Volumes with Known Cross Sections

Imagine a loaf of bread cut into thin slices. If you know the area of every slice, you can add them up to get the volume of the loaf. That is exactly how calculus finds volumes: slice the solid, find the area of one slice, and integrate. On this page, the solid stands on a flat region (its base), and every slice is a familiar shape like a square or a semicircle.

Key ideas

Volume = integral of cross-sectional area

If each slice perpendicular to the $x$-axis has area $A(x)$, for $a \le x \le b$, then

$$V = \int_a^b A(x)\, dx$$

A slice of thickness $dx$ has volume about $A(x)\, dx$, and the integral adds up all the slices. For slices perpendicular to the $y$-axis, it’s $V = \displaystyle\int_c^d A(y)\, dy$.

How these problems are set up

1. A base region is drawn in the $xy$-plane (it lies flat, like a floor).
2. Slices perpendicular to one axis stand straight up from the base. Each slice is a given shape whose size depends on where it is.
3. The key length $s$ (side, base, or diameter) is the length of the slice across the base region: top − bottom for slices perpendicular to the $x$-axis, or right − left for slices perpendicular to the $y$-axis. This is the same strip you used in area between curves.

The base lies flat; each cross section stands on a strip of the base, and the strip’s length sets its size.

Area formulas for common cross sections

If $s$ is the length of the slice across the base:

Cross section	Area $A$
Square with side $s$	$s^2$
Rectangle with base $s$ and height $h$	$s h$ (for example, $3s^2$ if $h = 3s$)
Equilateral triangle with side $s$	$\dfrac{\sqrt{3}}{4}s^2$
Isosceles right triangle with a leg $s$ on the base	$\dfrac{1}{2}s^2$
Isosceles right triangle with the hypotenuse $s$ on the base	$\dfrac{1}{4}s^2$
Semicircle with diameter $s$	$\dfrac{1}{2}\pi\left(\dfrac{s}{2}\right)^2 = \dfrac{\pi}{8}s^2$

Watch the semicircle: the strip is the diameter, so the radius is $\dfrac{s}{2}$.

AP free-response questions often ask for a cross-section volume as one part of a region question. Writing the integral with the right $A(x)$ and limits usually earns most of the credit.

Worked examples

Example 1: Squares

The base of a solid is the region under $y = \sqrt{x}$ and above the $x$-axis, for $0 \le x \le 4$. Cross sections perpendicular to the $x$-axis are squares. Find the volume.

Solution. The side of the square at $x$ is the strip length, $s = \sqrt{x} - 0 = \sqrt{x}$. So $A(x) = (\sqrt{x})^2 = x$.

$$V = \int_0^4 x\, dx = \Big[ \tfrac{x^2}{2} \Big]_0^4 = 8$$

Example 2: Semicircles

The base of a solid is the region between $y = 4 - x^2$ and the $x$-axis. Cross sections perpendicular to the $x$-axis are semicircles. Find the volume.

Solution. The parabola meets the $x$-axis at $x = \pm 2$. The diameter at $x$ is $s = 4 - x^2$, so

$$A(x) = \frac{\pi}{8}(4 - x^2)^2$$ $$\begin{aligned} V &= \frac{\pi}{8} \int_{-2}^{2} (16 - 8x^2 + x^4)\, dx = \frac{\pi}{8} \left[ 16x - \frac{8x^3}{3} + \frac{x^5}{5} \right]_{-2}^{2} \\ &= \frac{\pi}{8} \cdot 2\left( 32 - \frac{64}{3} + \frac{32}{5} \right) = \frac{\pi}{8} \cdot \frac{512}{15} = \frac{64\pi}{15} \approx 13.404 \end{aligned}$$

(The integrand is even, so the integral from $-2$ to $2$ is twice the integral from $0$ to $2$.)

Example 3: Equilateral triangles, perpendicular to the y-axis

The base of a solid is the region bounded by $y = x^2$ and $y = 4$. Cross sections perpendicular to the $y$-axis are equilateral triangles. Find the volume.

Solution. Slices perpendicular to the $y$-axis are horizontal strips, so work in $y$. At height $y$, the strip runs from $x = -\sqrt{y}$ to $x = \sqrt{y}$, so $s = 2\sqrt{y}$, for $0 \le y \le 4$.

$$A(y) = \frac{\sqrt{3}}{4}(2\sqrt{y})^2 = \frac{\sqrt{3}}{4} \cdot 4y = \sqrt{3}\, y$$ $$V = \int_0^4 \sqrt{3}\, y\, dy = \sqrt{3} \Big[ \tfrac{y^2}{2} \Big]_0^4 = 8\sqrt{3} \approx 13.856$$

Example 4: Rectangles between two curves

The base of a solid is the region between $y = x$ and $y = x^2$. Cross sections perpendicular to the $x$-axis are rectangles whose height is $3$ times the length of their base. Find the volume.

Solution. The curves meet at $x = 0$ and $x = 1$, with $y = x$ on top. The base of each rectangle is $s = x - x^2$, and the height is $3s$, so $A(x) = 3(x - x^2)^2$.

$$V = 3\int_0^1 (x^2 - 2x^3 + x^4)\, dx = 3\left( \frac{1}{3} - \frac{1}{2} + \frac{1}{5} \right) = 3 \cdot \frac{1}{30} = \frac{1}{10}$$

Common mistakes

Using the diameter as the radius. For semicircles, the strip is the diameter. $A = \dfrac{\pi}{8}s^2$, not $\dfrac{\pi}{2}s^2$.

Forgetting to square. The area of a square is $s^2$. Writing $\int s\, dx$ gives the area of the base, not a volume.

