AP Calculus BC Unit 8 Applications of Integration: Area and Volume via Slicing

Finding the Area Between Curves

What “area between curves” means (and why integrals show up)

When you find the area of a region in geometry, you usually rely on formulas like $A = \frac{1}{2}bh$ or $A = \pi r^2$. But in calculus, the boundary of a region is often made of curves rather than straight lines or simple circles. There usually isn’t a single “area formula” for the region, so you build the area from many thin pieces whose areas you can approximate.

The key idea is accumulation: if you slice a region into lots of thin strips, the total area is approximately the sum of the areas of those strips. Taking the limit as the strips become infinitely thin turns that sum into a definite integral.

The slicing principle: vertical vs. horizontal slices

A region between curves is typically described by two boundary curves plus two “endpoints” that close the region. To integrate, you choose a slicing direction.

• Vertical slices have thickness $dx$. Each slice’s area looks like a thin rectangle.
• Horizontal slices have thickness $dy$.

Your choice matters because it determines what expression represents the “height” or “width” of a slice.

Vertical slices: “top minus bottom”

If you slice vertically at a particular $x$ value, the slice runs from the lower curve to the upper curve. If the upper function is $y = f(x)$ and the lower is $y = g(x)$, then the slice height is $f(x) - g(x)$ (assuming $f(x) \ge g(x)$ on the interval).

So the area becomes:

$A = \int_a^b \big(f(x) - g(x)\big)\,dx$

This is the classic top minus bottom rule.

Why it works: each thin slice has approximate area $\big(f(x) - g(x)\big)\Delta x$, and the integral adds them all up exactly.

Horizontal slices: “right minus left”

If you slice horizontally at a particular $y$ value, the slice runs from the left boundary curve to the right boundary curve. In this case you typically express curves as $x$ in terms of $y$.

If the right boundary is $x = r(y)$ and the left boundary is $x = \ell(y)$, then the slice width is $r(y) - \ell(y)$, so:

$A = \int_c^d \big(r(y) - \ell(y)\big)\,dy$

This is the horizontal analogue: right minus left.

How to choose the “best” direction

You usually choose the direction that makes the integral simpler.

• If the region is naturally described with $y$ as functions of $x$ and you can identify a single “top” and “bottom” on one interval, vertical slices are simplest.
• If “top” and “bottom” switch roles (so you’d need multiple integrals in $dx$), horizontal slices might let you do it in one integral (or vice versa).
• If the boundary curves are given as $x$ in terms of $y$ already, horizontal slices might be more direct.

A common mistake is to decide “I’ll integrate with respect to $x$” before thinking about whether the region stays “top minus bottom” over the whole interval. Always sketch or at least mentally picture the region first.

Finding bounds: intersection points and endpoints

For a bounded region, the limits of integration come from where the boundary curves meet or where vertical/horizontal boundaries are given.

• With vertical slicing, you often solve $f(x) = g(x)$ to find the $x$-values where the curves intersect.
• With horizontal slicing, you often solve for intersections to find $y$-limits, and you need $x = r(y)$ and $x = \ell(y)$.

If a region is bounded by more than two curves, you may need to split it into pieces so each piece has a consistent top/bottom or right/left.

A note about absolute value

Sometimes you’ll see a formula like:

$A = \int_a^b |f(x) - g(x)|\,dx$

This is conceptually correct because area is never negative. But on AP-style problems, it’s usually better to determine which function is on top (or right) on each subinterval and then write piecewise integrals. That prevents algebra mistakes and forces you to think about the geometry.

Notation reference (common equivalent setups)

Worked Example 1: Area between two functions (vertical slices)

Find the area of the region bounded by $y = x$ and $y = x^2$.

Step 1: Understand the region. These curves intersect where $x = x^2$. The region between them is “closed” between intersection points.

Step 2: Find intersection points (bounds).

$x = x^2$

$x^2 - x = 0$

$x(x-1) = 0$

So the curves intersect at $x=0$ and $x=1$.

Step 3: Determine top and bottom. On $0 \le x \le 1$, the line $y=x$ is above $y=x^2$ (for example at $x=\tfrac{1}{2}$, $\tfrac{1}{2} > \tfrac{1}{4}$).

Step 4: Set up and compute the integral.

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

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

$A = \left(\frac{1}{2} - \frac{1}{3}\right) - 0$

$A = \frac{1}{6}$

Worked Example 2: When one integral is not enough (function order changes)

Find the area between $y = x^3$ and $y = x$ from $x=-1$ to $x=1$.

Why this is tricky: On part of the interval, $x^3$ is above $x$; on another part, it’s below. If you just do $\int_{-1}^1 (x - x^3)\,dx$ you might accidentally compute a signed area that cancels.

