How do I know whether to integrate with respect to x or y?

Look at the orientation of the cross sections. If they are perpendicular to the x-axis, the base segments are vertical and you integrate in x using A(x)\,dx. If they are perpendicular to the y-axis, the segments are horizontal and you integrate in y, so you must rewrite the boundary curves as functions of y.

What is the difference between this and the disc or washer method?

The disc and washer methods (taught in U8.9) apply when a region is revolved around an axis, producing circular cross sections automatically. Here the solid is not revolved—its cross sections can be squares, triangles, or semicircles as described in the problem. You must read the given shape and use its area formula rather than assuming circles.

How do I handle a semicircle when the base is the diameter versus the radius?

If the base segment is the diameter L, the radius is L/2, so the semicircle area is 1/2(L/2)^2=/8L^2. If the base segment is the radius L, the area is 1/2 L^2. Always read the wording carefully—this distinction changes the coefficient and is a common source of lost points.

Do I get partial credit if I set up the integral but make an arithmetic error?

Yes. On free-response questions, most points come from a correct integrand (the area function with the right shape coefficient and base length) and correct limits of integration. Writing the full integral before evaluating protects those points even if you slip during the final computation.

AP-CALCBC-8.7-8.8

U8.7 Volumes with Known Cross Sections

Master AP Calculus BC volumes with known cross sections: use V = ∫ A(x) dx for square, rectangle, triangle, and semicircle cross sections with clear worked steps.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U8.7 Volumes with Known Cross Sections, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start U8.7 Volumes with Known Cross Sections free

What this lesson covers

When a solid is not formed by revolution, you can still find its volume by slicing it into thin pieces perpendicular to an axis and adding up their volumes. Each slice is a thin slab whose face is a familiar shape—a square, rectangle, triangle, or semicircle—built on a segment between two curves. This lesson teaches you to translate a geometric description into an area function

A(x)

and integrate it.

This is one of the most reliable FRQ topics on the exam, and it rewards students who set up the area carefully before integrating. By the end you will confidently identify the "base" segment, express its length, plug it into the correct area formula, and evaluate the definite integral.

The Core Idea: Slicing a Solid

Imagine a solid sitting over a region in the

xy

-plane. If you slice the solid with planes perpendicular to the

x

-axis, each slice is a thin slab of thickness

dx

. The face of that slab has some area

A(x)

, and its volume is approximately

A(x)\,dx

. Adding up infinitely many slabs gives the exact volume:

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

Here

a

and

b

are the

x

-values where the solid begins and ends. If the slices are perpendicular to the

y

-axis instead, you integrate with respect to

y

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

The whole game is finding

A(x)

. The area depends on a length—usually the distance between two curves—that forms one edge (the "base") of the cross-sectional shape. Once you know that length in terms of

x

, you substitute it into the appropriate area formula for a square, triangle, semicircle, and so on.

A common misconception is confusing this with volumes of revolution. Here nothing is spinning; the shapes are simply stacked. The exam expects you to read the geometry, not assume a disc or washer. Always ask: perpendicular to which axis? What shape is the cross section? What is its base length?

Finding the Base Length

The base of each cross section is a segment lying in the plane region. If the region is bounded above by

y = f(x)

and below by

y = g(x)

, then for cross sections perpendicular to the

x

-axis the base length is

L(x) = f(x) - g(x),

the top curve minus the bottom curve. If the region lies between a single curve

y = f(x)

and the

x

-axis, then

L(x) = f(x)

.

When slices are perpendicular to the

y

-axis, you rewrite everything in terms of

y

: solve the curves for

x

, and the base length is (right curve) minus (left curve),

L(y) = x_{\text{right}} - x_{\text{left}}

.

A frequent error is integrating over the wrong variable. If the cross sections are perpendicular to the

x

-axis, your base runs vertically and your integral is

dx

. If perpendicular to the

y

-axis, the base runs horizontally and the integral is

dy

. Sketching the region and drawing one representative slice prevents this mistake. Label the slice's endpoints so you can see which curve is larger.

