Unit 8Multiple Integrals

Concept 4.97Built for May 2027 AP Exam

Multiple Integrals

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

Some problems need more than one integral, chained in sequence rather than added side by side. The standard case is a motion chain: integrate acceleration to get velocity, then integrate velocity to get position, adding a constant at each stage. A second case is a multi-part question where each part needs its own definite integral, and the value from one part cannot be reused for another because the limits or the integrand differ. In both cases the discipline is the same. Label each integral, keep its constant visible, and check that the output of one stage has the units the next stage expects.

On the AP Exam

AP Calculus AB tests this skill through the representations and reasoning described below.

What you’ll see: A motion problem that gives acceleration and asks for velocity and position, or a multi-part question where each part carries its own integral with different limits.
What you’ll do: Integrate once per stage, apply the given anchor at each stage, and label the result. For multi-part questions, start each part from the stem rather than from your previous answer.
Memory hook: One integral per stage, one constant per integral. Chained, not stacked.
Use it on the exam: For MCQs: Watch for a question that gives $a(t)$ and asks for the change in position. That is two integrations, and both constants need values. For FRQs: Write the constant as a named symbol, then solve for it from the given condition. Dropping it costs the anchor point in almost every released rubric.

Concept Review

Chained Integration

Figure 1 — Two integrations in sequence, each introducing its own constant.

Given $a(t)$ with $v(t_0) = v_0$ and $s(t_0) = s_0$ :

$v(t) = v_0 + \int_{t_0}^{t} a(u)\,du \qquad\qquad s(t) = s_0 + \int_{t_0}^{t} v(u)\,du$

Each line is an application of the net change theorem from topic 8.5. The dummy variable $u$ keeps the upper limit free, and the constants $v_0$ and $s_0$ are the anchors from topic 8.6. Two integrations, two anchors, no exceptions.

Three small panels left to right: acceleration 6t, velocity 2 plus 3t squared, position 5 plus 2t plus t cubed, with arrows between them labelled by integral signs

Why the Constants Cannot Be Dropped

An antiderivative is determined only up to a constant, so the first integration produces a family of velocity functions and the given $v_0$ selects one. Skipping that step and integrating again produces a position function that is wrong by a linear term, which is exactly the error that looks smallest and costs the most.

Stage	Result of integrating	Anchor needed	What the anchor fixes
first	velocity family	$v(t_0) = v_0$	vertical shift of $v$
second	position family	$s(t_0) = s_0$	vertical shift of $s$
third (rare)	accumulated displacement	none beyond the above	nothing new

Several Integrals in One Question

Multi-part FRQs often need several unrelated integrals. Part (a) may ask for an average value, part (b) for an area, and part (c) for a volume. Each begins from the stem, and each has its own limits.

The failure mode here is contamination: a student computes one integral in part (a), then reuses that number in part (b) where a different integrand was required. The graders score each part independently, so a contaminated part loses its own points even when the earlier part was perfect.

Chained Versus Stacked

Two integrals can be related in two different ways, and the wording tells you which.

Relationship	Signal in the question	Structure
chained	“find the velocity, then the position”	output of one integral feeds the next integrand
stacked (split)	“the region has two parts”	same integrand class, different limits, added
parallel	“find the area and the average value”	independent integrals, no shared data

Topic 8.9 covered the split case. This topic covers the chained case, and the parallel case is really just several topics appearing in one question.

Units Through the Chain

Each integration multiplies by one power of time, so units shift at every stage.

Quantity	Units	How obtained
acceleration	meters per second squared	given
velocity	meters per second	integrate acceleration once
position	meters	integrate velocity once
displacement	meters	subtract two positions

If the final answer carries the units of velocity, one integration is missing. That check catches the majority of chained-integration errors before any arithmetic is verified.

Common Misconception Check

Myth 1: “One constant serves the whole chain.” Truth: Each integration introduces its own constant, and each needs its own condition.
Myth 2: “You can integrate $a(t)$ twice directly to get $s(t)$ .” Truth: You can, but the result carries two unknown constants $C_1 t + C_2$ , and both must be fixed by conditions.
Myth 3: “The answer to part (a) can be reused in part (b).” Truth: Only when the integral is genuinely the same. Different limits mean a different number.