Step 1: Find where they intersect.

$x^3 = x$

$x^3 - x = 0$

$x(x^2-1) = 0$

So $x = -1, 0, 1$.

Step 2: Determine which is on top on each subinterval.

• On $(-1,0)$, test $x=-\tfrac{1}{2}$: $x^3=-\tfrac{1}{8}$ and $x=-\tfrac{1}{2}$, so $x^3 > x$.
• On $(0,1)$, test $x=\tfrac{1}{2}$: $x^3=\tfrac{1}{8}$ and $x=\tfrac{1}{2}$, so $x > x^3$.

Step 3: Add areas piecewise.

$A = \int_{-1}^0 (x^3 - x)\,dx + \int_0^1 (x - x^3)\,dx$

Compute one and use symmetry or compute both directly. Let’s compute both quickly:

$\int (x^3 - x)\,dx = \frac{x^4}{4} - \frac{x^2}{2}$

First part:

$\left[\frac{x^4}{4} - \frac{x^2}{2}\right]_{-1}^0 = \left(0-0\right) - \left(\frac{1}{4} - \frac{1}{2}\right) = \frac{1}{4}$

Second part:

$\int (x - x^3)\,dx = \frac{x^2}{2} - \frac{x^4}{4}$

$\left[\frac{x^2}{2} - \frac{x^4}{4}\right]_0^1 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$

So:

$A = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$

Exam Focus

• Typical question patterns:
  • “Find the area of the region enclosed by …” where you must first find intersections, then integrate top minus bottom (or right minus left).
  • A region described with multiple curves and a line like $x=2$ or $y=0$, requiring you to choose vertical vs. horizontal slicing.
  • A setup question: “Write (but do not evaluate) an integral expression for the area…” which tests bounds and the correct integrand.
• Common mistakes:
  • Forgetting to split the integral when the top/bottom function changes (leading to cancellation).
  • Mixing up right minus left with top minus bottom when switching to $dy$.
  • Using intersection points incorrectly (solving for $y$ when you need $x$-bounds, or vice versa).

Volumes with Cross Sections

What “volume with cross sections” means

Sometimes a solid isn’t a neat sphere or cylinder, but you still can describe it systematically: you know its base region (a 2D region in the plane) and you know what shape each cross section is when you slice the solid by planes perpendicular to an axis.

In volume by known cross sections, you build the solid from thin slices with thickness $dx$ or $dy$. Each slice has a cross-sectional area $A(x)$ or $A(y)$, so the volume is the integral of these areas.

The foundational formula is:

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

or, if slicing horizontally,

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

The variables $a$ and $b$ (or $c$ and $d$) represent the start and end of the solid along the slicing direction.

Why this method matters

This method is powerful because it handles many solids that are not “solids of revolution.” On AP problems, it also tests whether you can:

• interpret geometric information from a 2D base region,
• translate it into a side length, radius, or other dimension,
• use the correct area formula for the cross-sectional shape,
• integrate over the correct interval.

Conceptually, it’s the same accumulation idea as area between curves: you’re summing infinitely many tiny volumes.

The key workflow: base, slice, area function, integrate

1. Sketch the base region (even a rough sketch). Identify the slicing direction (perpendicular to the $x$-axis or $y$-axis).
2. Express the relevant slice dimension as a function of the slicing variable.
   • If slicing perpendicular to the $x$-axis, the slice is at a particular $x$, and you often need a vertical distance in the base region.
   • That distance is frequently $\text{top}(x) - \text{bottom}(x)$.
3. Convert that dimension into cross-sectional area using the given shape.
4. Integrate area over the interval.

A major source of errors is confusing “the distance in the base region” with “the cross-sectional area.” The distance is often a side length or diameter; you still must apply the correct area formula.

Common cross-section shapes and how to translate dimensions

Suppose the slice at $x$ has a segment length $s(x)$ in the base region. This segment is often the “width” of the base region at that $x$.

Here are common AP shapes:

• Square cross sections: if the side length is $s(x)$,

$A(x) = (s(x))^2$

• Rectangular cross sections: if one side is $s(x)$ and the other is a constant multiple, say height $k\,s(x)$,

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

• Equilateral triangle cross sections: if side length is $s(x)$, area is:

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

• Right isosceles triangle cross sections: if the leg is $s(x)$, area is:

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

• Semicircle cross sections: if diameter is $s(x)$, radius is $\frac{s(x)}{2}$, so:

$A(x) = \frac{1}{2}\pi\left(\frac{s(x)}{2}\right)^2 = \frac{\pi}{8}(s(x))^2$

Notice a pattern: many cross-sectional areas end up being a constant times $(s(x))^2$. The constant depends on the shape.

