Topic 8.8 Volumes with Cross Sections: Triangles and Semicircles: integrate the cross-sectional area to find volume when cross sections are triangles or semicircles.

What this topic is asking

The College Board (Topic 8.8) extends volume-by-cross-section to triangular and semicircular cross sections. The method is identical, integrate $A(x)$, but you must use the correct area formula for the shape, with its side or diameter read from the base region.

The cross-section area formulas

A worked triangular-cross-section volume

Diameter versus radius for semicircles

The semicircle case is where errors cluster. The base region's height usually gives the diameter, not the radius, so you must halve it before squaring: $r = \frac{d}{2}$ and $A = \frac{1}{2}\pi r^2 = \frac{\pi}{8}d^2$. Forgetting to halve treats the height as the radius and overstates the area fourfold. Read the problem carefully: it almost always says "diameter in the base", meaning the region's height is the diameter. Writing $r = \frac{\text{height}}{2}$ explicitly before substituting into $\frac{1}{2}\pi r^2$ prevents this slip.

The method is one template with different area formulas

Every volume-by-cross-section problem, square, rectangle, triangle, or semicircle, follows the same template: read the relevant length (side or diameter) from the base region as a function of the integration variable, plug it into the correct area formula for the shape, and integrate that area between the region's limits. Only the area formula changes from shape to shape. So the skill is twofold: measure the base region's extent correctly (height for $x$-perpendicular sections, width for $y$-perpendicular), and apply the matching area formula with its constant ($\frac{\sqrt{3}}{4}$, $\frac{1}{2}$, or $\frac{\pi}{8}$). With both right, the integral gives the volume in cubic units.

Deriving the equilateral-triangle constant

It is worth knowing where the $\frac{\sqrt{3}}{4}$ comes from, so you can reconstruct it if you forget. An equilateral triangle of side $s$ has height $\frac{\sqrt{3}}{2}s$, found by dropping a perpendicular and using the Pythagorean theorem on the half-triangle with hypotenuse $s$ and base $\frac{s}{2}$: the height is $\sqrt{s^2 - \frac{s^2}{4}} = \frac{\sqrt{3}}{2}s$. The area is then $\frac{1}{2}\cdot\text{base}\cdot\text{height} = \frac{1}{2}\cdot s\cdot\frac{\sqrt{3}}{2}s = \frac{\sqrt{3}}{4}s^2$. Reconstructing the constant this way guards against misremembering it, which matters because the area appears squared in $s$ and any constant error scales the whole volume. For an isosceles right triangle with legs $s$, the area $\frac{1}{2}s^2$ is just half of a square, which is easier to recall.