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Calculus

Volume of cylinder, cone and sphere by integration

The calculation of solid volumes via definite integration relies on the slicing method (Disk and Washer methods) rooted in Riemann sums. This topic covers these surface-of-revolution formulas.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade
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Definition and statement

Disk Method for Volume of Revolution
Let $f$ be continuous and non-negative on $[a, b]$. The volume $V$ of the solid generated by revolving the region bounded by $y = f(x)$, $y = 0$, $x = a$, and $x = b$ about the $x$-axis is given by $$V = \pi \int_{a}^{b} [f(x)]^2 \, dx$$.

📐 Geometric Interpretation

The solid is approximated by an infinite stack of thin cylindrical disks of radius $r = f(x)$ and infinitesimal thickness $\Delta x$. As $\Delta x \to 0$, the Riemann sum converges to the exact definite integral of the cross-sectional area $A(x) = \pi [f(x)]^2$.

⚙️ Algebraic & Structural Meaning

Translates a geometric limit problem into a standard single-variable Riemann integrable function, enabling exact evaluation via the Fundamental Theorem of Calculus.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition Domain / Interval Necessity What Fails If Condition Dropped
Continuity of generator function $f(x) [a, b] (Closed interval) Strictly Mandatory / Necessary If $f(x)$ is discontinuous, Riemann integrability of $[f(x)]^2$ may fail, rendering the area and volume undefined.
Non-negativity $f(x) \ge 0$ [a, b] (Closed interval) Strictly Mandatory for radius interpretation If $f(x)$ oscillates between positive and negative values without squaring, the net integral would cancel out, misrepresenting physical volume.
Rectifiability / Smoothness of boundaries (a, b) (Open interval) Essential for well-behaved solids Pathological boundaries (like non-measurable sets) prevent standard Riemann integration.

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Direct Volume of a Right Circular Cone

$$V = \frac{1}{3} \pi r^2 h$$

Rule: Generated by revolving the line $y = \frac{r}{h}x$ from $x = 0$ to $x = h$. Memorize the $\frac{1}{3}$ factor to skip integration in standard cone problems.

Application: Use this result in the stated type of problem.
02

Direct Volume of a Sphere

$$V = \frac{4}{3} \pi R^3$$

Rule: Generated by revolving the semicircle $y = \sqrt{R^2 - x^2}$ from $x = -R$ to $x = R$. Use symmetry: $2 \pi \int_{0}^{R} (R^2 - x^2) \, dx$.

03

Volume of a Cylinder

$$V = \pi r^2 h$$

Rule: Generated by revolving the constant function $y = r$ from $x = 0$ to $x = h$.

Application: Base case tested in foundational calculus sets.
04

Pappus's Centroid Theorem (First Theorem)

$$V = 2\pi \bar{y} A$$

Rule: The volume of a solid of revolution is equal to the product of the area $A$ of the generating region and the distance $2\pi \bar{y}$ traveled by its geometric centroid during a full rotation.

05

Washer Method Quick Setup

$$V = \pi \int_{a}^{b} \left( [R(x)]^2 - [r(x)]^2 \right) \, dx$$

Rule: For hollow solids (outer radius $R(x)$, inner radius $r(x)$), square outer and inner radii separately before subtracting.

Application: Appears in bounded region rotation problems (e.g., between a parabola and a line).
06

Parabolic Segment Volume (Archimedes' Formula)

$$V = \frac{1}{2} \text{(Cylinder Volume)}$$ or specific paraboloid bounds

Rule: Volume of a paraboloid generated by $y^2 = 4ax$ up to $x = h$ is $V = 2 \pi a h^2$.

Application: Direct time-saver for UP PGT calculus questions.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Squaring the Function vs. Squaring the Integral

Common Pitfall: Candidates mistakenly compute $\pi \left( \int f(x) \, dx \right)^2$ instead of the correct $\pi \int [f(x)]^2 \, dx$.

Counterexample: $f(x) = x$ on $[0, 1]$: $\left(\int_0^1 x dx\right)^2 = \frac{1}{4}$, whereas $\int_0^1 x^2 dx = \frac{1}{3}$.

Why it fails: The volume is the sum of disk volumes ($\pi r^2 \Delta x$), not the square of the accumulated area.

🛡️ Exam Defense: Always square the radius function inside the integrand before integrating.
⚠️ COMMON ERROR #02

Axis of Revolution Not the Coordinate Axis

Common Pitfall: Integrating with respect to $x$ when revolving about a line parallel to the $y$-axis, or failing to adjust radius for rotation about $y = k$ or $x = h$.

Counterexample: Revolving $y = \sqrt{x}$ about $y = 2$.

Why it fails: The radius is not $\sqrt{x}$, but $2 - \sqrt{x}$. Using $y$ directly leads to complete calculation failure.

🛡️ Exam Defense: Always draw a quick sketch and measure the perpendicular distance from the generating curve to the axis of rotation to establish the correct radius.