Slicing in the wrong direction. “Perpendicular to the $y$-axis” means horizontal strips, so integrate with $dy$ and use right − left.

Using a curve instead of the strip length. If the base is between two curves, $s$ is top − bottom, not just the top curve.

Mixing up the two triangle formulas. Check whether the leg or the hypotenuse lies in the base: $\dfrac{1}{2}s^2$ versus $\dfrac{1}{4}s^2$.

Squaring a difference incorrectly. $(4 - x^2)^2 = 16 - 8x^2 + x^4$, not $16 - x^4$. Expand carefully.

Practice

1. (Warm-up) Find the area of a semicircle whose diameter is $6$.

Solution

$$A = \frac{\pi}{8}(6)^2 = \frac{36\pi}{8} = \frac{9\pi}{2}$$

(Check: the radius is $3$, and half of $\pi(3)^2$ is $\dfrac{9\pi}{2}$.)

2. (Warm-up) The base of a solid is the region under $y = 2x$ and above the $x$-axis, for $0 \le x \le 3$. Cross sections perpendicular to the $x$-axis are squares. Find the volume.

Solution

$s = 2x$, so $A(x) = 4x^2$.

$$V = \int_0^3 4x^2\, dx = \Big[ \tfrac{4x^3}{3} \Big]_0^3 = 36$$

3. (Warm-up) Find the area of (a) an equilateral triangle with side $4$, (b) an isosceles right triangle with legs of length $4$.

Solution

(a) $\dfrac{\sqrt{3}}{4}(4)^2 = 4\sqrt{3}$.

(b) $\dfrac{1}{2}(4)(4) = 8$.

4. (Core) The base of a solid is the disc $x^2 + y^2 \le 9$. Cross sections perpendicular to the $x$-axis are squares. Find the volume.

Solution

At $x$, the strip runs from $y = -\sqrt{9 - x^2}$ to $y = \sqrt{9 - x^2}$, so $s = 2\sqrt{9 - x^2}$ and $A(x) = 4(9 - x^2)$.

$$V = \int_{-3}^{3} 4(9 - x^2)\, dx = 4\Big[ 9x - \tfrac{x^3}{3} \Big]_{-3}^{3} = 4(18 + 18) = 144$$

5. (Core) The base of a solid is the region under $y = e^{-x}$ and above the $x$-axis, for $0 \le x \le \ln 4$. Cross sections perpendicular to the $x$-axis are squares. Find the volume.

Solution

$A(x) = \left(e^{-x}\right)^2 = e^{-2x}$.

$$V = \int_0^{\ln 4} e^{-2x}\, dx = \left[ -\frac{e^{-2x}}{2} \right]_0^{\ln 4} = -\frac{1}{2} \cdot \frac{1}{16} + \frac{1}{2} = \frac{15}{32}$$

(Here $e^{-2\ln 4} = 4^{-2} = \tfrac{1}{16}$.)

6. (Core) The base of a solid is the region bounded by $x = 2y - y^2$ and the $y$-axis. Cross sections perpendicular to the $y$-axis are semicircles. Find the volume.

Solution

$2y - y^2 = 0$ at $y = 0$ and $y = 2$. The diameter at height $y$ is $s = 2y - y^2$.

$$V = \frac{\pi}{8} \int_0^2 (2y - y^2)^2\, dy = \frac{\pi}{8} \int_0^2 (4y^2 - 4y^3 + y^4)\, dy = \frac{\pi}{8}\left( \frac{32}{3} - 16 + \frac{32}{5} \right) = \frac{\pi}{8} \cdot \frac{16}{15} = \frac{2\pi}{15}$$

7. (Core) The base of a solid is the region between $y = \sqrt{x}$ and $y = \dfrac{x}{2}$. Cross sections perpendicular to the $x$-axis are isosceles right triangles with one leg in the base. Find the volume.

Solution

The curves meet at $x = 0$ and $x = 4$, with $\sqrt{x}$ on top. The leg is $s = \sqrt{x} - \dfrac{x}{2}$, so $A(x) = \dfrac{1}{2}\left( \sqrt{x} - \dfrac{x}{2} \right)^2 = \dfrac{1}{2}\left( x - x^{3/2} + \dfrac{x^2}{4} \right)$.

$$V = \frac{1}{2} \left[ \frac{x^2}{2} - \frac{2}{5}x^{5/2} + \frac{x^3}{12} \right]_0^4 = \frac{1}{2}\left( 8 - \frac{64}{5} + \frac{16}{3} \right) = \frac{1}{2} \cdot \frac{8}{15} = \frac{4}{15}$$

8. (Challenge) (Calculator active.) The base of a solid is the region bounded by $y = 3 - x^2$ and $y = e^x$. Cross sections perpendicular to the $x$-axis are squares. Find the volume.

Solution

Find the intersections with a calculator: $3 - x^2 = e^x$ at $x = a \approx -1.677$ and $x = b \approx 0.834$ (store them). At $x = 0$, $3 \gt 1$, so the parabola is on top.

$$V = \int_a^b \big( 3 - x^2 - e^x \big)^2\, dx \approx 6.358$$

9. (Challenge) A pyramid has a square base with side $L$ and height $H$. Use cross sections to show that its volume is $\dfrac{1}{3}L^2 H$.

Solution

Measure $y$ downward from the apex, so $0 \le y \le H$. Slices perpendicular to this axis are squares. By similar triangles, the side at distance $y$ from the apex is $\dfrac{L}{H}y$ (it grows from $0$ at the apex to $L$ at the base).

$$V = \int_0^H \left( \frac{L}{H}y \right)^2 dy = \frac{L^2}{H^2} \cdot \frac{H^3}{3} = \frac{1}{3}L^2 H$$