Topic-Specific Study Method

Label Every Stage

Borrowed idea: Chained integration is bookkeeping as much as calculus. Each stage introduces a constant, and each constant needs a condition from the stem. A velocity without its anchor is a family of functions. Integrating that family gives a position that drifts away by a linear term. No later step can repair the drift. Label the stages, write each constant as a symbol, and solve for it immediately. When a question holds several independent integrals, treat each one as a fresh start from the given data, because contamination between parts is scored as an error in the part where it appears.

Exam disguise	What you see on the page	What your brain must translate	Required execution
“given $a(t)$ and $v(0)$ , find $s(3)$ ”	two conditions, one target	two-stage chain	integrate twice, anchor twice
“find the change in velocity on $[0,4]$ ”	one condition only	single integral	integrate $a$ on $[0,4]$
“the particle is at rest at $t = 0$ ”	a hidden anchor	$v_0 = 0$	use it before the second stage
“find the area and the average value”	two separate tasks	parallel integrals	compute each from the stem
“how far has it travelled by $t = 5$ ”	a chained quantity	distance, not position	chain, then split at zeros

① Write the two stage equations before substituting anything. The structure prevents the missing-anchor error.

② Solve for each constant the moment it appears. A constant carried forward becomes two unknowns later.

③ Check the units at the end of the chain. Meters is right; meters per second means a missing stage.

Training & Error Diagnosis

Core Archetype (Simplest Form): A particle has acceleration $a(t) = 6t$ meters per second squared, initial velocity $v(0) = 2$ meters per second, and initial position $s(0) = 5$ meters. Find $s(2)$ .

Standard Execution:

Velocity. $v(t) = 2 + \int_0^t 6u\,du = 2 + 3t^{2}$ .

Position. $s(t) = 5 + \int_0^t \left(2 + 3u^{2}\right)du = 5 + 2t + t^{3}$ .

Evaluate. $s(2) = 5 + 4 + 8 = 17$ meters.

Check. The velocity grows quickly, so a position well beyond the initial $5$ meters is expected.

Exam Variation (Missing Anchor / Context Shift): A particle moves with velocity $v(t) = 3t^{2} - 12t$ meters per second and is at position $s = 4$ when $t = 1$ . Find the position at $t = 4$ and the total distance travelled on $1 \le t \le 4$ .

Solution: Integrating gives $s(t) = t^{3} - 6t^{2} + C$ , and $s(1) = 1 - 6 + C = 4$ forces $C = 9$ . Then $s(4) = 64 - 96 + 9 = -23$ , so the displacement on $[1,4]$ is $-27$ meters. For distance, note $v = 3t(t-4)$ , so $v = 0$ at $t = 4$ only, and $v < 0$ on $(1,4)$ . The distance is $\left\lvert \int_1^4 v\,dt \right\rvert = 27$ meters, which equals the magnitude of the displacement here because the velocity never reverses inside the open interval.

Common Traps & Failure Signals:

One constant (Signal: $s(t) = t^{3} + C$ with no $C t$ term from the velocity family): each stage needs its own anchor.
Anchor unused (Signal: a position answer with an unevaluated $C$ ): the condition in the stem exists to be used.
Part contamination (Signal: an area value reused inside a volume integral): each part restarts from the stem.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Constant Dropped: You integrated twice with only one constant of integration.
- Anchor Skipped: You carried a symbolic $C$ to the end of the problem and never used the condition the stem supplied, so the answer described a whole family of motions instead of the one the question named.
- Stage Blurred: You treated the velocity integral and the position integral as the same computation, which produces a number with velocity units where meters were required.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; Chained integration appears regularly in multiple choice, usually as an acceleration given with two conditions, but the released FRQs stop at one integration per part
10-Year Total Count: 0 of 9 Exams · 0 Parts; Motion FRQ parts since 2017 supply the velocity directly, which is why the chained version stays in the multiple-choice pool
FRQ Years Appeared: None in Released FRQs

Practice Examples

MCQ 1

One Stage

A particle has acceleration $a(t) = 6t$ and initial velocity $v(0) = 2$ . Its velocity at $t = 3$ is

MCQ 2

Two Stages

With $a(t) = 6t$ , $v(0) = 2$ , and $s(0) = 5$ , the position at $t = 2$ is

MCQ 3

Which Anchor Applies

A particle has $v(1) = 3$ and $s(1) = 0$ . Which expression gives $s(t)$ ?

