If a function has cross-sectional area A(x) perpendicular to the x-axis, its volume over [a,b] is:

If a function has cross-sectional area A(x) perpendicular to the x-axis, its volume over [a,b] is:

• A. ∫_a^b A(x) dx
• B. ∫_a^b x A(x) dx
• C. ∫_a^b A(x) dx^2
• D. ∫_a^b dA(x) dx

Answer: (A)

Think of building the solid by stacking thin slices perpendicular to the x-axis. Each slice has cross-sectional area A(x) and thickness dx, so its volume is A(x) dx. Adding up all these slices over the interval from a to b gives the total volume, which is V = ∫_a^b A(x) dx. This uses the fundamental idea of slicing: area times thickness summed continuously along x yields volume. The other expressions don’t match this idea. Multiplying by x would weight each slice by its position, giving a different quantity. Using dx^2 isn’t a standard infinitesimal for volume in this context. Using the derivative dA(x)/dx times dx would collapse to dA, which represents a change in area, not the actual volume formed by the slices.

Think of building the solid by stacking thin slices perpendicular to the x-axis. Each slice has cross-sectional area A(x) and thickness dx, so its volume is A(x) dx. Adding up all these slices over the interval from a to b gives the total volume, which is V = ∫_a^b A(x) dx. This uses the fundamental idea of slicing: area times thickness summed continuously along x yields volume.

The other expressions don’t match this idea. Multiplying by x would weight each slice by its position, giving a different quantity. Using dx^2 isn’t a standard infinitesimal for volume in this context. Using the derivative dA(x)/dx times dx would collapse to dA, which represents a change in area, not the actual volume formed by the slices.