8.8 Volumes with Cross Sections: Triangles and Semicircles

8.8 Volumes with Cross Sections: Triangles and Semicircles

Topic 8.7 built solids from square and rectangular slices and found their volumes with $V=\int_a^b A(x),dx$, where $A(x)$ is the area of the cross section at $x$. The same accumulation works for any slice shape whose area you can write down. This page uses triangles and semicircles, where the area formula brings in a factor of $\frac12$ or $\frac{\pi}{8}$ that is easy to lose.

A problem begins with a base region, usually bounded by curves. At each $x$-value, a slice through that region has a measurable length. The problem then specifies what two-dimensional shape is erected on that slice: a square, rectangle, triangle, semicircle, or another geometric figure.

The essential workflow is:

1. Identify the interval $[a,b]$.
2. Find the slice dimensions from the boundary curves.
3. Use the specified cross-sectional shape to write $A(x)$.
4. Accumulate those areas with $\int_a^b A(x),dx$.

This is different from a problem that merely asks for the area of a planar region. The base region supplies the dimensions of the slices; the specified cross sections determine the three-dimensional volume. Do not replace the stated shape with a rotation formula: use the cross-sectional geometry the problem gives.

Triangular cross sections from intersecting curves

For a triangle,

$$ A=\frac12 bh, $$

where $b$ is the base and $h$ is the perpendicular height. The base region determines one or both of these dimensions; the wording must tell you how the triangle is constructed.

Worked example: triangular cross sections

Consider the region bounded by $y=x$ and $y=x^2$ for $0\le x\le1$. Since $x\ge x^2$ on this interval, the vertical slice length is

$$ d(x)=x-x^2. $$

Suppose each cross section perpendicular to the $x$-axis is a right triangle whose base and height both equal this slice length. Thus,

$$ b(x)=d(x)=x-x^2, \qquad h(x)=d(x)=x-x^2. $$

The cross-sectional area becomes

$$ A(x)=\frac12(x-x^2)(x-x^2) =\frac12(x-x^2)^2. $$

The volume is therefore

$$ V=\int_0^1 \frac12(x-x^2)^2,dx =\frac1{60}. $$

Notice how the bounds $0$ and $1$ come from the intersections of the curves: solving $x=x^2$ gives $x=0$ and $x=1$. The graph supplies both the limits and the slice length.

Two other triangles are common on the exam. An equilateral triangle with side $d(x)$ has area $\frac{\sqrt3}{4}[d(x)]^2$, and an isosceles right triangle whose hypotenuse is $d(x)$ has area $\frac14[d(x)]^2$.

Semicircular cross sections

If the slice length is the diameter $d(x)$ of a semicircle, then its radius is

$$ r(x)=\frac{d(x)}{2}. $$

Using the semicircle area formula,

$$ A(x)=\frac12\pi r(x)^2 =\frac12\pi\left(\frac{d(x)}2\right)^2 =\frac{\pi}{8}[d(x)]^2. $$

For the same base region between $y=x$ and $y=x^2$ on $[0,1]$, if each cross section is a semicircle with diameter equal to the vertical slice length, then $d(x)=x-x^2$ and

$$ V=\int_0^1 \frac{\pi}{8}(x-x^2)^2,dx =\frac{\pi}{240}. $$

The factor $\frac{\pi}{8}$ matters: using $\pi[d(x)]^2$ would treat the diameter as the radius and produce a volume four times too large.

The suggested AP skill is 3.D — Apply an appropriate mathematical definition, theorem, or test. In these problems, justification means making the geometry visible in the equation: state the slice dimension, identify whether it is a side, length, base, height, or diameter, and then apply the correct area formula.

Misconception check — “The base region is already the volume.” A planar region has area, not volume. The region only provides the dimensions of infinitely many cross sections. Volume appears after those cross-sectional areas are accumulated with a definite integral.

Retrieval check: A vertical slice has length $d(x)$, and each cross section is a semicircle with diameter $d(x)$. What is the integrand? The answer is

$$ A(x)=\frac{\pi}{8}[d(x)]^2, $$

because the radius is half the diameter.