h π(r²/h²)x² dx = π(r²/h²)·h³/3 = πr²h/3
Solution: Let the vertex be the origin and the axis be the x-axis. The radius at position x is (r/h)x, so the cross-sectional area is S(x) = π(r/h)²x².
V = ∫_0^h π(r²/h²)x² dx = π(r²/h²)·h³/3 = πr²h/3
Key Point
- Express the cross-sectional area as a function of x → integrate
Practice Problems
- The volume of a square pyramid with a square base of side length a and height h.
Solution and Explanation
- Answer: a²h/3
S(x) = a²(x/h)²、V = ∫_0^h a²x²/h² dx = a²h/3。
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