FRQ 4

A Two-Stage Motion Problem

A particle moves along a line with acceleration $a(t) = 6t$ meters per second squared for $0 \le t \le 3$ . At $t = 0$ its velocity is $2$ meters per second and its position is $5$ meters.

(a) Find $v(t)$ .

(b) Find $s(t)$ and evaluate $s(3)$ .

(c) Find the total distance travelled on $0 \le t \le 3$ and explain why it equals the displacement here.

Cheatsheet

Core Reference

	Action	Failure if skipped
1	list the stages and the anchor for each	a constant is missing
2	integrate stage one	the velocity family is unavailable
3	solve for the first constant immediately	two unknowns accumulate
4	integrate stage two	the position family is unavailable
5	solve for the second constant	the answer stays symbolic
6	check units at the end	a missing stage survives

Sentences That Score

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Unit 8Multiple Integrals

Concept 4.97Built for May 2027 AP Exam

Multiple Integrals

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

AP Calculus AB tests this skill through the representations and reasoning described below.

What you’ll see: A motion problem that gives acceleration and asks for velocity and position, or a multi-part question where each part carries its own integral with different limits.
What you’ll do: Integrate once per stage, apply the given anchor at each stage, and label the result. For multi-part questions, start each part from the stem rather than from your previous answer.
Memory hook: One integral per stage, one constant per integral. Chained, not stacked.
Use it on the exam: For MCQs: Watch for a question that gives $a(t)$ and asks for the change in position. That is two integrations, and both constants need values. For FRQs: Write the constant as a named symbol, then solve for it from the given condition. Dropping it costs the anchor point in almost every released rubric.

Concept Review

Chained Integration

Figure 1 — Two integrations in sequence, each introducing its own constant.

Given $a(t)$ with $v(t_0) = v_0$ and $s(t_0) = s_0$ :

$v(t) = v_0 + \int_{t_0}^{t} a(u)\,du \qquad\qquad s(t) = s_0 + \int_{t_0}^{t} v(u)\,du$

Three small panels left to right: acceleration 6t, velocity 2 plus 3t squared, position 5 plus 2t plus t cubed, with arrows between them labelled by integral signs

Why the Constants Cannot Be Dropped

Stage	Result of integrating	Anchor needed	What the anchor fixes
first	velocity family	$v(t_0) = v_0$	vertical shift of $v$
second	position family	$s(t_0) = s_0$	vertical shift of $s$
third (rare)	accumulated displacement	none beyond the above	nothing new

Several Integrals in One Question

Multi-part FRQs often need several unrelated integrals. Part (a) may ask for an average value, part (b) for an area, and part (c) for a volume. Each begins from the stem, and each has its own limits.

Chained Versus Stacked

Two integrals can be related in two different ways, and the wording tells you which.

Relationship	Signal in the question	Structure
chained	“find the velocity, then the position”	output of one integral feeds the next integrand
stacked (split)	“the region has two parts”	same integrand class, different limits, added
parallel	“find the area and the average value”	independent integrals, no shared data

Topic 8.9 covered the split case. This topic covers the chained case, and the parallel case is really just several topics appearing in one question.

Units Through the Chain

Each integration multiplies by one power of time, so units shift at every stage.

Quantity	Units	How obtained
acceleration	meters per second squared	given
velocity	meters per second	integrate acceleration once
position	meters	integrate velocity once
displacement	meters	subtract two positions

If the final answer carries the units of velocity, one integration is missing. That check catches the majority of chained-integration errors before any arithmetic is verified.

Common Misconception Check

Myth 1: “One constant serves the whole chain.” Truth: Each integration introduces its own constant, and each needs its own condition.
Myth 2: “You can integrate $a(t)$ twice directly to get $s(t)$ .” Truth: You can, but the result carries two unknown constants $C_1 t + C_2$ , and both must be fixed by conditions.
Myth 3: “The answer to part (a) can be reused in part (b).” Truth: Only when the integral is genuinely the same. Different limits mean a different number.

