From a Base Region to a Solid
A solid can be described by its two-dimensional base and by the shape of each cross section cut perpendicular to a chosen axis. The base tells us where the slices occur and how long each slice is. The geometric description tells us the area of the cross section built on that slice.
Once the cross-sectional area is written as a function of position, the volume is found by integrating that area across the base. The important step is to use the correct length from the base as the dimension of the specified cross section.
The units are consistent: cross-sectional area has units of square units, and slice thickness has units of units, so the integral has cubic units. A volume integral requires both a correct area formula and bounds covering the entire solid.
Turning a Slice Length into an Area
For a vertical slice at \(x\), the length of the segment in the base is generally the upper boundary minus the lower boundary. For a horizontal slice at \(y\), it is generally the right boundary minus the left boundary. These lengths then become dimensions of the stated cross-sectional shape.
| Cross section | Length from the base | Cross-sectional area |
|---|---|---|
| Square | Side length \(s\) | \(A=s^2\) |
| Semicircle with diameter \(d\) | Diameter \(d\), so radius \(d/2\) | \(A=\frac{\pi}{8}d^2\) |
| Triangle with base \(b\) and perpendicular height \(h\) | The specified base and height | \(A=\frac12 bh\) |
| Equilateral triangle with side \(s\) | Side length \(s\) | \(A=\frac{\sqrt3}{4}s^2\) |
A triangle’s area cannot be determined from one length alone unless the geometric description supplies the missing information. For example, if the cross sections are equilateral, the base-region slice is a side of the triangle and the equilateral-triangle formula applies. If the description gives a base and perpendicular height instead, use both in \(A=\frac12 bh\).
Why the Area Integral Gives Volume
The integral formula follows by approximating the solid with thin slices. Each thin slice is approximately a short prism: its cross-sectional area is about \(A(x_i)\), and its thickness is about \(\Delta x\). Adding the approximate volumes gives a Riemann sum. As the slices become thinner, that sum approaches the integral of the cross-sectional area.
Divide \([a,b]\) into subintervals of widths \(\Delta x_i\), and choose a sample position \(x_i^*\) in each one.
Over a sufficiently thin subinterval, the area is approximately \(A(x_i^*)\), so its volume is approximately \(A(x_i^*)\Delta x_i\).
The sum approaches the exact volume as the largest thickness approaches zero.
This reasoning also explains why the integration variable must match the slice direction. Vertical slices have thickness \(\Delta x\), so their sum leads to \(dx\). Horizontal slices have thickness \(\Delta y\), so their sum leads to \(dy\).
Worked Example: Square Cross Sections
Worked Example: Square Cross Sections
The base is the region between \(y=x\) and \(y=0\) for \(0\le x\le 2\). Cross sections perpendicular to the \(x\)-axis are squares. Find the volume.
Step 1: Find a square’s side. At position \(x\), the vertical slice in the base runs from \(y=0\) to \(y=x\). Its length is \[ s(x)=x-0=x. \]
Step 2: Find the cross-sectional area. The area of a square is its side squared, so \[ A(x)=[s(x)]^2=x^2. \]
Step 3: Integrate over the full interval. \[ V=\int_0^2 x^2\,dx =\left[\frac{x^3}{3}\right]_0^2 =\frac{2^3}{3}-\frac{0^3}{3} =\frac83. \]
Check: The side length ranges from \(0\) to \(2\), so the cross-sectional area ranges from \(0\) to \(4\). A positive volume of \(\frac83\) is consistent with that range. Differentiating the antiderivative gives \(\frac{d}{dx}(x^3/3)=x^2=A(x)\), and the bounds have been applied as upper endpoint minus lower endpoint. The volume is \(\frac83\) cubic units.
Worked Example: Semicircular Cross Sections
Worked Example: Semicircular Cross Sections
The base is bounded by \(y=0\), \(y=2-x\), \(x=0\), and \(x=2\). Cross sections perpendicular to the \(x\)-axis are semicircles whose diameters lie in the base. Find the volume.
Step 1: Find the diameter. A vertical slice extends from \(y=0\) to \(y=2-x\), so its length, and therefore the semicircle’s diameter, is \[ d(x)=(2-x)-0=2-x. \]
Step 2: Express the area using the diameter. The radius is \(r(x)=d(x)/2=(2-x)/2\). Thus \[ A(x)=\frac12\pi[r(x)]^2 =\frac12\pi\left(\frac{2-x}{2}\right)^2 =\frac{\pi}{8}(2-x)^2. \]
Step 3: Integrate. \[ \begin{aligned} V&=\int_0^2 \frac{\pi}{8}(2-x)^2\,dx\\ &=\frac{\pi}{8}\int_0^2(2-x)^2\,dx\\ &=\frac{\pi}{8}\left[-\frac{(2-x)^3}{3}\right]_0^2\\ &=\frac{\pi}{8}\left(0-\left[-\frac{8}{3}\right]\right)\\ &=\frac{\pi}{3}. \end{aligned} \]
Check: The antiderivative is correct because \[ \frac{d}{dx}\left(-\frac{(2-x)^3}{3}\right)=(2-x)^2. \] The diameter decreases from \(2\) to \(0\), so the area decreases from \(\pi/2\) to \(0\). The positive result \(\pi/3\) is consistent with the nonnegative cross-sectional areas. The volume is \(\frac{\pi}{3}\) cubic units.