Area Formulas by Cross-Section Shape

Once you have the base length

L

, apply the correct area formula. Memorize these—they appear constantly on the FRQ.

Cross section	Area in terms of base $L$
Square (base is a side)	$A = L^2$
Semicircle (base is diameter)	$A = \frac{\pi}{8}L^2$
Semicircle (base is radius)	$A = \frac{\pi}{2}L^2$
Equilateral triangle (base is a side)	$A = \frac{\sqrt{3}}{4}L^2$
Isosceles right triangle (base is a leg)	$A = \frac{1}{2}L^2$
Isosceles right triangle (base is hypotenuse)	$A = \frac{1}{4}L^2$
Rectangle of height $h$	$A = L\cdot h$

The semicircle case trips students up: if the base is the diameter, the radius is

L/2

, so

A = \frac{1}{2}\pi(L/2)^2 = \frac{\pi}{8}L^2

. Read carefully whether the segment is a diameter or a radius.

For an equilateral triangle with side

L

, the height is

\frac{\sqrt{3}}{2}L

, so

A = \frac{1}{2}\cdot L\cdot\frac{\sqrt{3}}{2}L = \frac{\sqrt{3}}{4}L^2

. Deriving these on scratch paper is safer than memorizing if you forget.

How the Exam Tests This

On the AP exam this topic appears in both multiple-choice and free-response form. FRQ versions typically give you a region bounded by curves and describe cross sections in words: "Cross sections perpendicular to the

x

-axis are squares." You earn points for the correct integrand (the area function) and the correct limits, then for the evaluated value.

A calculator-active problem often wants a decimal answer, so set up the integral exactly and let the calculator evaluate. On the no-calculator section, the integrand is chosen so the antiderivative is clean, so watch for factors like

\frac{\sqrt{3}}{4}

\frac{\pi}{8}

that pull out front.

A classic partial-credit trap: writing

A(x) = (f(x)-g(x))

instead of squaring it, or forgetting the shape's coefficient. Another trap is mixing up the diameter versus radius for semicircles. Always write the area formula symbolically first, then substitute the base length. Show the integral with limits before computing; that setup earns most of the points even if arithmetic slips.

Key terms

Cross section.: The two-dimensional shape formed when a plane slices through a solid, oriented perpendicular to a chosen axis.
Area function $A(x)$ .: A formula giving the area of the cross section at position $x$ , obtained by substituting the base length into the shape's area formula.
Base length.: The length of the segment in the plane region that forms one edge of the cross section, typically top curve minus bottom curve, $f(x)-g(x)$ .
Slab (slice).: A thin piece of the solid of thickness $dx$ or $dy$ whose volume is approximately $A(x)\,dx$ .
Volume by cross sections.: The method $V=\int_a^b A(x)\,dx$ that sums the volumes of all thin slabs across the solid.
Perpendicular orientation.: The direction in which slices are taken; determines whether you integrate in $x$ or $y$ and how base length is expressed.

Worked example

Let

R

be the region bounded by

y=\sqrt{x}

, the

x

-axis, and the line

x=4

. A solid has base

R

, and cross sections perpendicular to the

x

-axis are squares. Find the volume of the solid.

First identify the base length. For each

x

from

0

4

, the region runs from the

x

-axis (

y=0

) up to

y=\sqrt{x}

. So the base length of each square is

L(x)=\sqrt{x}-0=\sqrt{x}.

Since the cross sections are squares with side equal to this base, the area is

A(x)=L(x)^2=(\sqrt{x})^2=x.

The solid extends from

x=0

x=4

, so

V=\int_0^4 A(x)\,dx=\int_0^4 x\,dx.

Evaluate the integral:

V=\left[\frac{x^2}{2}\right]_0^4=\frac{16}{2}-0=8.

The volume is

8

cubic units. Notice how squaring the base length made the integrand simple—this is typical of no-calculator problems. Had the cross sections been semicircles with the base as diameter, we would instead use

A(x)=\frac{\pi}{8}x

, giving

V=\frac{\pi}{8}\cdot 8=\pi

Practice questions

The base of a solid is the region bounded by

y=x^2

and

y=4

. Cross sections perpendicular to the

y

-axis are squares. Which integral gives the volume?

