Intuition The one-line ideaWhen you slice a region into tiny polar boxes (a wedge between two radii and two angles), each box is NOT d r d θ dr\,d\theta d r d θ = = r = = ==r== == r == d A = r d r d θ . dA = r\,dr\,d\theta. d A = r d r d θ . r r r not decoration — it is the area of the little curved box, derived below.
Cartesian boxes d x d y dx\,dy d x d y x x x y y y disk, annulus, sector, or anything with circular symmetry , those rectangles fight the geometry: the boundary x 2 + y 2 = a 2 x^2+y^2=a^2 x 2 + y 2 = a 2 ( r , θ ) (r,\theta) ( r , θ ) constant limits, so the integral becomes trivial.
The price you pay: the area element changes. That is the whole subtopic.
Definition Polar substitutionA point in the plane is written as
x = r cos θ , y = r sin θ , x = r\cos\theta, \qquad y = r\sin\theta, x = r cos θ , y = r sin θ , r ≥ 0 r\ge 0 r ≥ 0 distance from the origin and θ \theta θ x x x r = x 2 + y 2 r=\sqrt{x^2+y^2} r = x 2 + y 2 θ = arctan ( y / x ) \theta=\arctan(y/x) θ = arctan ( y / x )
Derivation First principles: the area of one polar boxTake a tiny region bounded by radii r r r r + d r r+dr r + d r θ \theta θ θ + d θ \theta+d\theta θ + d θ
The radial side has length d r dr d r
The arc side at radius r r r = r d θ = r\,d\theta = r d θ = = = × \times ×
For small increments the box is almost a rectangle with sides d r dr d r r d θ r\,d\theta r d θ d A ≈ ( d r ) ( r d θ ) = r d r d θ . dA \approx (dr)\,(r\,d\theta) = r\,dr\,d\theta. d A ≈ ( d r ) ( r d θ ) = r d r d θ . ( r + d r ) d θ (r+dr)d\theta ( r + d r ) d θ d r d θ ⋅ d r dr\,d\theta\cdot dr d r d θ ⋅ d r Hence the Jacobian factor is r r r ∎
Derivation First principles: the determinantThe general change-of-variables theorem says, for ( x , y ) = T ( u , v ) (x,y)=T(u,v) ( x , y ) = T ( u , v ) d x d y = ∣ det ∂ ( x , y ) ∂ ( u , v ) ∣ d u d v . dx\,dy = \left|\det \frac{\partial(x,y)}{\partial(u,v)}\right| du\,dv. d x d y = det ∂ ( u , v ) ∂ ( x , y ) d u d v . ( u , v ) = ( r , θ ) (u,v)=(r,\theta) ( u , v ) = ( r , θ ) x = r cos θ x=r\cos\theta x = r cos θ y = r sin θ y=r\sin\theta y = r sin θ
=\begin{vmatrix} \cos\theta & -r\sin\theta\\ \sin\theta & \;\;r\cos\theta\end{vmatrix}.$$
Expand:
$$\det J = \cos\theta\,(r\cos\theta) - (-r\sin\theta)(\sin\theta) = r\cos^2\theta + r\sin^2\theta = r.$$
So $|\det J| = r$ (since $r\ge 0$), confirming $dx\,dy = r\,dr\,d\theta$. ∎
Compute ∬ R 1 d A \displaystyle \iint_R 1\,dA ∬ R 1 d A x 2 + y 2 ≤ a 2 x^2+y^2\le a^2 x 2 + y 2 ≤ a 2
∫ 0 2 π ∫ 0 a 1 ⋅ r d r d θ . \int_0^{2\pi}\int_0^a 1\cdot r\,dr\,d\theta. ∫ 0 2 π ∫ 0 a 1 ⋅ r d r d θ .
Why r r r Because d A = r d r d θ dA=r\,dr\,d\theta d A = r d r d θ Why limits r : 0 → a r:0\to a r : 0 → a θ : 0 → 2 π \theta:0\to 2\pi θ : 0 → 2 π The disk is all radii up to a a a
= ∫ 0 2 π [ r 2 2 ] 0 a d θ = ∫ 0 2 π a 2 2 d θ = a 2 2 ⋅ 2 π = π a 2 . ✓ =\int_0^{2\pi}\!\left[\tfrac{r^2}{2}\right]_0^a d\theta=\int_0^{2\pi}\tfrac{a^2}{2}\,d\theta = \tfrac{a^2}{2}\cdot 2\pi = \pi a^2.\;\checkmark = ∫ 0 2 π [ 2 r 2 ] 0 a d θ = ∫ 0 2 π 2 a 2 d θ = 2 a 2 ⋅ 2 π = π a 2 . ✓
Forecast Forecast-then-verifyBefore computing, predict the answer. You know a disk's area is π a 2 \pi a^2 π a 2 r r r r r r ∫ 0 2 π ∫ 0 a d r d θ = 2 π a \int_0^{2\pi}\int_0^a dr\,d\theta = 2\pi a ∫ 0 2 π ∫ 0 a d r d θ = 2 π a
Let I = ∫ − ∞ ∞ e − x 2 d x I=\int_{-\infty}^{\infty}e^{-x^2}dx I = ∫ − ∞ ∞ e − x 2 d x I 2 = ∫ − ∞ ∞ ∫ − ∞ ∞ e − ( x 2 + y 2 ) d x d y . I^2=\!\int_{-\infty}^\infty\!\!\int_{-\infty}^\infty e^{-(x^2+y^2)}dx\,dy. I 2 = ∫ − ∞ ∞ ∫ − ∞ ∞ e − ( x 2 + y 2 ) d x d y .
