Volumes with Cross Sections: Squares and Rectangles: Study Guide (Calculus AB 8.7)

Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both.

A note on notation

This page has no equation renderer, so integrals are written in a compact form: ∫ (a to b) A(x) dx means the definite integral of A(x) from x = a to x = b. On paper, write a at the bottom of the integral sign and b at the top. [s(x)]² means the whole side length, squared. Keep the dx or dy: it tells the reader which way the slices go.

The idea: a solid is a stack of thin slabs

Picture a loaf of bread cut into slices. Each slice is thin, and its face has some area. The volume of the loaf is roughly the sum of (face area × thickness) over all the slices. Thinner slices give a better total.

Calculus makes this exact. Suppose a solid lies between the planes x = a and x = b, and the slice through the solid at position x has area A(x). Cut [a, b] into thin pieces of width Δx and pick a point xᵢ in each piece. The slab there is almost a prism, so

volume of one slab ≈ A(xᵢ) Δx

Adding the slabs gives a Riemann sum, Σ A(xᵢ) Δx. As Δx → 0, the sum becomes a definite integral:

Volume by cross sections. If the cross section at x, perpendicular to the x-axis, has area A(x) for a ≤ x ≤ b, and A is continuous, then the volume of the solid is ∫ (a to b) A(x) dx.

A quick check. If every slice has the same area, the solid is a prism and the formula should give “area × length”. For square slices of side 3 along a length of 5, the integral is ∫ (0 to 5) 9 dx = 45, which is exactly 3 × 3 × 5. The integral simply handles slices whose area changes.

The base and the side length

In this topic, the solid sits on a flat base: a region R in the xy-plane, like the regions you found areas of in Topics 8.4 to 8.6. Each cross section stands up out of the base, at right angles to it, and rests on a segment that runs across R.

• Slices perpendicular to the x-axis. The segment at x runs vertically across R, from the bottom curve to the top curve. Its length is s(x) = top − bottom. This is the same height you used for an area strip in Topic 8.4.
• Slices perpendicular to the y-axis. The segment at y runs horizontally across R, from the left curve to the right curve. Its length is s(y) = right − left, written in terms of y, as in Topic 8.5.

Then the shape of the cross section turns the length into an area:

Cross section	Its base edge	Area of the slice
Square	the segment, length s	A = s²
Rectangle with height h	the segment, length s	A = s × h
Rectangle whose height is k times its base	the segment, length s	A = s × ks = ks²
Rectangle with a fixed height h (a constant)	the segment, length s	A = hs, so V = h × (area of R)

Read the question carefully for the height of a rectangle. “Height equal to half the length of its base” gives A = s × s/2 = s²/2. “Height 4” gives A = 4s. “Height equal to x” gives A = x × s(x).

The method, step by step

1. Sketch the base and shade it. Mark the direction of the slices.
2. Choose the variable. Perpendicular to the x-axis: use x and dx. Perpendicular to the y-axis: use y and dy.
3. Find the limits: the first and last values of x (or y) in the base.
4. Find the side length s: top − bottom in terms of x, or right − left in terms of y. Test a point to get the order right.
5. Write the area of one cross section, A = s² for squares or A = s × h for rectangles.
6. Write the integral in full, V = ∫ A dx (or dy), then evaluate it. Give cubic units.

Worked example 1: square slices across x (no calculator)

Question. The base of a solid is the region enclosed by y = 2x and y = x² (Figure 1). Cross sections perpendicular to the x-axis are squares. Find the volume of the solid.

1. Limits. 2x = x² gives x(x − 2) = 0, so x = 0 or x = 2.
2. Top curve. At x = 1, 2x = 2 and x² = 1. The line is on top for 0 < x < 2.
3. Side length. s(x) = 2x − x².
4. Area of a slice. A(x) = (2x − x²)² = 4x² − 4x³ + x⁴.
5. Integral. V = ∫ (0 to 2) (2x − x²)² dx = ∫ (0 to 2) (4x² − 4x³ + x⁴) dx
6. Evaluate. The antiderivative is 4x³/3 − x⁴ + x⁵/5. At x = 2: 32/3 − 16 + 32/5 = (160 − 240 + 96)/15 = 16/15. At x = 0 it is 0.

Answer. V = 16/15 cubic units (about 1.067).