Topic-Specific Study Method

Label Every Stage

Exam disguise	What you see on the page	What your brain must translate	Required execution
“given $a(t)$ and $v(0)$ , find $s(3)$ ”	two conditions, one target	two-stage chain	integrate twice, anchor twice
“find the change in velocity on $[0,4]$ ”	one condition only	single integral	integrate $a$ on $[0,4]$
“the particle is at rest at $t = 0$ ”	a hidden anchor	$v_0 = 0$	use it before the second stage
“find the area and the average value”	two separate tasks	parallel integrals	compute each from the stem
“how far has it travelled by $t = 5$ ”	a chained quantity	distance, not position	chain, then split at zeros

① Write the two stage equations before substituting anything. The structure prevents the missing-anchor error.

② Solve for each constant the moment it appears. A constant carried forward becomes two unknowns later.

③ Check the units at the end of the chain. Meters is right; meters per second means a missing stage.

Training & Error Diagnosis

Standard Execution:

Velocity. $v(t) = 2 + \int_0^t 6u\,du = 2 + 3t^{2}$ .

Position. $s(t) = 5 + \int_0^t \left(2 + 3u^{2}\right)du = 5 + 2t + t^{3}$ .

Evaluate. $s(2) = 5 + 4 + 8 = 17$ meters.

Check. The velocity grows quickly, so a position well beyond the initial $5$ meters is expected.

Solution: Integrating gives $s(t) = t^{3} - 6t^{2} + C$ , and $s(1) = 1 - 6 + C = 4$ forces $C = 9$ . Then $s(4) = 64 - 96 + 9 = -23$ , so the displacement on $[1,4]$ is $-27$ meters. For distance, note $v = 3t(t-4)$ , so $v = 0$ at $t = 4$ only, and $v < 0$ on $(1,4)$ . The distance is $\left\lvert \int_1^4 v\,dt \right\rvert = 27$ meters, which equals the magnitude of the displacement here because the velocity never reverses inside the open interval.

Common Traps & Failure Signals:

One constant (Signal: $s(t) = t^{3} + C$ with no $C t$ term from the velocity family): each stage needs its own anchor.
Anchor unused (Signal: a position answer with an unevaluated $C$ ): the condition in the stem exists to be used.
Part contamination (Signal: an area value reused inside a volume integral): each part restarts from the stem.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Constant Dropped: You integrated twice with only one constant of integration.
- Anchor Skipped: You carried a symbolic $C$ to the end of the problem and never used the condition the stem supplied, so the answer described a whole family of motions instead of the one the question named.
- Stage Blurred: You treated the velocity integral and the position integral as the same computation, which produces a number with velocity units where meters were required.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; Chained integration appears regularly in multiple choice, usually as an acceleration given with two conditions, but the released FRQs stop at one integration per part
10-Year Total Count: 0 of 9 Exams · 0 Parts; Motion FRQ parts since 2017 supply the velocity directly, which is why the chained version stays in the multiple-choice pool
FRQ Years Appeared: None in Released FRQs

Practice Examples

MCQ 1

One Stage

A particle has acceleration $a(t) = 6t$ and initial velocity $v(0) = 2$ . Its velocity at $t = 3$ is

MCQ 2

Two Stages

With $a(t) = 6t$ , $v(0) = 2$ , and $s(0) = 5$ , the position at $t = 2$ is

MCQ 3

Which Anchor Applies

A particle has $v(1) = 3$ and $s(1) = 0$ . Which expression gives $s(t)$ ?

FRQ 4

A Two-Stage Motion Problem

A particle moves along a line with acceleration $a(t) = 6t$ meters per second squared for $0 \le t \le 3$ . At $t = 0$ its velocity is $2$ meters per second and its position is $5$ meters.

(a) Find $v(t)$ .

(b) Find $s(t)$ and evaluate $s(3)$ .

(c) Find the total distance travelled on $0 \le t \le 3$ and explain why it equals the displacement here.

