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Most students memorise the table of moments of inertia and hope the question uses a standard shape. Then a problem gives a rod whose density increases along its length, and the table is no help. This post shows how each standard value comes from one integral, works through the rod, ring, disc and solid sphere, and finishes with a non-uniform rod of the kind JEE likes.
Do you need to know the derivations?
NCERT's rotational motion chapter gives Table 6.1 of standard values and says plainly that "the derivations of these expressions are beyond the scope of this textbook". The entrance exams go further. The JEE (Advanced) 2026 brochure lists "moment of inertia of uniform bodies with simple geometrical shapes", and the JEE (Main) 2026 syllabus lists "values of moments of inertia for simple geometrical objects". Check the bulletin for your exam year. Knowing the integral lets you handle shapes and densities no table covers.
What is the recipe?
OpenStax University Physics builds the formula from the sum Σmr² for point masses: for a continuous body it becomes I = ∫r² dm. Four steps:
Choose a piece whose mass is all at the same distance from the axis.
Write its mass dm using the density: λ dx for a rod, σ dA for a sheet, ρ dV for a solid.
Write its distance r from the axis.
Set the limits so the pieces cover the whole body once.
This set-up is where most students stumble. Nguyen and Rebello (2011) interviewed 15 students in calculus-based physics. Most recognised that a problem needed an integral but failed to set it up, often because they did not understand what the small term stood for. Saying step 1 aloud fixes much of this. Our guide to the calculus you need for JEE physics gives more practice in building integrals.
Rod and ring
Rod about one end. Mass M, length L, so λ = M/L. A piece of length dx at distance x has dm = λ dx. Then I = ∫ from 0 to L of x²λ dx = λL³/3 = ML²/3. OpenStax does the same integral from −L/2 to L/2 to get ML²/12 about the centre.
Ring about its axis. Every bit of the ring is at distance R from the axis, so no integration is needed: I = R²∫dm = MR².
Disc
Slice a uniform disc of mass M and radius R into thin rings of radius r and width dr. The surface density is M/(πR²) and the ring's area is 2πr dr, so dm = (2M/R²) r dr. Each ring contributes r² dm:
I = ∫ from 0 to R of (2M/R²) r³ dr = (2M/R²)(R⁴/4) = MR²/2.
This matches row 4 of NCERT's table.
Solid sphere
Slice the sphere into thin discs perpendicular to the axis. A disc at height z has radius y, with y² = R² − z², and thickness dz. Its mass is dm = ρπ(R² − z²) dz, and from the disc result its moment of inertia is ½ dm y².
I = ½ρπ ∫ from −R to R of (R² − z²)² dz.
Expanding, ∫(R⁴ − 2R²z² + z⁴) dz from −R to R = 2R⁵(1 − 2/3 + 1/5) = 16R⁵/15. So I = ½ρπ × 16R⁵/15 = 8πρR⁵/15. With ρ = 3M/(4πR³), this becomes I = 2MR²/5, as in NCERT's table.
Worked example: a non-uniform rod
A rod of length 1.0 m has linear density λ = 6x kg/m, where x is measured from end A. Find its mass, centre of mass, and moment of inertia about A and about its centre of mass.
Mass: M = ∫ from 0 to 1 of 6x dx = 3x² from 0 to 1 = 3.0 kg.
Centre of mass: ∫x dm = ∫6x² dx = 2.0, so x_cm = 2.0/3.0 ≈ 0.67 m from A.
About A: I = ∫x² × 6x dx = (6/4)x⁴ from 0 to 1 = 1.5 kg m².
About the centre of mass, by the parallel axis theorem: 1.5 − 3.0 × (2/3)² = 1.5 − 1.33 ≈ 0.17 kg m².
Sense check: a uniform 3.0 kg rod of the same length has ML²/3 = 1.0 kg m² about its end. Our rod is heavier towards the far end, so its value about A is larger. For more on step 4, see our post on the parallel and perpendicular axis theorems.
Which mistakes cost marks?
Measuring r from the centre of a sphere instead of from the axis.
Forgetting that a disc's dm grows with r: it is (2M/R²) r dr, not (M/R) dr.
Wrong limits: −L/2 to L/2 about the centre, 0 to L about an end.
For non-uniform bodies, using the parallel axis theorem from the geometric centre rather than the centre of mass.
A practice routine
Derive the rod, disc and sphere results on a blank page this week, saying step 1 aloud each time.
Derive a thin spherical shell (answer 2MR²/3) by slicing it into rings.
Do two non-uniform rods, one with λ = kx and one with λ = k(L − x), and compare.
Check each result against NCERT's table or a dimensional check: every answer must be mass times length squared.
If a step in a derivation still feels like a jump, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, each with its own lesson.
Start with the disc tonight: write dm in words before you write it in symbols.