Worked Example 1: Squares built on a base region

The base of a solid is the region bounded by $y = \sqrt{x}$ and $y = 0$ from $x=0$ to $x=4$. Cross sections perpendicular to the $x$-axis are squares. Find the volume.

Step 1: Identify the slice segment length in the base.
At a given $x$, a vertical slice through the base runs from $y=0$ up to $y=\sqrt{x}$. So the segment length is:

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

Step 2: Convert to cross-sectional area.
Squares have area $s(x)^2$, so:

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

Step 3: Integrate from $0$ to $4$.

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

$V = \left[\frac{x^2}{2}\right]_0^4 = \frac{16}{2} = 8$

So the volume is $8$ cubic units.

What to notice: The hard-looking $\sqrt{x}$ simplified after squaring. That’s common with square or semicircle cross sections.

Worked Example 2: Semicircular cross sections (diameter from top-bottom)

The base is the region bounded by $y = 4 - x^2$ and $y = 0$. Cross sections perpendicular to the $x$-axis are semicircles whose diameters lie in the base. Find the volume.

Step 1: Determine the interval. The region meets $y=0$ when:

$4 - x^2 = 0$

$x^2 = 4$

So $x=-2$ to $x=2$.

Step 2: Diameter as a function of $x$. A vertical segment in the base goes from $0$ to $4-x^2$, so the diameter is:

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

Step 3: Area of a semicircle in terms of diameter.

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

Step 4: Integrate.

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

You could expand and integrate, or use symmetry because the integrand is even. Using symmetry:

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

Now expand:

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

So:

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

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

Evaluate at $2$:

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

Common denominator $15$:

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

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

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

So the bracket equals:

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

Thus:

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

What goes wrong most often (built into the geometry)

• Using radius when the problem gives diameter: if the cross section is a circle or semicircle and the base segment is the diameter, then $r = \frac{s(x)}{2}$.
• Forgetting to square: cross-sectional areas for squares and many triangles involve $(s(x))^2$.
• Slicing in the wrong direction: “perpendicular to the $x$-axis” means your thickness is $dx$, and the segment length usually comes from top minus bottom as functions of $x$.

Exam Focus

• Typical question patterns:
  • “A solid has base region bounded by … Cross sections perpendicular to the $x$-axis are squares/semicircles/triangles. Find the volume.”
  • “Write an integral expression for the volume” (they may not require evaluation).
  • Problems where you must decide whether to use $dx$ or $dy$ based on “perpendicular to” wording.
• Common mistakes:
  • Treating the slice length $s(x)$ as the area directly (forgetting the shape’s area formula).
  • Mixing up “perpendicular to $x$-axis” with “parallel to $x$-axis.”
  • Using the wrong bounds by not first finding where the base region starts and ends.

Volumes with Disc and Washer Methods

What makes a solid a “solid of revolution”

A solid of revolution is formed when you rotate a plane region around a line (an axis of rotation). The rotation sweeps out a 3D object. Two especially common cases are:

• rotating around the $x$-axis or $y$-axis,
• rotating around horizontal or vertical lines like $y = k$ or $x = h$.

The disc and washer methods come from the same slicing idea as before: a thin slice perpendicular to the axis of rotation becomes a circle (a disc) or a ring (a washer).

Disc method: when slices have no hole

If the region touches the axis of rotation, the cross section is a solid disc.

If you rotate around the $x$-axis and your radius at position $x$ is $R(x)$, then the cross-sectional area is:

$A(x) = \pi (R(x))^2$

and the volume is:

$V = \int_a^b \pi (R(x))^2\,dx$

Similarly, if you rotate around the $y$-axis using horizontal slices (so radius is a function of $y$):

$V = \int_c^d \pi (R(y))^2\,dy$

Why squaring happens: the area of a circle depends on radius squared, so even a small mistake in the radius expression can cause a big error.

Washer method: when slices have a hole

If the region does not touch the axis of rotation, rotating it creates a hole down the center. The cross section is a washer: an outer circle minus an inner circle.

If outer radius is $R(x)$ and inner radius is $r(x)$, then cross-sectional area is:

$A(x) = \pi (R(x))^2 - \pi (r(x))^2 = \pi\big((R(x))^2 - (r(x))^2\big)$

and volume:

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

A helpful memory aid is: Outside minus inside (for area), then integrate.

How to find radii correctly

The radii are distances from the axis of rotation.

• Rotating around the $x$-axis (which is $y=0$): radius is just the $y$-value of the curve (distance to $y=0$), so $R(x) = f(x)$ if $f(x) \ge 0$.
• Rotating around a horizontal line $y=k$: radius is vertical distance to $y=k$.
  • If curve is above the line, radius might look like $f(x) - k$.
  • If curve is below the line, radius might look like $k - f(x)$.
• Rotating around the $y$-axis (which is $x=0$): radius is the $x$-distance to that axis.
• Rotating around a vertical line $x=h$: radius is horizontal distance to $x=h$.

