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Calculus AB · Unit 8 · 8.8
AP Calculus · Applications of Integration
Volumes with Cross Sections: Triangles & Semicircles
Same integral every time. Draw the slice, name what $s$ measures, then pick the area formula.
key idea$V=\int_a^b A(x)\,dx\quad\text{— only }A\text{ changes}$
Picture the region
Every cross-section problem is the same integral. What changes from one problem to the next is the area formula of the slice — and the only way to pick the right one is to draw the slice. Do not skip the picture; on the AP exam the set-up earns the points.
One base region, two different slices. At each $x$ the vertical segment of length $s=f(x)$ is the base of a slice standing out of the page. For the triangle $s$ is a side, so $A=\tfrac{\sqrt3}{4}s^{2}$; for the semicircle $s$ is the diameter, so $A=\tfrac{\pi}{8}s^{2}$.[Figure: The same base, a different slice. The segment s(x) is now the diameter, so the radius is s/2 and the area is pi/2(s/2)2 — halving matters.]
The same base, a different slice. The segment $s(x)$ is now the diameter, so the radius is $s/2$ and the area is $\tfrac{\pi}{2}(s/2)^2$ — halving matters.Setting up the integral
Five moves, every time
- Draw the base region and shade it. Find where the boundaries meet — those are your limits $a$ and $b$.
- Draw one representative slice perpendicular to the stated axis. If slices are ⊥ to the $x$-axis, integrate in $x$; if ⊥ to the $y$-axis, integrate in $y$.
- Write $s$, the length of that slice, in terms of the integration variable. Between two curves, $s=\text{top}-\text{bottom}$ (or $\text{right}-\text{left}$).
- Convert $s$ to area using the shape's formula. This is the only step that differs between problems.
- Integrate: $V=\int_a^b A\,dx$. State units as cubic units.
| Cross-section shape | What $s$ is | Area $A$ in terms of $s$ |
|---|
| Square | side | $A=s^{2}$ |
| Rectangle, height $=ks$ | base | $A=ks^{2}$ |
| Equilateral triangle | side | $A=\dfrac{\sqrt3}{4}s^{2}$ |
| Isosceles right triangle, leg $=s$ | leg | $A=\dfrac12 s^{2}$ |
| Isosceles right triangle, hypotenuse $=s$ | hypotenuse | $A=\dfrac14 s^{2}$ |
| Semicircle | diameter | $A=\dfrac{\pi}{8}s^{2}$ |
| Quarter circle, radius $=s$ | radius | $A=\dfrac{\pi}{4}s^{2}$ |
Read the shape's description carefully — whether $s$ is a side, a leg, a hypotenuse or a diameter changes the constant.
Worked examples
EX 1
Equilateral-triangle cross sections
$A=\tfrac{\sqrt3}{4}s^{2}$The base is the region bounded by $y=\sqrt{x}$, the $x$-axis and $x=4$. Cross sections perpendicular to the $x$-axis are equilateral triangles. Find the volume.
1
Limits from the region
The region runs from $x=0$ to $x=4$.
2
Length of the slice
At a given $x$ the slice runs from the $x$-axis up to the curve, so $s=\sqrt{x}-0=\sqrt{x}$.
3
Slice area
For an equilateral triangle of side $s$: $A=\tfrac{\sqrt3}{4}s^{2}=\tfrac{\sqrt3}{4}\left(\sqrt{x}\right)^{2}=\tfrac{\sqrt3}{4}x$.
4
Integrate
\[V=\int_0^4 \tfrac{\sqrt3}{4}x\,dx=\tfrac{\sqrt3}{4}\cdot\tfrac{x^{2}}{2}\Big|_0^4=\tfrac{\sqrt3}{4}\cdot 8=2\sqrt3\]
5
Answer
$V=2\sqrt3\approx 3.46$ cubic units.
EX 2
Semicircular cross sections — watch the radius
$A=\tfrac{\pi}{8}s^{2}$Same base region. Now the cross sections perpendicular to the $x$-axis are semicircles with diameter in the base. Find the volume.