Worked Example: Equilateral Triangular Cross Sections
Worked Example: Equilateral Triangular Cross Sections
The base is the region between \(y=x\) and \(y=0\) for \(0\le x\le 2\). Cross sections perpendicular to the \(x\)-axis are equilateral triangles, with the base-region slice as one side of each triangle. Find the volume.
Step 1: Determine the side length. At \(x\), the vertical slice has length \[ s(x)=x-0=x. \]
Step 2: Express the area. The area of an equilateral triangle with side \(s\) is \(\frac{\sqrt3}{4}s^2\). Therefore, \[ A(x)=\frac{\sqrt3}{4}x^2. \]
Step 3: Integrate. \[ \begin{aligned} V&=\int_0^2\frac{\sqrt3}{4}x^2\,dx\\ &=\frac{\sqrt3}{4}\left[\frac{x^3}{3}\right]_0^2\\ &=\frac{\sqrt3}{4}\left(\frac83-0\right)\\ &=\frac{2\sqrt3}{3}. \end{aligned} \]
Check: Differentiating \(\frac{\sqrt3}{12}x^3\), the antiderivative used in the integral, gives \(\frac{\sqrt3}{4}x^2=A(x)\). The side length ranges from \(0\) to \(2\), so the area is nonnegative throughout. The volume is \(\frac{2\sqrt3}{3}\) cubic units.
When the Slice Length Changes Formula
Sometimes one pair of expressions does not describe the slice length across the whole interval. For instance, two curves may switch which one is above the other. The length must remain nonnegative, so identify the correct top-minus-bottom or right-minus-left expression on each part of the interval. If the area formula changes, split the integral at the point where it changes.
Worked Example: Square Sides from Crossing Boundaries
The base is between \(y=x\) and \(y=2-x\) for \(0\le x\le 2\). Cross sections perpendicular to the \(x\)-axis are squares. Find the volume.
The curves meet where \(x=2-x\), so \(2x=2\) and \(x=1\). On \(0\le x\le 1\), the upper curve is \(2-x\), so the square’s side is \((2-x)-x=2-2x\). On \(1\le x\le 2\), the upper curve is \(x\), so the side is \(x-(2-x)=2x-2\). The side is squared to obtain each area: \[ A(x)= \begin{cases} (2-2x)^2,&0\le x\le 1,\\ (2x-2)^2,&1\le x\le 2. \end{cases} \]
Therefore, \[ \begin{aligned} V&=\int_0^1(2-2x)^2\,dx+\int_1^2(2x-2)^2\,dx\\ &=4\int_0^1(1-x)^2\,dx+4\int_1^2(x-1)^2\,dx\\ &=4\left[\frac{(1-x)^3}{-3}\right]_0^1 +4\left[\frac{(x-1)^3}{3}\right]_1^2\\ &=4\left(0-\left[-\frac13\right]\right) +4\left(\frac13-0\right)\\ &=\frac43+\frac43=\frac83. \end{aligned} \]
Check: The antiderivatives differentiate to \(4(1-x)^2=(2-2x)^2\) and \(4(x-1)^2=(2x-2)^2\), respectively. Each side length is zero at \(x=1\), where the boundary curves meet, and positive on either side. The calculated volume, \(\frac83\) cubic units, is positive.
Common Pitfall: Using the Wrong Geometric Length
A cross-section’s area depends on the dimensions specified by its shape. For a semicircle, if the base-region slice is the diameter \(d\), its radius is \(d/2\). Treating \(d\) as the radius makes the area four times too large.
For example, if the diameter is \(2-x\), the incorrect calculation \[ A_{\mathrm{wrong}}(x)=\frac12\pi(2-x)^2 \] uses the diameter as the radius. The correct calculation is \[ A(x)=\frac12\pi\left(\frac{2-x}{2}\right)^2 =\frac{\pi}{8}(2-x)^2. \] The incorrect area is four times the correct area, so its volume integral would also be four times too large.
Another common error is using \(A=s^2\) for a triangular cross section, or applying the equilateral-triangle formula when the problem does not state that the triangles are equilateral. Read the geometric description carefully and use exactly the dimensions it supplies.
A Reliable Setup Routine
Before evaluating an integral, make sure each part of the setup answers a specific question:
- Which axis is perpendicular to the cross sections?
- What is the slice’s position variable and interval?
- What is the length of the slice in the base at that position?
- How does that length determine the specified shape’s area?
- Does the boundary order or area formula change, requiring a split?
Writing \(A(x)\) or \(A(y)\) explicitly before integrating is a useful check. It makes the geometry visible and helps prevent missing a square, using a diameter as a radius, or omitting a needed interval split.
Check Your Understanding
For each question, identify the slice length and explain how it determines the area or volume setup.
- A base region has vertical slice length \(3-x\) for \(0\le x\le 3\). If the cross sections perpendicular to the \(x\)-axis are squares, what is \(A(x)\), and what integral gives the volume?
- A semicircle’s diameter is \(d(y)=2y\) for \(0\le y\le 1\). Write its cross-sectional area as a function of \(y\).
- For equilateral triangular cross sections with side length \(s(x)=x+1\), write the area function.
- Why must a volume integral use \(dx\) for slices perpendicular to the \(x\)-axis?
- Two boundary curves cross inside the base interval. What should you check before writing a single expression for the slice length?
