8.8 Volumes with Cross Sections: Triangles and Semicircles

To find the volume of a solid whose base is a region in the plane and whose cross sections are triangles or semicircles, integrate the cross-sectional area function: $V = \int_a^b A(x)\, dx$. The trick is writing the side length or radius in terms of the curves that bound the base, usually as the top function minus the bottom function. For AP Calculus, name the cross-section shape before choosing the area formula.

Triangle Cross Section Formulas

For AP Calculus cross-section volume problems, the formula depends on the shape named in the prompt. If each cross section is an equilateral triangle with side length $s$, use $A=\frac{\sqrt{3}}{4}s^2$. If each cross section is an isosceles right triangle with legs $s$, use $A=\frac{1}{2}s^2$.

The side length usually comes from the distance across the base region. For slices perpendicular to the $x$-axis, that distance is often $\text{top curve}-\text{bottom curve}$. Once you have the area formula, volume is the definite integral of area across the interval.

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Why This Matters for the AP Calculus Exam

Volumes with known cross sections show up in both AB and BC, where Unit 8 carries a solid chunk of exam weight. You set up a definite integral from a geometric description, which tests whether you can connect a picture to a correct area function and correct bounds. On free-response questions, you often only need to set up the integral correctly with clear notation, so being precise with your area formula and limits is what earns you the work. The same skill carries over to disc and washer volumes later in the unit.

Key Takeaways

• Volume of a solid with known cross sections perpendicular to the x-axis is $V = \int_a^b A(x)\, dx$, where $A(x)$ is the cross-section area and $dx$ is the slice thickness.
• For cross sections perpendicular to the y-axis, integrate $A(y)$ with respect to $y$ between y-bounds.
• The side length or diameter of each cross section usually equals the distance between the bounding curves: $\text{top} - \text{bottom}$.
• Equilateral triangle cross section: $A = \frac{\sqrt{3}}{4}s^2$. Right isosceles triangle (legs $s$): $A = \frac{1}{2}s^2$. Semicircle (radius $r$): $A = \frac{1}{2}\pi r^2$.
• For a semicircle, the diameter equals top minus bottom, so the radius is half that distance.
• Find your limits of integration from where the bounding curves intersect.

Solids with Cross Sections: Quick Review

To find the volume of a solid with known cross sections, use

where $A(x)$ is the area of a cross section (a two-dimensional shape) perpendicular to the x-axis on the interval $[a,b]$, and $dx$ is the slice thickness. This is the same idea as squares and rectangles in Topic 8.7, just with new shapes. The plan is always the same: write $A(x)$, find the bounds, then integrate.

Triangular Cross Sections

The area formula depends on the type of triangle, so $A(x)$ changes based on the triangle shape. Two common cases are equilateral and right isosceles triangles.

Equilateral triangles. The area of an equilateral triangle is $\frac{\sqrt{3}}{4}s^2$, where $s$ is a side length. So the volume is

Right isosceles triangles. The area of a right isosceles triangle is $\frac{1}{2}s^2$, where $s$ is the length of the two equal legs. So the volume is

Semicircular Cross Sections

A full circle has area $\pi r^2$, where $r$ is the radius. A semicircle is half of that, so use $\frac{1}{2}\pi r^2$ for $A(x)$. The volume of a solid with semicircular cross sections is

Watch the difference between diameter and radius here. The distance between the bounding curves usually gives the diameter, so the radius is half of that.

Worked Example

Once you have an area formula, you still need $s$ or $r$ in terms of $x$. Here is the full process.

Suppose a region bounded by $y = x^2$ and $y = \sqrt{x}$ forms the base of a solid with cross sections taken perpendicular to the x-axis. What integral gives the volume if the cross sections are equilateral triangles? What about semicircles?

Start by sketching the curves so you can see the region and identify your bounds. Let $h(x) = \sqrt{x}$ be the upper curve and $g(x) = x^2$ be the lower curve. The region between them is the base of the solid, and each cross section stands up off that base.

Identifying Bounds

The two curves intersect at $x = 0$ and $x = 1$, so the bounds are $[0, 1]$. You can confirm this algebraically by setting the functions equal:

Equilateral Triangle

Side length. One side of each triangle equals the distance between the curves, so $s = h(x) - g(x) = \sqrt{x} - x^2$.

Area function. Plug $s$ into the equilateral area formula:

Volume integral. Putting the bounds and $A(x)$ into $V = \int_a^b A(x)\, dx$ gives

Evaluating:

Semicircle

Radius. The diameter of each semicircle equals the distance between the curves, $h(x) - g(x)$. The radius is half the diameter, so

Area function. Plug $r$ into the semicircle area formula:

Volume integral.

Evaluating:

How to Use This on the AP Calculus Exam

Problem Solving

1. Sketch the base region and label the bounding curves. Identify which is on top.

2. Find the bounds by setting the curves equal and solving for the intersection points.

3. Write the side length, base, or diameter as $\text{top} - \text{bottom}$ in terms of $x$ (or $y$ if the slices are perpendicular to the y-axis).

4. Plug that expression into the correct area formula to build $A(x)$.

5. Set up $V = \int_a^b A(x)\, dx$ and evaluate, or leave it set up if the question only asks for the integral.

Free Response

When a question describes a solid and asks for volume, a correctly set up integral with the right area formula and bounds shows your reasoning. Use clear notation and keep the area formula matched to the named cross section. Some free-response questions give you a region built from a graph or table, and the cross-section area might even be defined directly by a function instead of a standard shape, so read the description carefully before choosing a formula.

Common Trap

Mixing up diameter and radius for semicircles is the most frequent error. The curve gap gives the diameter, so always divide by 2 before squaring.

Common Misconceptions

• Using the gap as the radius for semicircles. The distance between curves is the diameter. The radius is half that, so $r = \frac{\text{top} - \text{bottom}}{2}$.
• Forgetting to square the side length. Triangle and circle area formulas square the side or radius. Square the whole expression $(\sqrt{x} - x^2)$, not just one term.
• Mixing up triangle area formulas. $\frac{\sqrt{3}}{4}s^2$ is only for equilateral triangles. A right isosceles triangle with legs $s$ uses $\frac{1}{2}s^2$. Use the formula that matches the shape named in the problem.
• Confusing cross-section volume with disc or washer volume. Cross-section solids are not made by revolving a region. You build the area function directly from the named 2D shape, so there is no automatic $\pi r^2$ unless the cross section is itself a circle or semicircle.
• Integrating with the wrong variable. If the cross sections are perpendicular to the y-axis, write $A(y)$ and integrate with respect to $y$ using y-bounds, not $x$.
• Picking the wrong intersection points as bounds. Always solve for where the bounding curves meet, and check that the curve you call "top" really is above the other across the whole interval.

zontal slices, use $\text{right}-\text{left}$.

What is the semicircle cross section formula?

A semicircle has area $A=\frac{1}{2}\pi r^2$. If the gap between curves gives the diameter, divide by 2 first to get the radius before squaring.

Are cross-section volumes the same as washer or disk volumes?

No. Cross-section volumes use a named shape for each slice, like a triangle or semicircle. Disk and washer problems come from rotating a region around an axis.

Related AP Calculus Guides

• Unit 8 Overview: Applications of Integration
• 8.1 Finding the Average Value of a Function on an Interval
• 8.7 Volumes with Cross Sections: Squares and Rectangles
• 8.2 Connecting Position, Velocity, and Acceleration of Functions Using Integrals
• 8.4 Finding the Area Between Curves Expressed as Functions of x
• 8.3 Using Accumulation Functions and Definite Integrals in Applied Contexts