AP Calculus AB/BC · Unit 8 · Topic 8.7
Volumes with Cross Sections: Squares and Rectangles
Build a cross-sectional area from a base distance and integrate it.
1. Topic Focus
Apply definite integrals to average value, motion, net change, area, volume, and BC arc length or distance problems.
This topic: Build a cross-sectional area from a base distance and integrate it.
2. Key Relationship
Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.
3. Visual Connection
4. Worked Example
For square slices with side f−g, use A(x)=[f(x)−g(x)]².
Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.
5. Concept Development
The slicing principle
Imagine cutting a solid into thin parallel slabs. If a slice perpendicular to the \(x\)-axis has cross-sectional area \(A(x)\) and thickness \(\Delta x\), then its volume is approximately \(A(x_i^*)\Delta x\). Adding slices and taking a limit gives
For slices perpendicular to the \(y\)-axis, use
This is not necessarily a solid of revolution. The problem directly specifies the shape of each cross section.
Connect the base region to a cross section
The given two-dimensional base determines one dimension of every cross section. For vertical slices, the base segment is
For horizontal slices, it is
This segment is a length inside the base plane. Use the shape description to convert it into the full cross-sectional area before integrating.
Square and rectangle area models
| Cross section | Dimensions | Area |
|---|---|---|
| Square | side \(s\) | \(A=s^2\) |
| Rectangle, height \(k\) times base | base \(s\), height \(ks\) | \(A=ks^2\) |
| Rectangle, fixed height \(h\) | base \(s\), height \(h\) | \(A=hs\) |
| Rectangle, variable height \(H(x)\) | base \(s(x)\), height \(H(x)\) | \(A=s(x)H(x)\) |
The phrase “height is twice the base” produces \(A=s(2s)=2s^2\). It does not produce \(2s\), which is still only a length.
A reliable workflow
- Sketch the planar base and mark the interval over which the solid extends.
- Use “perpendicular to the \(x\)-axis” or “perpendicular to the \(y\)-axis” to choose the slicing variable.
- Find the segment length in the base: top minus bottom or right minus left.
- Translate that length into the stated square or rectangle area formula.
- Integrate the area over the complete interval.
- Check that the result is positive and has cubic units.
Changing base boundaries
If the top, bottom, right, or left boundary changes, the segment function and cross-sectional area may also change. Split the volume integral at those transition points:
Squaring a signed difference can hide an incorrect boundary choice, so still identify the geometric length carefully.
Numerical and geometric checks
If only a table of segment lengths is given, calculate the cross-sectional areas first and then approximate the integral of those areas. Do not apply the trapezoidal rule to side lengths and square the final result. If every cross section has the same area \(A_0\), the formula becomes \(V=A_0(b-a)\), agreeing with the ordinary prism formula.
Units
If \(s(x)\) and all cross-sectional dimensions are measured in centimeters, then \(A(x)\) has units \(\mathrm{cm^2}\). Multiplication by \(dx\), measured in centimeters, gives volume in \(\mathrm{cm^3}\).
6. Detailed Worked Example and Error Check
Example 1: Square cross sections
The base lies between \(y=\sqrt{x}\) and \(y=x^2\) on \([0,1]\), and cross sections perpendicular to the \(x\)-axis are squares. The side length is
so
Example 2: Rectangle height proportional to its base
The base lies between \(y=1\) and \(y=x^2\) for \(0\le x\le1\). If each rectangle's height is twice its base, then
Therefore
Example 3: Rectangle with fixed height
The base is bounded by \(y=4-x^2\) and \(y=0\), and every perpendicular rectangle has fixed height \(3\). The bounds are \(-2\le x\le2\), and
Thus
A fixed height leads to a first power of the base segment, not its square.
Example 4: Variable rectangular height
The base lies between \(y=x\) and \(y=x^2\) on \([0,1]\), and each rectangle has height \(H(x)=x\). Since \(x\ge x^2\),
Hence
Example 5: Slices perpendicular to the y-axis
The base lies between \(x=y^2\) and \(x=4\), and horizontal cross sections are squares. Their side is \(4-y^2\), with \(-2\le y\le2\). Therefore
Because the slices are perpendicular to the \(y\)-axis, both the area function and bounds use \(y\).
