Difficulty: Intermediate | Prerequisites: Calc 1 (definite integrals, area between curves), basic algebra (solving equations with fractional exponents).
Volumes by cross-sections is one of the central applications of integration in Calc 2. The idea is that if you know the shape of every cross-sectional slice of a solid, you can integrate the area of those slices to find the total volume. This topic extends the "area between curves" work from Calc 1 into three dimensions. The problem set's theater example is typical: you are given a base region and told what shape the cross-sections are, and you must set up and evaluate the integral.
To find the volume of a solid with known cross-sections, express the cross-sectional area A(y) (or A(x)) as a function of the slicing variable, then integrate. The theater problem requires you to find the base width from the bounding curves, define the rectangle dimensions, and integrate. A sub-problem asks you to split a region into equal areas, which means solving an equation involving an integral.
Cross-sectional area function A(y)
The area of the slice of a solid at position y (or x). For the theater problem, each cross-section perpendicular to the y-axis is a rectangle, so A(y) = base × height, where both depend on y. Think of it as: if you could cut the solid with a knife at position y, A(y) is the area of the cut face.
Base region
The 2D region that forms the "floor" of the solid. Here it is bounded by y = x⁴ and y = 25. At a given y-value, the width of the base (in the x-direction) is 2x = 2y^{1/4}, since x⁴ = y gives x = y^{1/4} (and by symmetry the base extends from -y^{1/4} to y^{1/4}).
Equal-area division
Finding a value y = A that splits a region into two parts of equal area. You set the integral from the lower bound to A equal to half the total area and solve for A.
The base is the region bounded by y = x⁴ and y = 25. Since x⁴ = y, we have x = ±y^{1/4}. The width of the base at height y is 2y^{1/4}.
Cross-sections perpendicular to the y-axis are rectangles whose base is 4 units shorter than its height. If the base of the rectangle equals the width of the region, then base = 2y^{1/4} and height = 2y^{1/4} + 4.
The cross-sectional area is:
A(y) = 2y^{1/4} · (2y^{1/4} + 4) = 4y^{1/2} + 8y^{1/4}
The "floor" is the base region. Its total area (integrating with respect to y) is:
Total area = ∫₀²⁵ 2y^{1/4} dy = 2 · [y^{5/4} / (5/4)]₀²⁵ = 2 · (4/5) · 25^{5/4}
Since 25 = 5², we get 25^{5/4} = (5²)^{5/4} = 5^{5/2} = 25√5.
Total area = (8/5) · 25√5 = 40√5.
Half the area = 20√5.
Set the integral from 0 to A equal to 20√5:
∫₀ᴬ 2y^{1/4} dy = 20√5
(8/5) · A^{5/4} = 20√5
A^{5/4} = (100√5) / 8 = (25√5)/2
A = [(25√5)/2]^{4/5}
Simplify: 25√5 = 5^{5/2}, so (25√5)/2 = 5^{5/2}/2.
A = [5^{5/2}/2]^{4/5} = 5² / 2^{4/5} = 25 / 2^{4/5}
This is the exact value of A.
The total volume is:
V = ∫₀²⁵ A(y) dy = ∫₀²⁵ (4y^{1/2} + 8y^{1/4}) dy
= 4 · [2y^{3/2}/3]₀²⁵ + 8 · [4y^{5/4}/5]₀²⁵
= (8/3) · 25^{3/2} + (32/5) · 25^{5/4}
25^{3/2} = 125, and 25^{5/4} = 25√5 ≈ 55.9017.
V = (8/3)(125) + (32/5)(25√5) = 1000/3 + 160√5 ≈ 333.33 + 357.77 ≈ 691.10
The volume for y ≥ A is ∫ₐ²⁵ (4y^{1/2} + 8y^{1/4}) dy, computed similarly with A = 25/2^{4/5} ≈ 14.35.
The percentage is (volume for y ≥ A) / (total volume) × 100.