Comparison

Concept / TheoremHypotheses / AssumptionsKey Formula / OutcomeGeometric MeaningExam Identification Clue
Disk MethodContinuous $f(x) \ge 0$ on $[a,b]$$V = \pi \int_a^b [f(x)]^2 \, dx$Sticking solid disks perpendicular to axisRotation about coordinate axis with no gap
Washer MethodContinuous outer $R(x)$, inner $r(x)$$V = \pi \int_a^b (R^2 - r^2) \, dx$Stacking annular rings (hollow disks)Region bounded between two curves rotated
Shell MethodContinuous $f(x)$ on $[a,b]$, $0 \le a < b$$V = 2\pi \int_a^b x f(x) \, dx$Nested cylindrical shellsRotation about $y$-axis using function of $x$
Pappus's TheoremBounded planar area $A$, centroid distance $\bar{y}$$V = 2\pi \bar{y} A$Centroid path multiplied by areaSymmetric geometric shapes like circles/triangles rotated

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the volume of the solid generated by revolving the region bounded by $y = \sqrt{x}$, $y = 0$, and $x = 4$ about the $x$-axis.

(A) $4\pi$ cubic units(B) $8\pi$ cubic units(C) $16\pi$ cubic units(D) $2\pi$ cubic units
✓ Correct Answer: (B)
📚 Method A: Formal Textbook Derivation
Step 1: Identify radius $r(x) = \sqrt{x}$ and interval $[0, 4]$. Step 2: Apply Disk formula $V = \pi \int_{0}^{4} (\sqrt{x})^2 \, dx$. Step 3: Evaluate integral $\pi \int_{0}^{4} x \, dx = \pi \left[ \frac{x^2}{2} \right]_0^4 = \frac{16\pi}{2} = 8\pi$.
⚡ Method B: 25-Second Exam Shortcut
Recognize the parabola $y^2 = x$ up to $x=4$; volume of paraboloid slice $V = \frac{1}{2} \pi r^2 h = \frac{1}{2} \pi (2)^2 (4) = 8\pi$.
💡 Tip: Direct substitution test in UP TGT; ensure limits are properly squared.
CANONICAL PROBLEM 02 Dual-Method Breakdown

Find the volume of a sphere of radius $R$ generated by revolving the circle $x^2 + y^2 = R^2$ about the $x$-axis.

(A) $\frac{4}{3}\pi R^3$(B) $\frac{2}{3}\pi R^3$(C) $2\pi R^3$(D) $\frac{4}{5}\pi R^3$
✓ Correct Answer: (A)
📚 Method A: Formal Textbook Derivation
Step 1: Upper semicircle equation is $y = \sqrt{R^2 - x^2}$. Step 2: Set up integral $V = 2\pi \int_{0}^{R} (R^2 - x^2) \, dx$. Step 3: Integrate term by term: $2\pi \left[ R^2 x - \frac{x^3}{3} \right]_0^R = 2\pi \left( R^3 - \frac{R^3}{3} \right) = \frac{4}{3}\pi R^3$.
⚡ Method B: 25-Second Exam Shortcut
Standard derived result; recall $\frac{4}{3}\pi R^3$ instantly for time management.
💡 Tip: Memorize standard derivation steps as direct theoretical questions appear in PGT exams.

Practice questions

Choose an answer to check it and see the solution:

Q1
MCQ 1: The volume of the solid generated by revolving the region bounded by $y = x^2$ and $y = x$ about the $x$-axis is:
A $\frac{\pi}{15}$
B $\frac{2\pi}{15}$
C $\frac{\pi}{3}$
D $\frac{4\pi}{15}$
Derivation & Analysis:

Intersection points are $x = 0$ and $x = 1$. Using the washer method: $V = \pi \int_{0}^{1} \left(x^2 - (x^2)^2\right) dx = \pi \int_{0}^{1} (x^2 - x^4) dx = \pi \left[ \frac{x^3}{3} - \frac{x^5}{5} \right]_0^1 = \pi \left(\frac{1}{3} - \frac{1}{5}\right) = \frac{2\pi}{15}$.

Q2
MCQ 2: What is the volume of a cone generated by revolving the line $y = 3x$ from $x = 0$ to $x = 2$ about the $x$-axis?
A $6\pi$
B $12\pi$
C $18\pi$
D $24\pi$
Derivation & Analysis:

Using $V = \pi \int_{0}^{2} (3x)^2 dx = \pi \int_{0}^{2} 9x^2 dx = 9\pi \left[ \frac{x^3}{3} \right]_0^2 = 3\pi (8) = 24\pi$. Alternatively, radius $r = 6$, height $h = 2$, $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (36)(2) = 24\pi$.

Q3
MCQ 3: The volume generated by revolving the curve $y = \sin x$ from $x = 0$ to $x = \pi$ about the $x$-axis is:
A $\frac{\pi^2}{2}$
B $\pi^2$
C $\frac{\pi}{2}$
D $2\pi^2$
Derivation & Analysis:

Using $V = \pi \int_{0}^{\pi} \sin^2 x \, dx = \pi \int_{0}^{\pi} \frac{1 - \cos(2x)}{2} dx = \frac{\pi}{2} \left[ x - \frac{\sin(2x)}{2} \right]_0^\pi = \frac{\pi^2}{2}$.

