AP Calculus AB Unit 8 Notes: Finding Volumes Using Integration

Volumes with Cross Sections: Squares and Rectangles

What this method is

When you find volume by cross sections, you imagine a 3D solid sliced into many thin “slabs” perpendicular to an axis (usually the $x$-axis or $y$-axis). Each slab has a cross-sectional face area and a small thickness. If you add up the volumes of all these thin slabs, you approximate the solid’s volume. Taking the limit as the thickness goes to zero turns this into an integral.

The key idea is simple:

• Volume of one thin slice $\approx$ (area of cross section) $\times$ (thickness)
• Total volume $\approx$ sum of slice volumes
• Exact volume comes from an integral

The fundamental setup is:

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

Here, $A(x)$ is the area of the cross section at position $x$, and $dx$ represents a tiny thickness in the $x$ direction. If your slices are perpendicular to the $y$-axis, you use:

$V=\int_c^d A(y)\,dy$

Why it matters

This method lets you compute volumes of solids that are not formed by simple rotation. Instead, the solid is defined by a 2D region (a “base”) and a rule describing the shape of each cross section (square, rectangle, triangle, semicircle, etc.). On AP Calculus AB, this is a major application of integration because it tests whether you can:

• Translate geometry into an area function $A(x)$ or $A(y)$
• Choose correct bounds and the correct variable
• Build and evaluate a definite integral

How it works (step-by-step)

To set up a cross-sections volume problem, you usually follow this chain:

1. Identify the base region in the plane (bounded by curves/lines).
2. Decide whether cross sections are perpendicular to the $x$-axis (use $dx$) or perpendicular to the $y$-axis (use $dy$).
3. Write the dimension of the cross section in terms of the variable.
   • Often this is a “width” in the base region such as (top minus bottom) or (right minus left).
4. Use geometry to write area $A(x)$ or $A(y)$.
5. Integrate over the correct interval.

A common pattern: if the base is bounded vertically (top curve and bottom curve) and slices are perpendicular to the $x$-axis, then the cross-sectional “side length” is:

$s(x)=y_{\text{top}}(x)-y_{\text{bottom}}(x)$

Squares

If each cross section is a square, and the side length is $s(x)$, then:

$A(x)=\left(s(x)\right)^2$

So the volume is:

$V=\int_a^b \left(s(x)\right)^2\,dx$

Rectangles

If each cross section is a rectangle, you must be told how its dimensions relate to the base. Typically one side is the base width $s(x)$ and the other side is a constant multiple of that width.

If the rectangle’s height is $k\,s(x)$, then:

$A(x)=s(x)\cdot k\,s(x)=k\left(s(x)\right)^2$

So:

$V=\int_a^b k\left(s(x)\right)^2\,dx$

Example 1 (Square cross sections)

Base region: bounded by $y=\sqrt{x}$ and $y=0$ from $x=0$ to $x=4$. Cross sections perpendicular to the $x$-axis are squares.

Reasoning: For a fixed $x$, the vertical segment in the base goes from $0$ up to $\sqrt{x}$, so the square’s side length is:

$s(x)=\sqrt{x}-0=\sqrt{x}$

Area of cross section:

$A(x)=\left(\sqrt{x}\right)^2=x$

Volume:

$V=\int_0^4 x\,dx$

Evaluate:

$V=\left[\frac{x^2}{2}\right]_0^4$

$V=\frac{16}{2}-0=8$

So the volume is $8$ cubic units.

Example 2 (Rectangular cross sections with a ratio)

Base region: bounded by $y=2-x$ and $y=0$ from $x=0$ to $x=2$. Cross sections perpendicular to the $x$-axis are rectangles whose height is twice the base segment in the region.

For a given $x$, the vertical segment length is:

$s(x)=(2-x)-0=2-x$

Rectangle height is $2s(x)$, so area is:

$A(x)=s(x)\cdot 2s(x)=2(2-x)^2$

Volume:

$V=\int_0^2 2(2-x)^2\,dx$

Compute by substitution or expansion. Expand:

$2(2-x)^2=2(4-4x+x^2)=8-8x+2x^2$

So:

$V=\int_0^2 (8-8x+2x^2)\,dx$

$V=\left[8x-4x^2+\frac{2}{3}x^3\right]_0^2$

$V=16-16+\frac{16}{3}=\frac{16}{3}$

Exam Focus

• Typical question patterns
  • “Find the volume of the solid with base bounded by ___ and cross sections perpendicular to the $x$-axis are squares/rectangles.”
  • “Cross sections are rectangles with height (or width) equal to a constant multiple of the base segment.”
  • Given a diagram of a region, you’re asked to build $A(x)$ or $A(y)$.
• Common mistakes
  • Using the wrong segment: mixing up $y_{\text{top}}-y_{\text{bottom}}$ versus $x_{\text{right}}-x_{\text{left}}$.
  • Squaring too early or incorrectly: for squares, area is $s(x)^2$, not $2s(x)$ or $s(x)$.
  • Wrong bounds: integrating over $y$-values when using $dx$ (or vice versa).

