Intuition The one-line ideaA triple integral ∭ E f d V \iiint_E f\,dV ∭ E f d V adds up the value of f f f d V dV d V E E E (1) chop the region into boxes, (2) pick coordinates that make the boxes and the region easy to describe. ==Cartesian= straight walls=, ==cylindrical= round in one plane + straight axis=, ==spherical= round in all directions=.
Definition Triple integral (Riemann definition)Partition a solid E E E N N N Δ V k \Delta V_k Δ V k ( x k ∗ , y k ∗ , z k ∗ ) (x_k^*,y_k^*,z_k^*) ( x k ∗ , y k ∗ , z k ∗ ) ∭ E f ( x , y , z ) d V = lim N → ∞ ∑ k = 1 N f ( x k ∗ , y k ∗ , z k ∗ ) Δ V k . \iiint_E f(x,y,z)\,dV \;=\; \lim_{N\to\infty}\sum_{k=1}^{N} f(x_k^*,y_k^*,z_k^*)\,\Delta V_k. ∭ E f ( x , y , z ) d V = lim N → ∞ ∑ k = 1 N f ( x k ∗ , y k ∗ , z k ∗ ) Δ V k .
WHAT it computes:
If f = 1 f=1 f = 1 ∭ E d V = Volume ( E ) \iiint_E dV = \text{Volume}(E) ∭ E d V = Volume ( E )
If f = ρ ( x , y , z ) f=\rho(x,y,z) f = ρ ( x , y , z ) ∭ E ρ d V = Mass \iiint_E \rho\,dV = \text{Mass} ∭ E ρ d V = Mass
General f f f
WHY it converges to something useful: each term f ⋅ Δ V k f\cdot\Delta V_k f ⋅ Δ V k
HOW to set the limits (the only skill that matters):
Innermost (z z z freeze x , y x,y x , y enters E E E z = z 1 ( x , y ) z=z_1(x,y) z = z 1 ( x , y ) exits at z = z 2 ( x , y ) z=z_2(x,y) z = z 2 ( x , y ) x , y x,y x , y Middle (y y y project the shadow of E E E x y xy x y y y y y 1 ( x ) , y 2 ( x ) y_1(x),y_2(x) y 1 ( x ) , y 2 ( x ) Outer (x x x the constant range [ x 1 , x 2 ] [x_1,x_2] [ x 1 , x 2 ]
Common mistake "Why are my outer limits functions of the inner variable?"It feels natural to write ∫ 0 x … d x \int_0^x \dots dx ∫ 0 x … d x outer limits must be constants , and each limit may only depend on variables outside it. Reason: you integrate inside-out; by the time you hit the outer integral, all other variables are gone, so a number is all that's left.
Worked example Volume of the tetrahedronx , y , z ≥ 0 , x + y + z ≤ 1 x,y,z\ge 0,\; x+y+z\le 1 x , y , z ≥ 0 , x + y + z ≤ 1 Step 1 — inner z z z Arrow up: enters z = 0 z=0 z = 0 z = 1 − x − y z=1-x-y z = 1 − x − y Why? The slanted face is x + y + z = 1 x+y+z=1 x + y + z = 1 ∫ 0 1 − x − y d z = 1 − x − y \int_0^{1-x-y} dz = 1-x-y ∫ 0 1 − x − y d z = 1 − x − y Step 2 — middle y y y