Volumes with Cross Sections: Triangles and Semicircles
Same idea, different slice shapes
- The volume recipe $V=\int_a^b A(x)\,dx$ works for any cross-section shape — you just need $A(x)$.
- Beyond squares, the AP loves triangular and semicircular cross sections.
- The only new step is the right area formula for that shape, written in terms of the base.
- Get $A(x)$ from geometry, then integrate as before.
Triangular cross sections
- For an equilateral triangle of side $s$: area $=\dfrac{\sqrt3}{4}s^2$.
- For a right isosceles triangle with legs $s$: area $=\dfrac12 s^2$; a general triangle: $\tfrac12\,\text{base}\times\text{height}$.
- Use the base length $s(x)$ (the gap the slice spans) in the correct triangle-area formula.
- $A(x)=(\text{shape constant})\cdot s(x)^2$, then integrate.
The base whose slices vary
y = a·√x
The same base region gives different volumes depending on slice shape — only the area constant times $s(x)^2$ changes.
The area of an equilateral triangle with side $s$ is...
Equilateral triangle: $\frac{\sqrt3}{4}s^2$.
Match each cross-section shape to its area constant (times $s^2$).
Each area is its constant times the base squared.
Semicircular cross sections
- A semicircle with diameter $s$ has radius $\dfrac{s}{2}$, so its area is $\dfrac12\pi\Big(\dfrac{s}{2}\Big)^2=\dfrac{\pi}{8}s^2$.
- The base of the slice is the diameter, so halve it to get the radius before squaring.
- $A(x)=\dfrac{\pi}{8}\big(s(x)\big)^2$.
- Watch that factor: diameter $\to$ radius is a divide-by-two that's easy to skip.
A semicircular cross section has its diameter on the base $s(x)$. Its radius is...
Base is the diameter, so radius is half.
For a semicircle whose diameter is the base $s$, you must halve $s$ to get the radius before squaring.
Radius $=s/2$; forgetting this is a common error.
It's always "area in terms of the base"
- Every one of these is $A(x)=(\text{constant depending on shape})\times s(x)^2$.
- Square $\to 1$; equilateral triangle $\to \tfrac{\sqrt3}{4}$; semicircle $\to \tfrac{\pi}{8}$; right isosceles $\to \tfrac12$.
- Find the base $s(x)$ from the region, plug into the shape's area, and integrate.
- The calculus is identical; only the leading constant changes.
Base under $y=\sqrt x$ on $[0,4]$, semicircular slices: $A(x)=\tfrac{\pi}{8}x$. Then $V=$
$\tfrac{\pi}{8}\int_0^4 x\,dx=\tfrac{\pi}{8}\cdot8=\pi$.
Every cross-section volume is $\int_a^b A(x)\,dx$, where $A(x)$ = (shape constant) $\times s(x)$ ____.
Areas scale with the base squared.
For a semicircle, the slice's base is the diameter, so the radius is $\frac{s}{2}$ — don't square the whole base as if it were the radius. And match the triangle formula to the type stated (equilateral vs. right isosceles); they have different constants. The base $s(x)$ is still the gap the slice spans.
Base is the region under $y=\sqrt x$ on $[0,4]$; cross sections are semicircles with diameter on the base.
- Diameter $s(x)=\sqrt x$, so radius $=\dfrac{\sqrt x}{2}$.
- $A(x)=\dfrac{\pi}{8}(\sqrt x)^2=\dfrac{\pi}{8}x$.
- $V=\displaystyle\int_0^4 \dfrac{\pi}{8}x\,dx=\dfrac{\pi}{8}\cdot 8 = \pi$.
For triangular or semicircular cross sections, use $V=\int_a^b A(x)\,dx$ with the shape's area: equilateral triangle $\frac{\sqrt3}{4}s^2$, right isosceles $\frac12 s^2$, semicircle $\frac{\pi}{8}s^2$ (base $=$ diameter, so radius $=\frac s2$). Find the base $s(x)$ and integrate.