A frequent misconception is to treat the function value itself as the radius even when rotating around $y=k$ or $x=h$. Always compute “distance to the axis,” not “distance to zero.”

Choosing $dx$ vs. $dy$ for discs and washers

Discs/washers use slices perpendicular to the axis of rotation.

• If you rotate around a horizontal line (like $y=0$ or $y=3$), then perpendicular slices are vertical, thickness $dx$.
• If you rotate around a vertical line (like $x=0$ or $x=2$), then perpendicular slices are horizontal, thickness $dy$.

That means sometimes you must rewrite a function as $x$ in terms of $y$ to use discs/washers comfortably.

Worked Example 1: Disc method 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.

Step 1: Identify the radius. Rotating around $y=0$, the radius is the distance from the curve to the $x$-axis:

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

This is a disc (no hole) because the region touches the axis.

Step 2: Set up the volume integral.

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

Simplify:

$(\sqrt{x})^2 = x$

So:

$V = \pi\int_0^4 x\,dx = \pi\left[\frac{x^2}{2}\right]_0^4 = 8\pi$

Worked Example 2: Washer method around a horizontal line

Find the volume of the solid formed by rotating the region between $y = x$ and $y = x^2$ on $0 \le x \le 1$ about the line $y=2$.

Step 1: Visualize the axis location. The line $y=2$ is above both curves on $[0,1]$ (since both are between 0 and 1 there). Rotating around a line above the region creates washers.

Step 2: Determine outer vs. inner radius. The radii are distances down from $y=2$.

• The outer radius goes from $y=2$ to the lower curve (farther away). On $[0,1]$, the lower curve is $y=x^2$, so:

$R(x) = 2 - x^2$

• The inner radius goes from $y=2$ to the upper curve (closer). The upper curve is $y=x$, so:

$r(x) = 2 - x$

Step 3: Set up the washer integral.

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

At this point you could expand and integrate.

Common check: The integrand should be nonnegative because $R(x) \ge r(x)$. Here, $2-x^2 \ge 2-x$ on $[0,1]$ because $x \ge x^2$.

Worked Example 3: Rotating around the $y$-axis using $dy$

Rotate the region bounded by $x = y^2$ and $x = 4$ from $y=-2$ to $y=2$ about the $y$-axis. Find the volume.

Step 1: Choose discs/washers orientation. The axis of rotation is vertical (the $y$-axis), so discs/washers use horizontal slices, thickness $dy$.

Step 2: Determine radii as distances to $x=0$.
At a given $y$:

• Outer boundary is $x=4$, so outer radius is $R(y) = 4$.
• Inner boundary is $x=y^2$, so inner radius is $r(y) = y^2$.

Step 3: Washer integral.

$V = \int_{-2}^{2} \pi\big((4)^2 - (y^2)^2\big)\,dy$

Simplify:

$V = \int_{-2}^{2} \pi(16 - y^4)\,dy$

Use symmetry (even function):

$V = 2\pi\int_{0}^{2} (16 - y^4)\,dy$

Integrate:

$V = 2\pi\left[16y - \frac{y^5}{5}\right]_0^2 = 2\pi\left(32 - \frac{32}{5}\right)$

$V = 2\pi\cdot \frac{128}{5} = \frac{256\pi}{5}$

What goes wrong most often in disc/washer problems

• Wrong radius expressions: Students write $R(x)=f(x)$ when rotating around $y=k$, but the radius is a distance to $y=k$.
• Swapping outer and inner: If you accidentally make $r(x)$ larger than $R(x)$, the integrand becomes negative. A quick reasonableness check is: washers should give nonnegative cross-sectional area.
• Wrong variable and bounds: Rotating around the $y$-axis often pushes you toward $dy$ with radii in terms of $y$. If you insist on $dx$, you might need a different method (like shells, which is a different topic) or you may end up doing unnecessary algebra.

Exam Focus

• Typical question patterns:
  • “Find the volume when the region bounded by … is rotated about the $x$-axis / $y$-axis / $y=k$ / $x=h$.”
  • “Set up (but do not evaluate) an integral for the volume” emphasizing correct radii and bounds.
  • Multi-step regions where you must first find intersection points to get limits before setting up washers.
• Common mistakes:
  • Using the curve itself as the radius when the axis is shifted (forgetting the distance-to-axis idea).
  • Forgetting to square radii in $\pi R^2$ and $\pi r^2$.
  • Integrating with respect to the wrong variable (using $dx$ when the washer radii are naturally functions of $y$ for rotation about a vertical axis).