Shadow on x y xy x y x , y ≥ 0 , x + y ≤ 1 x,y\ge0,\,x+y\le1 x , y ≥ 0 , x + y ≤ 1 y y y 0 → 1 − x 0\to 1-x 0 → 1 − x Why? Set z = 0 z=0 z = 0 ∫ 0 1 − x ( 1 − x − y ) d y = [ ( 1 − x ) y − y 2 2 ] 0 1 − x = ( 1 − x ) 2 2 \int_0^{1-x}(1-x-y)\,dy = \Big[(1-x)y-\tfrac{y^2}{2}\Big]_0^{1-x}=\tfrac{(1-x)^2}{2} ∫ 0 1 − x ( 1 − x − y ) d y = [ ( 1 − x ) y − 2 y 2 ] 0 1 − x = 2 ( 1 − x ) 2 Step 3 — outer x x x x : 0 → 1 x:0\to1 x : 0 → 1 ∫ 0 1 ( 1 − x ) 2 2 d x = 1 2 ⋅ 1 3 = 1 6 \int_0^1 \tfrac{(1-x)^2}{2}\,dx = \tfrac12\cdot\tfrac{1}{3}=\boxed{\tfrac16} ∫ 0 1 2 ( 1 − x ) 2 d x = 2 1 ⋅ 3 1 = 6 1 1 6 \tfrac16 6 1
d V dV d V ( x , y , z ) (x,y,z) ( x , y , z ) ( u , v , w ) (u,v,w) ( u , v , w ) d u d v d w du\,dv\,dw d u d v d w warped parallelepiped. Its volume = (volume of the box) × (local stretch). That stretch is ∣ J ∣ |J| ∣ J ∣ Jacobian determinant .
x = r cos θ , y = r sin θ , z = z , r ≥ 0 , 0 ≤ θ < 2 π . x=r\cos\theta,\quad y=r\sin\theta,\quad z=z,\qquad r\ge0,\ 0\le\theta<2\pi. x = r cos θ , y = r sin θ , z = z , r ≥ 0 , 0 ≤ θ < 2 π .
Use it when: the region or integrand has rotational symmetry about an axis (cylinders, cones, paraboloids).
Worked example Volume of cylinderr ≤ a r\le a r ≤ a 0 ≤ z ≤ h 0\le z\le h 0 ≤ z ≤ h
=\int_0^{2\pi}\!\!d\theta\int_0^a r\,dr\int_0^h dz
=2\pi\cdot\tfrac{a^2}{2}\cdot h=\pi a^2 h.\ \checkmark$$
**Why split into a product?** Limits are all constants and the integrand factors, so Fubini lets us multiply 1D integrals.
x = ρ sin ϕ cos θ , y = ρ sin ϕ sin θ , z = ρ cos ϕ , x=\rho\sin\phi\cos\theta,\quad y=\rho\sin\phi\sin\theta,\quad z=\rho\cos\phi, x = ρ sin ϕ cos θ , y = ρ sin ϕ sin θ , z = ρ cos ϕ , ρ ≥ 0 \rho\ge0 ρ ≥ 0 0 ≤ ϕ ≤ π 0\le\phi\le\pi 0 ≤ ϕ ≤ π + z +z + z 0 ≤ θ < 2 π 0\le\theta<2\pi 0 ≤ θ < 2 π z z z
Common mistake Forgettingsin ϕ \sin\phi sin ϕ sin θ \sin\theta sin θ feels like by analogy with cylindrical you only need one extra factor. Fix: there are two angular arc-lengths in spherical. The latitude circles shrink to points at the poles (ϕ = 0 , π \phi=0,\pi ϕ = 0 , π sin ϕ → 0 \sin\phi\to0 sin ϕ → 0 ϕ \phi ϕ θ \theta θ
Worked example Volume of a ball of radiusa a a
=\underbrace{\int_0^{2\pi}d\theta}_{2\pi}\;\underbrace{\int_0^\pi\sin\phi\,d\phi}_{2}\;\underbrace{\int_0^a\rho^2 d\rho}_{a^3/3}