1
Same limits, same $s$
Nothing about the base changed: $x$ from $0$ to $4$, and $s=\sqrt{x}$.
2
Identify what $s$ measures
The problem says the diameter lies in the base. So the radius is $r=\tfrac{s}{2}=\tfrac{\sqrt{x}}{2}$ — not $\sqrt{x}$.
3
Slice area
$A=\tfrac12\pi r^{2}=\tfrac12\pi\left(\tfrac{\sqrt x}{2}\right)^{2}=\tfrac{\pi x}{8}$.
4
Integrate
\[V=\int_0^4 \frac{\pi x}{8}\,dx=\frac{\pi}{8}\cdot\frac{x^{2}}{2}\Big|_0^4=\frac{\pi}{8}\cdot 8=\pi\]
5
Answer
$V=\pi\approx 3.14$ cubic units. Had you used $r=\sqrt{x}$ you would have got $4\pi$ — four times too big.
EX 3
Slices between two curves
$s=\text{top}-\text{bottom}$The base is bounded by $y=x$ and $y=x^{2}$. Cross sections perpendicular to the $x$-axis are squares. Find the volume.
1
Find where the curves meet
$x=x^{2}\Rightarrow x(x-1)=0$, so $x=0$ and $x=1$. These are the limits.
2
Decide which curve is on top
On $(0,1)$, e.g. at $x=\tfrac12$: $y=x$ gives $0.5$, $y=x^{2}$ gives $0.25$. So $y=x$ is the top.
3
Length of the slice
$s=x-x^{2}$.
4
Slice area and integral
Squares: $A=s^{2}=(x-x^{2})^{2}=x^{2}-2x^{3}+x^{4}$. \[V=\int_0^1\!\left(x^{2}-2x^{3}+x^{4}\right)dx=\tfrac13-\tfrac12+\tfrac15\]
5
Answer
$V=\tfrac{10-15+6}{30}=\tfrac{1}{30}$ cubic units.
EX 4
Match the formula to the shape
discriminationThree problems share the base region bounded by $y=4-x^{2}$ and the $x$-axis, with slices perpendicular to the $x$-axis. Write $A$ for each: (a) squares, (b) isosceles right triangles with the hypotenuse in the base, (c) semicircles.
1
Common set-up
The curve meets the axis at $x=\pm 2$, so limits are $-2$ to $2$, and $s=4-x^{2}$ in all three.
2
(a) Squares
$A=s^{2}=(4-x^{2})^{2}$.
3
(b) Hypotenuse in the base
If the hypotenuse is $s$, each leg is $s/\sqrt2$, so $A=\tfrac12\cdot\tfrac{s}{\sqrt2}\cdot\tfrac{s}{\sqrt2}=\tfrac{s^{2}}{4}=\tfrac{(4-x^{2})^{2}}{4}$.
4
(c) Semicircles
$s$ is the diameter, so $A=\tfrac{\pi}{8}s^{2}=\tfrac{\pi}{8}(4-x^{2})^{2}$.
5
The point
One region, one $s$, one set of limits — three different constants in front of the same $s^{2}$. That constant is the whole problem.
Watch out
Diameter versus radius
“Semicircle with diameter in the base” means $r=s/2$. Squaring the wrong one multiplies your answer by $4$. This is the single most common error on this topic.
Leg versus hypotenuse
For an isosceles right triangle, $A=\tfrac12 s^{2}$ if $s$ is a leg, but $A=\tfrac14 s^{2}$ if $s$ is the hypotenuse. Read which one lies in the base.
Perpendicular to which axis?
Slices perpendicular to the $y$-axis force you to rewrite the boundaries as functions of $y$ and integrate $dy$. Changing the variable is not optional.
This is not a rotation problem
There is no revolution here and no $\pi r^{2}$ washer unless the slice genuinely is a disc. Do not reach for the disc/washer formula out of habit.