Cheatsheet

Core Reference

	Action	Failure if skipped
1	list the stages and the anchor for each	a constant is missing
2	integrate stage one	the velocity family is unavailable
3	solve for the first constant immediately	two unknowns accumulate
4	integrate stage two	the position family is unavailable
5	solve for the second constant	the answer stays symbolic
6	check units at the end	a missing stage survives

Sentences That Score

TOPIC REVIEW • NEXT STEP IN YOUR LEARNING LOOP

Ready to check your understanding?

Strengthen this topic with AP-style practice, or continue to next topic.

Start Topic Practice Next Topic: 8.11 Volume = $\int_{a}^{b} A(x) dx$, where $A(x)$ is area of cross-section

Unit 8Multiple Integrals

Concept 4.97Built for May 2027 AP Exam

Multiple Integrals

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

AP Calculus AB tests this skill through the representations and reasoning described below.

What you’ll see: A motion problem that gives acceleration and asks for velocity and position, or a multi-part question where each part carries its own integral with different limits.
What you’ll do: Integrate once per stage, apply the given anchor at each stage, and label the result. For multi-part questions, start each part from the stem rather than from your previous answer.
Memory hook: One integral per stage, one constant per integral. Chained, not stacked.
Use it on the exam: For MCQs: Watch for a question that gives $a(t)$ and asks for the change in position. That is two integrations, and both constants need values. For FRQs: Write the constant as a named symbol, then solve for it from the given condition. Dropping it costs the anchor point in almost every released rubric.

Concept Review

Chained Integration

Figure 1 — Two integrations in sequence, each introducing its own constant.

Given $a(t)$ with $v(t_0) = v_0$ and $s(t_0) = s_0$ :

$v(t) = v_0 + \int_{t_0}^{t} a(u)\,du \qquad\qquad s(t) = s_0 + \int_{t_0}^{t} v(u)\,du$

Three small panels left to right: acceleration 6t, velocity 2 plus 3t squared, position 5 plus 2t plus t cubed, with arrows between them labelled by integral signs

Why the Constants Cannot Be Dropped

Stage	Result of integrating	Anchor needed	What the anchor fixes
first	velocity family	$v(t_0) = v_0$	vertical shift of $v$
second	position family	$s(t_0) = s_0$	vertical shift of $s$
third (rare)	accumulated displacement	none beyond the above	nothing new

Several Integrals in One Question

Multi-part FRQs often need several unrelated integrals. Part (a) may ask for an average value, part (b) for an area, and part (c) for a volume. Each begins from the stem, and each has its own limits.

Chained Versus Stacked

Two integrals can be related in two different ways, and the wording tells you which.

Relationship	Signal in the question	Structure
chained	“find the velocity, then the position”	output of one integral feeds the next integrand
stacked (split)	“the region has two parts”	same integrand class, different limits, added
parallel	“find the area and the average value”	independent integrals, no shared data

Topic 8.9 covered the split case. This topic covers the chained case, and the parallel case is really just several topics appearing in one question.

Units Through the Chain

Each integration multiplies by one power of time, so units shift at every stage.

Quantity	Units	How obtained
acceleration	meters per second squared	given
velocity	meters per second	integrate acceleration once
position	meters	integrate velocity once
displacement	meters	subtract two positions

If the final answer carries the units of velocity, one integration is missing. That check catches the majority of chained-integration errors before any arithmetic is verified.

Common Misconception Check

Myth 1: “One constant serves the whole chain.” Truth: Each integration introduces its own constant, and each needs its own condition.
Myth 2: “You can integrate $a(t)$ twice directly to get $s(t)$ .” Truth: You can, but the result carries two unknown constants $C_1 t + C_2$ , and both must be fixed by conditions.
Myth 3: “The answer to part (a) can be reused in part (b).” Truth: Only when the integral is genuinely the same. Different limits mean a different number.

Topic-Specific Study Method

Label Every Stage

Exam disguise	What you see on the page	What your brain must translate	Required execution
“given $a(t)$ and $v(0)$ , find $s(3)$ ”	two conditions, one target	two-stage chain	integrate twice, anchor twice
“find the change in velocity on $[0,4]$ ”	one condition only	single integral	integrate $a$ on $[0,4]$
“the particle is at rest at $t = 0$ ”	a hidden anchor	$v_0 = 0$	use it before the second stage
“find the area and the average value”	two separate tasks	parallel integrals	compute each from the stem
“how far has it travelled by $t = 5$ ”	a chained quantity	distance, not position	chain, then split at zeros

① Write the two stage equations before substituting anything. The structure prevents the missing-anchor error.