=2\pi\cdot2\cdot\tfrac{a^3}{3}=\tfrac{4}{3}\pi a^3.\ \checkmark$$
**Why this is the 80/20 example:** it simultaneously checks all three factors and lands on the famous $\frac43\pi a^3$.
∭ E z d V \iiint_E z\,dV ∭ E z d V ρ ≤ a , z ≥ 0 \rho\le a,\ z\ge0 ρ ≤ a , z ≥ 0 z = ρ cos ϕ z=\rho\cos\phi z = ρ cos ϕ z ≥ 0 ⇒ ϕ ∈ [ 0 , π / 2 ] z\ge0\Rightarrow \phi\in[0,\pi/2] z ≥ 0 ⇒ ϕ ∈ [ 0 , π /2 ] ∫ 0 2 π ∫ 0 π / 2 ∫ 0 a ( ρ cos ϕ ) ρ 2 sin ϕ d ρ d ϕ d θ \int_0^{2\pi}\!\!\int_0^{\pi/2}\!\!\int_0^a (\rho\cos\phi)\,\rho^2\sin\phi\,d\rho\,d\phi\,d\theta ∫ 0 2 π ∫ 0 π /2 ∫ 0 a ( ρ cos ϕ ) ρ 2 sin ϕ d ρ d ϕ d θ Step why: integrand becomes ρ 3 cos ϕ sin ϕ \rho^3\cos\phi\sin\phi ρ 3 cos ϕ sin ϕ
=2\pi\cdot\tfrac12\cdot\tfrac{a^4}{4}=\tfrac{\pi a^4}{4}.$$
Used $\int_0^{\pi/2}\sin\phi\cos\phi\,d\phi=\tfrac12$.
Region / symmetry
Best coords
d V dV d V
Box, planes, slanted faces
Cartesian
d x d y d z dx\,dy\,dz d x d y d z
Circle/cylinder/cone/paraboloid about an axis
Cylindrical
r d r d θ d z r\,dr\,d\theta\,dz r d r d θ d z
Sphere, ball, x 2 + y 2 + z 2 x^2+y^2+z^2 x 2 + y 2 + z 2
Spherical
ρ 2 sin ϕ d ρ d ϕ d θ \rho^2\sin\phi\,d\rho\,d\phi\,d\theta ρ 2 sin ϕ d ρ d ϕ d θ
Common mistake Using Cartesian on a sphereIt feels safe (no Jacobian to remember). But limits like z = ± a 2 − x 2 − y 2 z=\pm\sqrt{a^2-x^2-y^2} z = ± a 2 − x 2 − y 2 x 2 + y 2 + z 2 x^2+y^2+z^2 x 2 + y 2 + z 2 ρ = a \rho=a ρ = a
Recall Test yourself (hide answers)
What is d V dV d V
Where does the r r r
Why ρ 2 sin ϕ \rho^2\sin\phi ρ 2 sin ϕ ρ 2 \rho^2 ρ 2
Why must outer limits be constants?
Recall Feynman: explain to a 12-year-oldImagine filling a shape with tiny sugar cubes and counting how much "stuff" (sweetness) is in each cube. Add them all up — that's a triple integral. If the shape is a box, use square cubes (Cartesian). If it's a tin can, slice it into curved tiles that fan out — the tiles far from the center are wider , so we multiply by how far out they are (r r r shorter near the top and bottom poles, so we multiply by sin ϕ \sin\phi sin ϕ
d V dV d V "Cylinders carry ONE r r r ρ 2 sin ϕ \rho^2\sin\phi ρ 2 sin ϕ
Spherical mantra: "rho-squared sine-phi" — two ρ \rho ρ sin ϕ \sin\phi sin ϕ kills the poles .
What is d V dV d V What is d V dV d V r d r d θ d z r\,dr\,d\theta\,dz r d r d θ d z What is d V dV d V ρ 2 sin ϕ d ρ d ϕ d θ \rho^2\sin\phi\,d\rho\,d\phi\,d\theta ρ 2 sin ϕ d ρ d ϕ d θ Why does the cylindrical Jacobian equal r r r A polar pixel has area
r d r d θ r\,dr\,d\theta r d r d θ (arc length
r d θ r\,d\theta r d θ times width
d r dr d r ); times
d z dz d z gives volume.