Check. The base has area ∫ (0 to 2) (2x − x²) dx = 4/3, and the widest segment is only 1 unit long (at x = 1), so every square has area at most 1. The volume must be less than 1 × 2 = 2, and it is. A common wrong answer is ∫ (0 to 2) [(2x)² − (x²)²] dx = 64/15, from squaring each function separately.

Worked example 2: rectangles across y (no calculator)

Question. The base of a solid is the region enclosed by x = y² and x = y + 2. Cross sections perpendicular to the y-axis are rectangles whose height is half the length of the side that lies in the base. Find the volume.

1. Direction. Slices are perpendicular to the y-axis, so work in y with dy. Each segment across the base is horizontal.
2. Limits. y² = y + 2 gives y² − y − 2 = 0, so (y − 2)(y + 1) = 0 and y = −1 or y = 2.
3. Right and left. At y = 0, the line gives x = 2 and the parabola gives x = 0, so x = y + 2 is on the right.
4. Side length. s(y) = (y + 2) − y² = y + 2 − y².
5. Area of a slice. Base s, height s/2, so A(y) = s × s/2 = (y + 2 − y²)²/2.
6. Integral. V = ∫ (−1 to 2) (y + 2 − y²)²/2 dy = (1/2) ∫ (−1 to 2) (y⁴ − 2y³ − 3y² + 4y + 4) dy
7. Evaluate. An antiderivative of the bracket is y⁵/5 − y⁴/2 − y³ + 2y² + 4y. At y = 2 it is 32/5. At y = −1 it is −17/10. The difference is 32/5 + 17/10 = 81/10. Halve it: 81/20.

Answer. V = 81/20 cubic units (4.05).

Why not dx? Slicing across y is what the question describes. A vertical segment in this base would sometimes end on the lower half of the parabola and sometimes on the line, so x-slices would need two integrals and give the volume of a different solid.

Worked example 3: a calculator-active volume

Question. R is the region in the first quadrant bounded by y = 4 − x, y = e^(x/2) and the y-axis. R is the base of a solid whose cross sections perpendicular to the x-axis are squares. Find the volume. (Calculator allowed.)

1. Limits. The left edge is the y-axis, x = 0. The right edge is where 4 − x = e^(x/2). This needs a calculator: x = B ≈ 1.682. Store B unrounded.
2. Top curve. At x = 0, 4 − 0 = 4 and e⁰ = 1, so y = 4 − x is on top.
3. Side and area. s(x) = 4 − x − e^(x/2), so A(x) = (4 − x − e^(x/2))².
4. Integral. Write it before you press keys: V = ∫ (0 to B) (4 − x − e^(x/2))² dx
5. Evaluate with the stored B: V ≈ 5.510 cubic units.

On paper, show the setup with the rounded limit, “∫ (0 to 1.682) (4 − x − e^(x/2))² dx = 5.510”, but use the stored value in the calculation.

Sense check. The largest slice is at x = 0, where s = 3 and A = 9, and the slices shrink to 0 at x = B. The volume must be less than 9 × 1.682 ≈ 15.1, and 5.510 is well below that.

Common misconceptions

• Squaring each function separately. The side is f − g; the area is (f − g)². Writing f² − g² is the washer pattern from Topic 8.10, and it gives the wrong number here (64/15 instead of 16/15 in Worked example 1).
• Forgetting to square. ∫ s(x) dx is the area of the base, not the volume.
• Adding π. Square and rectangular slices have no π in their areas.
• Mixing variables. Slices perpendicular to the y-axis need s in terms of y, y-limits and dy. Do not leave an x inside a dy integral.
• Treating the side as a y-value. The side is a length across the base (top − bottom), not just the top curve, unless the bottom is the x-axis.
• Misreading the rectangle’s height. “Height half the base” means A = s²/2, not s/2. Write A = (base)(height) with both parts in terms of s.
• Rounding the limit early on a calculator question, or leaving out the integral expression and writing only the decimal.
• Thinking the solid is the region turned around an axis. Here the solid stands on the base. Revolving a region is the disc method in Topic 8.9.

Where this leads

Topic 8.8 keeps the same integral, V = ∫ A dx, and changes only the shape of the slice: triangles and semicircles. Topic 8.9 then makes the slices circles by revolving a region around an axis. Continue with Volumes with Cross Sections: Triangles and Semicircles, or look back at Topic 8.6 for the base regions. Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.