② Solve for each constant the moment it appears. A constant carried forward becomes two unknowns later.

③ Check the units at the end of the chain. Meters is right; meters per second means a missing stage.

Training & Error Diagnosis

Standard Execution:

Velocity. $v(t) = 2 + \int_0^t 6u\,du = 2 + 3t^{2}$ .

Position. $s(t) = 5 + \int_0^t \left(2 + 3u^{2}\right)du = 5 + 2t + t^{3}$ .

Evaluate. $s(2) = 5 + 4 + 8 = 17$ meters.

Check. The velocity grows quickly, so a position well beyond the initial $5$ meters is expected.

Solution: Integrating gives $s(t) = t^{3} - 6t^{2} + C$ , and $s(1) = 1 - 6 + C = 4$ forces $C = 9$ . Then $s(4) = 64 - 96 + 9 = -23$ , so the displacement on $[1,4]$ is $-27$ meters. For distance, note $v = 3t(t-4)$ , so $v = 0$ at $t = 4$ only, and $v < 0$ on $(1,4)$ . The distance is $\left\lvert \int_1^4 v\,dt \right\rvert = 27$ meters, which equals the magnitude of the displacement here because the velocity never reverses inside the open interval.

Common Traps & Failure Signals:

One constant (Signal: $s(t) = t^{3} + C$ with no $C t$ term from the velocity family): each stage needs its own anchor.
Anchor unused (Signal: a position answer with an unevaluated $C$ ): the condition in the stem exists to be used.
Part contamination (Signal: an area value reused inside a volume integral): each part restarts from the stem.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Constant Dropped: You integrated twice with only one constant of integration.
- Anchor Skipped: You carried a symbolic $C$ to the end of the problem and never used the condition the stem supplied, so the answer described a whole family of motions instead of the one the question named.
- Stage Blurred: You treated the velocity integral and the position integral as the same computation, which produces a number with velocity units where meters were required.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; Chained integration appears regularly in multiple choice, usually as an acceleration given with two conditions, but the released FRQs stop at one integration per part
10-Year Total Count: 0 of 9 Exams · 0 Parts; Motion FRQ parts since 2017 supply the velocity directly, which is why the chained version stays in the multiple-choice pool
FRQ Years Appeared: None in Released FRQs

Practice Examples

MCQ 1

One Stage

A particle has acceleration $a(t) = 6t$ and initial velocity $v(0) = 2$ . Its velocity at $t = 3$ is

MCQ 2

Two Stages

With $a(t) = 6t$ , $v(0) = 2$ , and $s(0) = 5$ , the position at $t = 2$ is

MCQ 3

Which Anchor Applies

A particle has $v(1) = 3$ and $s(1) = 0$ . Which expression gives $s(t)$ ?

FRQ 4

A Two-Stage Motion Problem

A particle moves along a line with acceleration $a(t) = 6t$ meters per second squared for $0 \le t \le 3$ . At $t = 0$ its velocity is $2$ meters per second and its position is $5$ meters.

(a) Find $v(t)$ .

(b) Find $s(t)$ and evaluate $s(3)$ .

(c) Find the total distance travelled on $0 \le t \le 3$ and explain why it equals the displacement here.

Cheatsheet

Core Reference

	Action	Failure if skipped
1	list the stages and the anchor for each	a constant is missing
2	integrate stage one	the velocity family is unavailable
3	solve for the first constant immediately	two unknowns accumulate
4	integrate stage two	the position family is unavailable
5	solve for the second constant	the answer stays symbolic
6	check units at the end	a missing stage survives

Sentences That Score

TOPIC REVIEW • NEXT STEP IN YOUR LEARNING LOOP

Ready to check your understanding?

Strengthen this topic with AP-style practice, or continue to next topic.

Start Topic Practice Next Topic: 8.11 Volume = $\int_{a}^{b} A(x) dx$, where $A(x)$ is area of cross-section