Why square it? The product of two 1-D integrals becomes a 2-D integral over the whole plane, and x 2 + y 2 x^2+y^2 x 2 + y 2
Switch to polar over all of R 2 \mathbb{R}^2 R 2 r : 0 → ∞ r:0\to\infty r : 0 → ∞ θ : 0 → 2 π \theta:0\to2\pi θ : 0 → 2 π x 2 + y 2 = r 2 x^2+y^2=r^2 x 2 + y 2 = r 2 d A = r d r d θ dA=r\,dr\,d\theta d A = r d r d θ I 2 = ∫ 0 2 π ∫ 0 ∞ e − r 2 r d r d θ . I^2=\int_0^{2\pi}\!\int_0^\infty e^{-r^2}\,r\,dr\,d\theta. I 2 = ∫ 0 2 π ∫ 0 ∞ e − r 2 r d r d θ .
Why this is now easy: the extra r r r exactly what you need for the substitution u = r 2 u=r^2 u = r 2 d u = 2 r d r du=2r\,dr d u = 2 r d r
∫ 0 ∞ e − r 2 r d r = 1 2 ∫ 0 ∞ e − u d u = 1 2 . \int_0^\infty e^{-r^2}r\,dr=\tfrac12\int_0^\infty e^{-u}du=\tfrac12. ∫ 0 ∞ e − r 2 r d r = 2 1 ∫ 0 ∞ e − u d u = 2 1 . I 2 = 2 π ⋅ 1 2 = π ⇒ I = π . I^2 = 2\pi\cdot\tfrac12=\pi \;\Rightarrow\; I=\sqrt{\pi}. I 2 = 2 π ⋅ 2 1 = π ⇒ I = π .
Without the Jacobian r r r impossible in closed form . With it, the answer drops out in three lines. The r r r
∬ R ( x 2 + y 2 ) d A \displaystyle \iint_R (x^2+y^2)\,dA ∬ R ( x 2 + y 2 ) d A 1 ≤ r ≤ 2 1\le r\le 2 1 ≤ r ≤ 2 0 ≤ θ ≤ π / 2 0\le\theta\le \pi/2 0 ≤ θ ≤ π /2
Replace x 2 + y 2 = r 2 x^2+y^2=r^2 x 2 + y 2 = r 2 d A = r d r d θ dA=r\,dr\,d\theta d A = r d r d θ ∫ 0 π / 2 ∫ 1 2 r 2 ⋅ r d r d θ = ∫ 0 π / 2 ∫ 1 2 r 3 d r d θ . \int_0^{\pi/2}\!\int_1^2 r^2\cdot r\,dr\,d\theta=\int_0^{\pi/2}\!\int_1^2 r^3\,dr\,d\theta. ∫ 0 π /2 ∫ 1 2 r 2 ⋅ r d r d θ = ∫ 0 π /2 ∫ 1 2 r 3 d r d θ .
Why r 3 r^3 r 3 The integrand r 2 r^2 r 2 r r r
∫ 1 2 r 3 d r = [ r 4 4 ] 1 2 = 16 − 1 4 = 15 4 , ∫ 0 π / 2 d θ = π 2 . \int_1^2 r^3\,dr=\left[\tfrac{r^4}{4}\right]_1^2=\tfrac{16-1}{4}=\tfrac{15}{4},\qquad \int_0^{\pi/2}d\theta=\tfrac{\pi}{2}. ∫ 1 2 r 3 d r = [ 4 r 4 ] 1 2 = 4 16 − 1 = 4 15 , ∫ 0 π /2 d θ = 2 π . Answer = 15 4 ⋅ π 2 = 15 π 8 . \text{Answer}=\tfrac{15}{4}\cdot\tfrac{\pi}{2}=\frac{15\pi}{8}. Answer = 4 15 ⋅ 2 π = 8 15 π .
Common mistake Mistake 1: Writingd A = d r d θ dA = dr\,d\theta d A = d r d θ r r r Why it feels right: in Cartesian you just multiply the two differentials, d x d y dx\,dy d x d y d r d θ dr\,d\theta d r d θ seems universal.
