Build volume from cross-sectional area

Try

Imagine stacking thin circular disks along an axis. A disk with radius r has cross-sectional area πr². Multiplying that area by a small thickness estimates the slice’s volume. Explain why integrating radius alone cannot give volume: both the geometry and the units would be wrong.

Predict

A ring-shaped cross section has outer radius 3 and inner radius 1. Predict whether its area is π(3−1)² or π(3²−1²). Draw the two complete disks and remove the smaller one. A washer’s width is not the radius of a replacement disk.

Experiment

Rotate the region 0≤y≤x, 0≤x≤2 about the x-axis. Each vertical strip makes a disk of radius x. At selected locations, compare radius with cross-sectional area.

x	Radius	Area
0	0	0
1	1	π
2	2	4π

The volume is π∫₀²x²dx=8π/3. Independently recognize the solid as a cone of height 2 and base radius 2: the geometric formula gives the same value. This check compares two methods rather than repeating the integral arithmetic.

Understand

For cross-sectional area A(x) perpendicular to the x direction, volume is ∫A(x)dx under the usual integrability assumptions. Rotations are one special way to determine A. Disk area is πR²; washer area is π(R²−r²), with outer radius R at least inner radius r and both nonnegative distances from the rotation axis.

Shifted axes require shifted distances. Rotating the region between y=0 and y=x on [0,1] about y=−1 gives outer radius x+1 and inner radius 1. The integrand is π[(x+1)²−1], not πx². Integration yields 4π/3. The region stays above the axis, so each vertical strip generates one nonoverlapping washer. If a strip crosses the axis, determine the swept disk from the farthest boundary; blindly assigning signed inner radii can subtract a nonexistent hole.

For rotation about the y-axis, horizontal slices often produce convenient washers. Between x=y² and x=2y for 0≤y≤2, the outer radius is 2y and inner radius y². The volume is π∫₀²(4y²−y⁴)dy=64π/15. The order of the radii follows from y²≤2y on that interval, not from which formula happens to look larger.

Not every cross section is circular. A solid with square sections of side 2−x on [0,2] has volume ∫₀²(2−x)²dx=8/3. Equilateral triangular sections of side x have area √3 x²/4, obtained from the triangle’s altitude. On [0,3] their volume is 9√3/4. Rectangular, triangular and semicircular sections each require their own area formula before integration.

This general slicing method assumes the sections describe the intended solid without overlap. A two-dimensional drawing is a guide, not proof that a particular cross section is circular or square. The stated construction supplies that information. Cross-sectional area times thickness has cubic units, providing a useful dimensional check.

Apply

For each practice task, draw the axis and a perpendicular slice, label every radius as a distance and derive the area expression. Keep π and radicals exact in written work. The rational checker asks for a coefficient only where the prompt explicitly defines that answer format.

Continue

Next use strips parallel to the axis to create shells. Changing the slicing direction can simplify the algebra, but it also changes the geometric factors in the integrand.