Volume by slicing:
V = ∫ₐᵇ A(y) dy, where A(y) is the cross-sectional area at position y
For this problem:
Base width at height y: w(y) = 2y^{1/4}
Rectangle base = w(y), height = w(y) + 4
A(y) = w(y) · (w(y) + 4) = 4y^{1/2} + 8y^{1/4}
Power rule for integration:
∫ yⁿ dy = y^{n+1}/(n + 1) + C (for n ≠ -1)
Volumes by cross-sections is exactly how architects and engineers compute the volume of irregularly shaped structures, from theaters to dams to aircraft fuselages. Any time you know the shape of each slice, this technique gives you the total volume.
Students often integrate with respect to the wrong variable. If the cross-sections are perpendicular to the y-axis, you integrate with respect to y, and the limits are y-values.
When the problem says "base is 4 units shorter than its height," students sometimes set height = base - 4 instead of height = base + 4. Read carefully: shorter than height means base = height - 4, or equivalently height = base + 4.
For the equal-area part, students sometimes try to divide the x-interval in half instead of the area. Dividing the area means solving an integral equation, not just finding the midpoint of the x or y range.
Fractional exponents trip students up. Remember that y^{1/4} is the fourth root of y, and when you integrate y^{1/4} you get y^{5/4}/(5/4) = (4/5)y^{5/4}.
⚠️ Cross-section volume problems are a staple of Calc 2 exams. Expect to be given a base region and a cross-section shape (squares, rectangles, semicircles, equilateral triangles) and asked to set up and evaluate the integral.
⚠️ The "equal area" sub-problem tests whether you can set up an equation involving an integral and solve it. This combines integration skill with algebraic manipulation of fractional exponents.
⚠️ Show all steps when simplifying expressions like 25^{5/4}. Examiners want to see that you know how to handle non-integer exponents.
True or false: If cross-sections are perpendicular to the x-axis, you integrate with respect to y. (Answer: False. Perpendicular to the x-axis means you integrate with respect to x.)
Fill in the blank: If y = x⁴, then x = y^{_____}. (Answer: 1/4)
True or false: To divide a region into two equal areas, you find the midpoint of the integration interval. (Answer: False. You solve for the point where the accumulated area equals half the total.)
Fill in the blank: ∫ y^{1/4} dy = y^{_____} / (5/4) + C. (Answer: 5/4)
Q: A solid has a base bounded by y = x² and y = 4. Cross-sections perpendicular to the y-axis are squares. Set up the volume integral.
A: At height y, the base width is 2√y (since x = ±√y). The cross-section is a square with side 2√y. A(y) = (2√y)² = 4y. Volume = ∫₀⁴ 4y dy = [2y²]₀⁴ = 32.
Q: For the theater problem (base bounded by y = x⁴ and y = 25, rectangular cross-sections with base 4 shorter than height), write out A(y).
A: Width = 2y^{1/4}. Base of rectangle = 2y^{1/4}. Height = 2y^{1/4} + 4. So A(y) = 2y^{1/4}(2y^{1/4} + 4) = 4y^{1/2} + 8y^{1/4}.
Q: Find the exact value of A such that y = A splits the base region (y = x⁴, y = 25) into two equal areas.
A: Total area = (8/5) · 25^{5/4} = 40√5. Half = 20√5. Solve (8/5)A^{5/4} = 20√5 to get A = 25/2^{4/5}.
This topic builds directly on area between curves (Calc 1) and connects to volumes of revolution (disk/washer/shell methods), which are a special case where the cross-sections are circles or washers. The equal-area sub-problem connects to the idea of the mean value theorem for integrals and to probability distributions, where you find the median of a distribution by solving a similar equation.
Volumes by cross-sections, slicing method, cross-sectional area, volume integral, known cross-sections, rectangular cross-sections, equal area division, base region bounded by curves, y = x⁴ theater problem, Calc 2 volumes, NYU Calculus II PS2, fractional exponents integration