Why it's wrong: r r r θ \theta θ different units (r r r θ \theta θ d r d θ dr\,d\theta d r d θ length , not an area . The arc length grows with r r r Fix: Always carry the Jacobian: d A = r d r d θ dA = r\,dr\,d\theta d A = r d r d θ π a 2 \pi a^2 π a 2
Common mistake Mistake 2: Puttingr r r ∬ x d A = ∬ x r d r d θ \iint x\,dA = \iint x\, r\,dr\,d\theta ∬ x d A = ∬ x r d r d θ x x x Fix: Convert everything : x → r cos θ x\to r\cos\theta x → r cos θ ∬ x d A = ∬ ( r cos θ ) r d r d θ \iint x\,dA=\iint (r\cos\theta)\,r\,dr\,d\theta ∬ x d A = ∬ ( r cos θ ) r d r d θ
Common mistake Mistake 3: Wrong order / limits depending on each otherIf a boundary like r = 2 cos θ r = 2\cos\theta r = 2 cos θ inner r r r θ \theta θ Fix: Sketch the region first; let r r r as a function of θ \theta θ , then θ \theta θ
Recall Feynman: explain to a 12-year-oldImagine cutting a pizza. Near the center the slices are skinny little triangles; near the crust the same-angle slice is a big fat piece. So a "slice" doesn't have a fixed size — it gets bigger the farther out you go. When we add up area in circles, each tiny tile far from the center is bigger , and "how much bigger" is just how far out you are — that's the r r r r r r
r r r "Arc grows with radius: r d θ r\,d\theta r d θ The curved side of the box is an arc, arc = = = × \times × d r dr d r r r r "Polar area Really needs r r r
What is the area element in polar coordinates? d A = r d r d θ dA = r\,dr\,d\theta d A = r d r d θ Why is there an extra factor of r r r The arc length of the box's curved side is
r d θ r\,d\theta r d θ , so the box area is
d r ⋅ r d θ = r d r d θ dr\cdot r\,d\theta = r\,dr\,d\theta d r ⋅ r d θ = r d r d θ ; it stretches with distance from the origin.
What are the polar substitutions for x x x y y y x = r cos θ , y = r sin θ x=r\cos\theta,\; y=r\sin\theta x = r cos θ , y = r sin θ Compute the Jacobian determinant for ( x , y ) = ( r cos θ , r sin θ ) (x,y)=(r\cos\theta, r\sin\theta) ( x , y ) = ( r cos θ , r sin θ ) det ( cos θ − r sin θ sin θ r cos θ ) = r \det\begin{pmatrix}\cos\theta & -r\sin\theta\\ \sin\theta & r\cos\theta\end{pmatrix}=r det ( cos θ sin θ − r sin θ r cos θ ) = r Why does d r d θ dr\,d\theta d r d θ r r r r r r is a length and
θ \theta θ is dimensionless, so
d r d θ dr\,d\theta d r d θ is a length, not an area; you need the extra length factor
r r r .
Evaluate ∫ 0 2 π ∫ 0 a r d r d θ \int_0^{2\pi}\int_0^a r\,dr\,d\theta ∫ 0 2 π ∫ 0 a r d r d θ π a 2 \pi a^2 π a 2 (area of a disk).
How does the r r r It supplies the
r r r needed for
u = r 2 u=r^2 u = r 2 ,
d u = 2 r d r du=2r\,dr d u = 2 r d r , turning
∫ e − r 2 r d r \int e^{-r^2}r\,dr ∫ e − r 2 r d r into
1 2 ∫ e − u d u \tfrac12\int e^{-u}du 2 1 ∫ e − u d u .
What is ∫ − ∞ ∞ e − x 2 d x \int_{-\infty}^\infty e^{-x^2}dx ∫ − ∞ ∞ e − x 2 d x π \sqrt{\pi} π (via squaring and polar coordinates).
For the circle r = 2 cos θ r=2\cos\theta r = 2 cos θ r r r No — the outer
r r r -limit depends on
θ \theta θ :
r r r runs
0 → 2 cos θ 0\to 2\cos\theta 0 → 2 cos θ .
area approx dr times r d theta
Circular symmetry disk annulus sector
Polar coordinates r theta
x = r cos theta, y = r sin theta
Constant integration limits
Tiny polar box, sides dr and r d theta
Master formula for polar integral
Disk area gives pi a squared
Intuition Hinglish mein samjhoDekho, jab hum double integral polar coordinates mein karte hain, toh sabse important cheez hai yeh chhota sa factor r r r d r d θ dr\,d\theta d r d θ d x d y dx\,dy d x d y r r r d A = r d r d θ dA = r\,dr\,d\theta d A = r d r d θ
Iski do tarah se proof hoti hai. Geometric: chhote box ka ek side d r dr d r = r d θ = r\,d\theta = r d θ = = = × \times × r r r x = r cos θ x=r\cos\theta x = r cos θ y = r sin θ y=r\sin\theta y = r sin θ r cos 2 θ + r sin 2 θ = r r\cos^2\theta + r\sin^2\theta = r r cos 2 θ + r sin 2 θ = r
Yeh kyun matter karta hai? Kyunki bahut saare integrals jo Cartesian mein impossible lagte hain, polar mein 3 line mein ho jaate hain. Best example — Gaussian integral ∫ e − x 2 d x = π \int e^{-x^2}dx = \sqrt{\pi} ∫ e − x 2 d x = π r r r u = r 2 u=r^2 u = r 2 r r r x , y x,y x , y π a 2 \pi a^2 π a 2 r r r