Q4
MCQ 4: The volume of the solid formed by revolving the region bounded by $y = e^x$, $y = 0$, $x = 0$, and $x = 1$ about the $x$-axis is:
A $\pi(e^2 - 1)$
B $\frac{\pi}{2}(e^2 - 1)$
C $\pi e^2$
D $\frac{\pi}{2} e^2$
Derivation & Analysis:

Using $V = \pi \int_{0}^{1} (e^x)^2 dx = \pi \int_{0}^{1} e^{2x} dx = \pi \left[ \frac{e^{2x}}{2} \right]_0^1 = \frac{\pi}{2}(e^2 - 1)$.

Q5
MCQ 5: If the region bounded by $y = \sqrt{x}$ and $y = x$ is rotated about the $x$-axis, the volume is:
A $\frac{\pi}{6}$
B $\frac{\pi}{3}$
C $\frac{\pi}{2}$
D $\frac{\pi}{5}$
Derivation & Analysis:

Intersection points are $0$ and $1$. $V = \pi \int_{0}^{1} \left((\sqrt{x})^2 - x^2\right) dx = \pi \int_{0}^{1} (x - x^2) dx = \pi \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1 = \pi \left(\frac{1}{6}\right) = \frac{\pi}{6}$.

Q6
MCQ 6: The volume of the cylinder generated by revolving $y = r$ from $x = 0$ to $x = h$ about the $x$-axis is given by:
A $\frac{1}{3}\pi r^2 h$
B $\pi r^2 h$
C $2\pi r^2 h$
D $\frac{4}{3}\pi r^2 h$
Derivation & Analysis:

Using $V = \pi \int_{0}^{h} r^2 dx = \pi r^2 [x]_0^h = \pi r^2 h$.

Q7
MCQ 7: Which of the following integrals represents the volume of a sphere of radius $R$ generated by revolving $x^2 + y^2 = R^2$?
A $\pi \int_{-R}^{R} (R^2 - x^2) dx$
B $2\pi \int_{-R}^{R} \sqrt{R^2 - x^2} dx$
C $\pi \int_{-R}^{R} \sqrt{R^2 - x^2} dx$
D $2\pi \int_{0}^{R} \sqrt{R^2 - x^2} dx$
Derivation & Analysis:

By the disk method, $y^2 = R^2 - x^2$, so $V = \pi \int_{-R}^{R} (R^2 - x^2) dx$.

Q8
MCQ 8: Assertion (A): The volume of a solid of revolution can be negative if the curve lies below the $x$-axis. Reason (R): Integration accumulates signed areas.
A Both (A) and (R) are true and (R) is the correct explanation of (A).
B Both (A) and (R) are true but (R) is NOT the correct explanation.
C (A) is true but (R) is false.
D Both (A) and (R) are false.
Derivation & Analysis:

Since the radius function is squared ($[f(x)]^2 \ge 0$), the volume is always non-negative regardless of whether the curve lies above or below the $x$-axis. Thus both Assertion and Reason are false.

Q9
MCQ 9: The volume generated by revolving the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ about the $x$-axis (forming a prolate/oblate spheroid) is:
A $\frac{4}{3}\pi a b^2$
B $\frac{4}{3}\pi a^2 b$
C $\frac{4}{3}\pi a b$
D $\pi a b^2$
Derivation & Analysis:

Using $y^2 = b^2 \left(1 - \frac{x^2}{a^2}\right)$, $V = 2\pi \int_{0}^{a} b^2 \left(1 - \frac{x^2}{a^2}\right) dx = 2\pi b^2 \left[ x - \frac{x^3}{3a^2} \right]_0^a = 2\pi b^2 \left(a - \frac{a}{3}\right) = \frac{4}{3}\pi a b^2$.

Q10
MCQ 10: If a solid is generated by revolving $y = \frac{1}{x}$ from $x = 1$ to $x = \infty$ about the $x$-axis (Gabriel's Horn), its volume is:
A Infinite
B $\pi$ cubic units
C $2\pi$ cubic units
D $\frac{\pi}{2}$
Derivation & Analysis:

Using improper integration: $V = \pi \int_{1}^{\infty} \frac{1}{x^2} dx = \lim_{t \to \infty} \pi \left[ -\frac{1}{x} \right]_1^t = \pi \left(0 - (-1)\right) = \pi$.

Common questions

The disk method sums individual cylindrical disk volumes given by $\pi r^2 h$. Here, the radius $r$ is the height of the curve $f(x)$, and the thickness $h$ is the infinitesimal element $\dx$. Squaring ensures that negative function values contribute positively to physical volume.

The Disk method is used for solid figures generated by a single bounding curve touching or forming the axis of rotation (no internal hollow space). The Washer method is used when there is a gap or hole between the axis of rotation and the generating region, requiring the subtraction of the inner radius squared from the outer radius squared.

Always sketch the region roughly, clearly identify intersection points by equating curves, and verify whether the rotation is about the $x$-axis, $y$-axis, or a shifted line before setting up the integral limits.

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