Volumes with Cross Sections: Triangles and Semicircles

What changes compared to squares/rectangles

The overall structure stays the same:

$V=\int A(\text{slice variable})\,d(\text{slice variable})$

The main difference is that the cross-sectional area formula now comes from triangle or semicircle geometry. That means you must correctly translate the base segment length into a triangle’s base/height or a semicircle’s radius.

Why it matters

Triangle and semicircle cross sections are common on AP problems because they test whether you can combine:

• A geometric area formula you already know
• With a calculus setup (integral with correct variable and bounds)

These problems often look “hard” at first, but they become routine if you consistently identify the segment length and then plug it into the right area formula.

How it works

Triangular cross sections

For a triangle, the general area formula is:

$A=\frac{1}{2}bh$

In cross-section problems, you are often told something like:

• cross sections are right triangles
• the base is the segment in the base region
• the height equals the base (or equals some multiple of the base)

If the segment length is $s(x)$ and the triangle height is $k\,s(x)$, then:

$A(x)=\frac{1}{2}s(x)\cdot k\,s(x)=\frac{k}{2}\left(s(x)\right)^2$

A particularly common special case: equilateral triangles. If an equilateral triangle has side length $s$, its area is:

$A=\frac{\sqrt{3}}{4}s^2$

So if the cross section is an equilateral triangle with side $s(x)$:

$A(x)=\frac{\sqrt{3}}{4}\left(s(x)\right)^2$

Semicircular cross sections

The area of a full circle is:

$A=\pi r^2$

A semicircle is half of that:

$A=\frac{1}{2}\pi r^2$

In many AP problems, the given segment is the diameter of the semicircle. If the diameter is $s(x)$, then:

$r(x)=\frac{s(x)}{2}$

So the cross-sectional area becomes:

$A(x)=\frac{1}{2}\pi\left(\frac{s(x)}{2}\right)^2$

This simplifies to:

$A(x)=\frac{\pi}{8}\left(s(x)\right)^2$

A big conceptual point: for semicircles, students often forget the radius is half the diameter. Writing $r(x)=s(x)/2$ explicitly prevents that.

Example 1 (Right triangles)

Base region: bounded by $y=x$ and $y=0$ from $x=0$ to $x=3$. Cross sections perpendicular to the $x$-axis are right triangles whose legs are both the vertical segment in the base region.

For fixed $x$, the segment length is:

$s(x)=x-0=x$

Each right triangle has legs $s(x)$ and $s(x)$, so area is:

$A(x)=\frac{1}{2}s(x)s(x)=\frac{1}{2}x^2$

Volume:

$V=\int_0^3 \frac{1}{2}x^2\,dx$

$V=\frac{1}{2}\left[\frac{x^3}{3}\right]_0^3$

$V=\frac{1}{2}\cdot 9=\frac{9}{2}$

Example 2 (Semicircles)

Base region: bounded by $y=4-x^2$ and $y=0$ from $x=-2$ to $x=2$. Cross sections perpendicular to the $x$-axis are semicircles with diameter equal to the vertical segment in the base.

For a given $x$, the diameter is:

$s(x)=(4-x^2)-0=4-x^2$

So the semicircle area is:

$A(x)=\frac{\pi}{8}(4-x^2)^2$

Volume:

$V=\int_{-2}^2 \frac{\pi}{8}(4-x^2)^2\,dx$

You could expand and integrate, but notice the integrand is even, which makes evaluation easier:

$V=2\int_0^2 \frac{\pi}{8}(4-x^2)^2\,dx$

Simplify the constant:

$V=\frac{\pi}{4}\int_0^2 (4-x^2)^2\,dx$

Expand:

$(4-x^2)^2=16-8x^2+x^4$

So:

$V=\frac{\pi}{4}\int_0^2 (16-8x^2+x^4)\,dx$

Integrate:

$V=\frac{\pi}{4}\left[16x-\frac{8}{3}x^3+\frac{1}{5}x^5\right]_0^2$

Evaluate at $2$:

$16(2)-\frac{8}{3}(8)+\frac{1}{5}(32)=32-\frac{64}{3}+\frac{32}{5}$

So:

$V=\frac{\pi}{4}\left(32-\frac{64}{3}+\frac{32}{5}\right)$

If you want a single fraction:

$32=\frac{480}{15}$

$\frac{64}{3}=\frac{320}{15}$

$\frac{32}{5}=\frac{96}{15}$

So the bracket is:

$\frac{480-320+96}{15}=\frac{256}{15}$

Therefore:

$V=\frac{\pi}{4}\cdot\frac{256}{15}=\frac{64\pi}{15}$

Exam Focus

• Typical question patterns
  • “Cross sections perpendicular to the $x$-axis are equilateral triangles/right triangles/semicircles.”
  • The segment is explicitly stated as “the distance between the curves” (you must write it as top minus bottom or right minus left).
  • A semicircle problem where the segment is the diameter (very common).
• Common mistakes
  • Forgetting triangle-specific constants: using $s^2$ instead of $\frac{1}{2}s^2$ for right triangles or $\frac{\sqrt{3}}{4}s^2$ for equilateral triangles.
  • For semicircles, using $s(x)$ as the radius instead of $s(x)/2$.
  • Integrating with respect to the wrong variable when the cross sections are described as “perpendicular to the $y$-axis.”

Volumes with Disc Method

What the disc method is

The disc method is a volume-by-slicing method for solids formed by rotating a region around an axis (like the $x$-axis or $y$-axis). Each thin slice perpendicular to the axis of rotation becomes a solid disc (a cylinder with very small thickness).

If you rotate a curve around an axis, the cross sections are circles. The “disc” is what you get when the region touches the axis of rotation, meaning there is no hole in the middle.

Why it matters

Rotational volume is one of the most classic applications of integration. It connects algebraic representations (functions and regions) to geometric solids. On AP Calculus AB, disc-method problems test whether you can:

• Recognize the axis of rotation and the orientation of slices
• Identify the radius correctly (distance from curve to axis)
• Use cross-sectional area $\pi r^2$ and integrate with correct bounds

How it works

The volume of a thin disc of thickness $dx$ is approximately:

$dV\approx \pi\left(r(x)\right)^2dx$

Adding all discs gives:

$V=\int_a^b \pi\left(r(x)\right)^2\,dx$

You use $dx$ when the slices are perpendicular to the $x$-axis (which is typical when rotating around a horizontal line like $y=0$). If rotating around a vertical line like $x=0$, you often use $dy$:

$V=\int_c^d \pi\left(r(y)\right)^2\,dy$

Finding the radius (most important skill)

The radius is a distance from the axis of rotation to the curve (or boundary of the region).

• If rotating around the $x$-axis (which is $y=0$), and the region goes from $y=0$ to $y=f(x)$, then $r(x)=f(x)$.
• If rotating around the line $y=k$, and the curve is $y=f(x)$, then the radius is:

$r(x)=\lvert f(x)-k\rvert$

In many AB problems, the geometry makes the radius nonnegative on the interval, so you can often drop the absolute value after checking.

Example 1 (Rotate around the $x$-axis)

Find the volume when the region under $y=\sqrt{x}$ from $x=0$ to $x=4$ is rotated about the $x$-axis.

Here, the radius is the distance from $y=0$ to $y=\sqrt{x}$:

$r(x)=\sqrt{x}$

Disc volume integral:

$V=\int_0^4 \pi(\sqrt{x})^2\,dx$

Simplify:

$V=\int_0^4 \pi x\,dx$

Evaluate:

$V=\pi\left[\frac{x^2}{2}\right]_0^4$

$V=\pi\cdot 8=8\pi$

Example 2 (Rotate around a horizontal line $y=1$)

Find the volume when the region between $y=x$ and $y=1$ from $x=0$ to $x=1$ is rotated about $y=1$.

Because the region touches the axis of rotation at $y=1$, the cross section is a disc (no hole). The radius is the distance from the curve $y=x$ up to $y=1$:

$r(x)=1-x$

So:

$V=\int_0^1 \pi(1-x)^2\,dx$

Integrate:

$V=\pi\left[\frac{(1-x)^3}{-3}\right]_0^1$

Evaluate endpoints carefully:

At $x=1$, $(1-x)^3=0$. At $x=0$, $(1-x)^3=1$. So:

$V=\pi\left(0-\left(-\frac{1}{3}\right)\right)=\frac{\pi}{3}$

A common check: volume should be positive; your bounds and signs should reflect that.

Exam Focus

• Typical question patterns
  • “Find the volume when the region under $y=f(x)$ on $[a,b]$ is rotated about the $x$-axis.”
  • Rotation about a shifted horizontal/vertical line, like $y=k$ or $x=h$.
  • Sometimes the problem is given as “bounded by” multiple curves; you must pick which curve determines the radius.
• Common mistakes
  • Using diameter instead of radius (disc method always uses radius in $\pi r^2$).
  • Forgetting to subtract the axis shift: radius should be distance to the axis, like $f(x)-k$.
  • Choosing $dx$ when rotating around a vertical line without rewriting in terms of $y$ (disc/washer typically uses slices perpendicular to the axis of rotation).