$\int_0^4 (2\sqrt{y})^2\,dy$
$\int_0^4 (\sqrt{y})^2\,dy$
$\int_{-2}^{2} (x^2)^2\,dx$
$\int_0^4 \left(4-y\right)^2\,dy$

Answer: $\int_0^4 (2\sqrt{y})^2\,dy$

Because slices are perpendicular to the

y

-axis, integrate in

y

. Solving

y=x^2

gives

x=\pm\sqrt{y}

, so the horizontal base runs from

-\sqrt{y}

\sqrt{y}

, a length of

2\sqrt{y}

. The region spans

y=0

y=4

. Squares give

A=L^2=(2\sqrt{y})^2

, so the volume is

\int_0^4 (2\sqrt{y})^2\,dy

The base of a solid is the region between

y=1-x^2

and the

x

-axis. Cross sections perpendicular to the

x

-axis are equilateral triangles. Set up and evaluate the volume.

Answer: $V=\frac{8\sqrt{3}}{15}$

The base length is

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

, and the region spans

x=-1

x=1

. For an equilateral triangle with side

L

A(x)=\frac{\sqrt{3}}{4}(1-x^2)^2

. So

V=\frac{\sqrt{3}}{4}\int_{-1}^{1}(1-x^2)^2\,dx

. Expand:

(1-x^2)^2=1-2x^2+x^4

. Its antiderivative is

x-\frac{2}{3}x^3+\frac{1}{5}x^5

. Evaluated from

-1

1

by symmetry equals

2\left(1-\frac{2}{3}+\frac{1}{5}\right)=2\cdot\frac{8}{15}=\frac{16}{15}

. Thus

V=\frac{\sqrt{3}}{4}\cdot\frac{16}{15}=\frac{8\sqrt{3}}{15}

A solid has base the region bounded by

y=x

and

y=x^2

. Cross sections perpendicular to the

x

-axis are semicircles whose diameters lie in the base. Which expression is the area function

A(x)

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

Answer: $\frac{\pi}{8}(x-x^2)^2$

[0,1]

the line

y=x

lies above

y=x^2

, so the base (diameter) length is

L=x-x^2

. For a semicircle whose diameter is

L

, the radius is

L/2

and area is

\frac{1}{2}\pi(L/2)^2=\frac{\pi}{8}L^2

. Substituting gives

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

FAQ

How do I know whether to integrate with respect to x or y?: Look at the orientation of the cross sections. If they are perpendicular to the $x$ -axis, the base segments are vertical and you integrate in $x$ using $A(x)\,dx$ . If they are perpendicular to the $y$ -axis, the segments are horizontal and you integrate in $y$ , so you must rewrite the boundary curves as functions of $y$ .
What is the difference between this and the disc or washer method?: The disc and washer methods (taught in U8.9) apply when a region is revolved around an axis, producing circular cross sections automatically. Here the solid is not revolved—its cross sections can be squares, triangles, or semicircles as described in the problem. You must read the given shape and use its area formula rather than assuming circles.
How do I handle a semicircle when the base is the diameter versus the radius?: If the base segment is the diameter $L$ , the radius is $L/2$ , so the semicircle area is $\frac{1}{2}\pi(L/2)^2=\frac{\pi}{8}L^2$ . If the base segment is the radius $L$ , the area is $\frac{1}{2}\pi L^2$ . Always read the wording carefully—this distinction changes the coefficient and is a common source of lost points.
Do I get partial credit if I set up the integral but make an arithmetic error?: Yes. On free-response questions, most points come from a correct integrand (the area function with the right shape coefficient and base length) and correct limits of integration. Writing the full integral before evaluating protects those points even if you slip during the final computation.

Learn this with a teacher, not a page

The Crimsora tutor teaches U8.7 Volumes with Known Cross Sections live — explaining on a whiteboard, asking you questions, and adapting to where you get stuck.

Try a live session free