Why does spherical d V dV d V sin ϕ \sin\phi sin ϕ The latitude-circle arc has radius
ρ sin ϕ \rho\sin\phi ρ sin ϕ , so its arc length is
ρ sin ϕ d θ \rho\sin\phi\,d\theta ρ sin ϕ d θ ; this shrinks to 0 at the poles.
In spherical coords, what are the ranges of ρ , ϕ , θ \rho,\phi,\theta ρ , ϕ , θ ρ ≥ 0 \rho\ge0 ρ ≥ 0 ,
0 ≤ ϕ ≤ π 0\le\phi\le\pi 0 ≤ ϕ ≤ π (polar, from
+ z +z + z ),
0 ≤ θ < 2 π 0\le\theta<2\pi 0 ≤ θ < 2 π (azimuth).
Volume of a ball of radius a a a ∫ 0 2 π ∫ 0 π ∫ 0 a ρ 2 sin ϕ d ρ d ϕ d θ = 4 3 π a 3 \int_0^{2\pi}\!\int_0^\pi\!\int_0^a \rho^2\sin\phi\,d\rho\,d\phi\,d\theta=\frac43\pi a^3 ∫ 0 2 π ∫ 0 π ∫ 0 a ρ 2 sin ϕ d ρ d ϕ d θ = 3 4 π a 3 .
Volume of cylinder radius a a a h h h ∫ 0 2 π ∫ 0 a ∫ 0 h r d z d r d θ = π a 2 h \int_0^{2\pi}\!\int_0^a\!\int_0^h r\,dz\,dr\,d\theta=\pi a^2 h ∫ 0 2 π ∫ 0 a ∫ 0 h r d z d r d θ = π a 2 h .
How do you choose the innermost (z z z Freeze
x , y x,y x , y , shoot a vertical arrow: it enters at
z 1 ( x , y ) z_1(x,y) z 1 ( x , y ) and exits at
z 2 ( x , y ) z_2(x,y) z 2 ( x , y ) .
Why must the outermost integration limits be constants? You integrate inside-out; by the outer step all other variables are gone, so only numbers remain.
When should you prefer spherical coordinates? When the region is a sphere/ball or the integrand contains
x 2 + y 2 + z 2 = ρ 2 x^2+y^2+z^2=\rho^2 x 2 + y 2 + z 2 = ρ 2 .
General change-of-variables formula for d V dV d V d V = ∣ det ∂ ( x , y , z ) / ∂ ( u , v , w ) ∣ d u d v d w dV=\left|\det\partial(x,y,z)/\partial(u,v,w)\right|\,du\,dv\,dw d V = ∣ det ∂ ( x , y , z ) / ∂ ( u , v , w ) ∣ d u d v d w .
What does ∭ E 1 d V \iiint_E 1\,dV ∭ E 1 d V Cartesian-to-spherical: what is z z z z = ρ cos ϕ z=\rho\cos\phi z = ρ cos ϕ .
Triple integral iiint f dV
Vertical arrow enter exit
Intuition Hinglish mein samjhoTriple integral ka matlab simple hai: kisi 3D shape ke andar har chhote cube ka "amount" (jaise density ya bas volume) nikaalo aur sabko jod do. Jaise 1D mein area aur 2D mein double integral, waise hi 3D mein hum poore solid ko tiny boxes mein todte hain, har box ka contribution f ⋅ d V f\cdot dV f ⋅ d V f = 1 f=1 f = 1 volume mil jaata hai.
Asli skill hai sahi coordinate choose karna. Agar shape ke walls seedhe hain (box, slanted planes) to Cartesian best — d V = d x d y d z dV=dx\,dy\,dz d V = d x d y d z z z z constant hone chahiye, yeh galti sabse zyada hoti hai. Agar shape kisi axis ke around round hai (cylinder, cone) to cylindrical use karo, aur yaad rakho ek extra r r r