Volumes with Washer Method

What the washer method is

The washer method is the rotational-volume version of cross sections when the rotated region does not touch the axis of rotation. Rotating creates a “hole” through the center, so each slice looks like a washer (a disc with a circular hole).

Geometrically, a washer is the region between two circles:

• Outer radius $R$
• Inner radius $r$

So the washer area is:

$A=\pi R^2-\pi r^2$

Why it matters

Many realistic regions sit between two curves, and when rotated they form hollow shapes (like pipes, cups, and shells of material). On AP Calculus AB, the washer method is important because it combines:

• Interpreting “between curves” as outer minus inner distances
• Correctly identifying which boundary is farther from the axis (outer radius)
• Building a single integral that subtracts areas

It’s also a frequent place where small sign or geometry errors lead to big point losses—so it rewards careful setup.

How it works

If using vertical slices of thickness $dx$, the washer-volume integral is:

$V=\int_a^b \pi\left(R(x)\right)^2-\pi\left(r(x)\right)^2\,dx$

It’s common to factor out $\pi$:

$V=\pi\int_a^b \left(\left(R(x)\right)^2-\left(r(x)\right)^2\right)\,dx$

If the axis of rotation is vertical and you use horizontal slices of thickness $dy$:

$V=\pi\int_c^d \left(\left(R(y)\right)^2-\left(r(y)\right)^2\right)\,dy$

Outer radius vs inner radius

A reliable way to avoid confusion is to think “distance to the axis.” For a given slice:

• $R$ is the farther distance from the axis to the region.
• $r$ is the nearer distance from the axis to the region.

If rotating around $y=0$:

• Outer radius often comes from the “top” curve
• Inner radius often comes from the “bottom” curve

But if the axis is shifted (like $y=2$), “top” and “bottom” can swap roles in terms of distance. Always measure distance to the axis.

Notation reference (common AP setups)

Example 1 (Between two curves, rotate about the $x$-axis)

Find the volume when the region between $y=\sqrt{x}$ and $y=x$ from $x=0$ to $x=1$ is rotated about the $x$-axis.

First, determine which function is on top on $[0,1]$. For $0<x<1$, $\sqrt{x}>x$, so:

• Outer radius: $R(x)=\sqrt{x}$
• Inner radius: $r(x)=x$

Washer integral:

$V=\pi\int_0^1 \left((\sqrt{x})^2-(x)^2\right)\,dx$

Simplify:

$V=\pi\int_0^1 (x-x^2)\,dx$

Integrate:

$V=\pi\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1$

$V=\pi\left(\frac{1}{2}-\frac{1}{3}\right)=\frac{\pi}{6}$

Example 2 (Rotate around a shifted line $y=2$)

Find the volume when the region between $y=x$ and $y=0$ from $x=0$ to $x=2$ is rotated about $y=2$.

Here the axis is above the region. For a fixed $x$:

• The farther boundary from $y=2$ is $y=0$ (distance $2-0=2$).
• The nearer boundary is $y=x$ (distance $2-x$).

So:

$R(x)=2$

$r(x)=2-x$

Washer integral:

$V=\pi\int_0^2 \left((2)^2-(2-x)^2\right)\,dx$

Simplify inside:

$(2)^2-(2-x)^2=4-(4-4x+x^2)=4x-x^2$

So:

$V=\pi\int_0^2 (4x-x^2)\,dx$

Integrate:

$V=\pi\left[2x^2-\frac{x^3}{3}\right]_0^2$

$V=\pi\left(8-\frac{8}{3}\right)=\pi\cdot\frac{16}{3}=\frac{16\pi}{3}$

A helpful sanity check: rotating around $y=2$ produces a fairly large “donut-like” volume compared to rotating around the axis at the region, so a value like $16\pi/3$ is plausible.

Exam Focus

• Typical question patterns
  • “Region between $y=f(x)$ and $y=g(x)$ rotated about the $x$-axis” (classic washer).
  • Rotation about a horizontal/vertical line not equal to an axis, like $y=k$ or $x=h$.
  • You may be asked to set up the integral only (no evaluation) or to compute an exact value.
• Common mistakes
  • Swapping radii: using the closer curve as $R$ and the farther as $r$ (this can even make the integrand negative).
  • Forgetting to square radii: the area uses $R^2$ and $r^2$, not just $R-r$.
  • Not rewriting bounds when switching to $dy$: if you integrate with respect to $y$, your bounds must be $y$-values and your radii must be written as functions of $y$.