Example 6: A circular base with square sections
The base is the disk \(x^2+y^2\le4\). A vertical segment across the disk has length
Square cross sections give
Example 7: A parameterized family
The base is between \(y=k\) and \(y=x^2\), where \(k>0\), and perpendicular sections are squares. Intersections occur at \(x=\pm\sqrt{k}\), so
The exponent \(5/2\) reflects two length dimensions from the square area and one from the integration interval.
Example 8: Approximate from tabular side lengths
Square sections have side lengths \(s(0)=2\), \(s(1)=4\), and \(s(3)=1\). Their areas are \(4,16,1\). A trapezoidal estimate is
Applying the trapezoidal rule to \(2,4,1\) and then squaring would not approximate \(\int s(x)^2dx\).
Common errors
- Integrating a side length instead of a cross-sectional area.
- Using \(A=s\) rather than \(A=s^2\) for squares.
- Interpreting “height twice the base” as area \(2s\) instead of \(2s^2\).
- Using top minus bottom when slices are perpendicular to the \(y\)-axis.
- Choosing bounds from the wrong axis.
- Squaring an entire numerical approximation instead of approximating the area function.
- Reporting square rather than cubic units.
7. AP Reasoning Routine
Sketch and label the region, decide whether slices are vertical or horizontal, write a nonnegative geometric quantity, and split bounds when the geometry changes.
- Identify the representation and requested quantity.
- State the rule or theorem and verify its conditions.
- Keep exact values until the final requested approximation.
- Interpret sign, units, interval, and context.
Find each volume using the slicing method. Define the cross-sectional area before integrating.
(a) The base is between \(y=2x\) and \(y=x^2\) for \(0\le x\le2\). Cross sections perpendicular to the \(x\)-axis are squares.
(b) The base is between \(y=1\) and \(y=x^2\) for \(0\le x\le1\). Cross sections are squares.
(c) Use the base from part (b), but let each rectangular height be three times its base.
(d) Use the base from part (b), but let every rectangular cross section have fixed height \(2\).
(e) The base is between \(y=\sqrt x\) and \(y=x^2\) on \([0,1]\). Rectangular cross sections have height \(x\).
(f) The base is between \(x=y^2\) and \(x=4\). Cross sections perpendicular to the \(y\)-axis are squares.
(g) The base is the disk \(x^2+y^2\le1\). Cross sections perpendicular to the \(x\)-axis are squares.
(h) Square cross sections have side lengths \(s(0)=1\), \(s(2)=3\), and \(s(5)=2\). Use the trapezoidal rule to estimate the volume.
(i) If all lengths are measured in meters, explain why \(\int_a^bA(x)dx\) has cubic-meter units.
(j) The bottom boundary is \(g(x)\), while the top boundary is \(f(x)\) on \([a,c]\) and \(h(x)\) on \([c,b]\). Write a volume integral for square cross sections.
Check the solution
(a) \(s(x)=2x-x^2\), so \(V=\int_0^2(2x-x^2)^2dx=16/15\).
(b) \(s(x)=1-x^2\), so \(V=\int_0^1(1-x^2)^2dx=8/15\).
(c) \(A(x)=3(1-x^2)^2\), so \(V=3(8/15)=8/5\).
(d) \(A(x)=2(1-x^2)\), so \(V=2\int_0^1(1-x^2)dx=4/3\).
(e) \(A(x)=x(\sqrt x-x^2)=x^{3/2}-x^3\), so \(V=\int_0^1(x^{3/2}-x^3)dx=3/20\).
(f) \(s(y)=4-y^2\) for \(-2\le y\le2\), so \(V=\int_{-2}^{2}(4-y^2)^2dy=512/15\).
(g) \(s(x)=2\sqrt{1-x^2}\), so \(V=\int_{-1}^{1}4(1-x^2)dx=16/3\).
(h) The areas are \(1,9,4\). Thus \(V\approx2(1+9)/2+3(9+4)/2=59/2\).
(i) \(A(x)\) has units \(\mathrm{m^2}\), and \(dx\) contributes meters, so the integral has units \(\mathrm{m^3}\).
(j) \(V=\int_a^c[f(x)-g(x)]^2dx+\int_c^b[h(x)-g(x)]^2dx\).