Intuition The big pictureA double integral ∬ R f d A \iint_R f\,dA ∬ R f d A R R R no matter how you sweep through it . You can sweep ==vertical strips first (integrate y y y x x x or horizontal strips first (integrate x x x y y y one direction may be impossible or ugly, the other easy .
Intuition Two motivations
The inner integral has no elementary antiderivative. e.g. ∫ e x 2 d x \int e^{x^2}\,dx ∫ e x 2 d x ∫ x e x 2 d x \int x e^{x^2}\,dx ∫ x e x 2 d x The region is awkward in one orientation (needs splitting into pieces), but a single clean description in the other.
The key fact we exploit (Fubini's theorem):
∬ R f ( x , y ) d A = ∫ x = a b ∫ y = g 1 ( x ) g 2 ( x ) f d y d x = ∫ y = c d ∫ x = h 1 ( y ) h 2 ( y ) f d x d y \iint_R f(x,y)\,dA = \int_{x=a}^{b}\!\!\int_{y=g_1(x)}^{g_2(x)}\!\! f\,dy\,dx = \int_{y=c}^{d}\!\!\int_{x=h_1(y)}^{h_2(y)}\!\! f\,dx\,dy ∬ R f ( x , y ) d A = ∫ x = a b ∫ y = g 1 ( x ) g 2 ( x ) f d y d x = ∫ y = c d ∫ x = h 1 ( y ) h 2 ( y ) f d x d y
for f f f R R R
Definition Type I vs Type II regions
A Type I (vertically simple) region: a ≤ x ≤ b a\le x\le b a ≤ x ≤ b g 1 ( x ) ≤ y ≤ g 2 ( x ) g_1(x)\le y\le g_2(x) g 1 ( x ) ≤ y ≤ g 2 ( x ) vertical strip ; its ends ride on curves y = g 1 y=g_1 y = g 1 y = g 2 y=g_2 y = g 2
A Type II (horizontally simple) region: c ≤ y ≤ d c\le y\le d c ≤ y ≤ d h 1 ( y ) ≤ x ≤ h 2 ( y ) h_1(y)\le x\le h_2(y) h 1 ( y ) ≤ x ≤ h 2 ( y ) horizontal strip ; its ends ride on x = h 1 x=h_1 x = h 1 x = h 2 x=h_2 x = h 2
Read the given limits and write the four inequalities they encode.Sketch the region R R R Decide the new outer variable and find its overall min/max → these are the new outer constants.For a fixed value of the new outer variable, find where the strip enters and exits R R R
Rewrite the integral with swapped d x d y dx\,dy d x d y
Intuition Building it from a sumA double integral is a limit of ∑ i , j f ( x i , y j ) Δ x Δ y \sum_{i,j} f(x_i,y_j)\,\Delta x\,\Delta y ∑ i , j f ( x i , y j ) Δ x Δ y R R R rows then columns or columns then rows . The total is identical. Each iterated integral is just one order of summing the same tiles . So the limits must describe the same set of tiles R R R
Concretely, take the triangle R : 0 ≤ x ≤ 1 , 0 ≤ y ≤ x R:\ 0\le x\le 1,\ 0\le y\le x R : 0 ≤ x ≤ 1 , 0 ≤ y ≤ x
Tiles described by columns: pick x ∈ [ 0 , 1 ] x\in[0,1] x ∈ [ 0 , 1 ] y y y 0 → x 0\to x 0 → x ∫ 0 1 ∫ 0 x f d y d x \int_0^1\!\int_0^x f\,dy\,dx ∫ 0 1 ∫ 0 x f d y d x Same tiles by rows: pick y ∈ [ 0 , 1 ] y\in[0,1] y ∈ [ 0 , 1 ] x x x x = y x=y x = y x = 1 x=1 x = 1 ∫ 0 1 ∫ y 1 f d x d y \int_0^1\!\int_y^1 f\,dx\,dy ∫ 0 1 ∫ y 1 f d x d y
The boundary y = x y=x y = x other variable (x = y x=y x = y is the order change.
Evaluate I = ∫ 0 1 ∫ x 1 e y 2 d y d x \displaystyle I=\int_0^1\!\int_x^1 e^{y^2}\,dy\,dx I = ∫ 0 1 ∫ x 1 e y 2 d y d x
Worked example Step by stepStep 1 — Read limits. Outer: 0 ≤ x ≤ 1 0\le x\le 1 0 ≤ x ≤ 1 x ≤ y ≤ 1 x\le y\le 1 x ≤ y ≤ 1 Why? These say: for each x x x y y y y = x y=x y = x y = 1 y=1 y = 1
Step 2 — Sketch. Region: above y = x y=x y = x y = 1 y=1 y = 1 x = 0 x=0 x = 0 ( 0 , 0 ) , ( 0 , 1 ) , ( 1 , 1 ) (0,0),(0,1),(1,1) ( 0 , 0 ) , ( 0 , 1 ) , ( 1 , 1 ) Why this step? ∫ e y 2 d y \int e^{y^2}dy ∫ e y 2 d y no elementary form, so we MUST swap.
Step 3 — New outer variable y y y Overall y y y 0 ≤ y ≤ 1 0\le y\le 1 0 ≤ y ≤ 1 Why? Lowest point has y = 0 y=0 y = 0 y = 1 y=1 y = 1
Step 4 — New inner x x x Fix y y y x = 0 x=0 x = 0 y = x y=x y = x x = y x=y x = y 0 ≤ x ≤ y 0\le x\le y 0 ≤ x ≤ y Why? For fixed y y y x = 0 x=0 x = 0 x = y x=y x = y
Step 5 — Rewrite & evaluate.
I = ∫ 0 1 ∫ 0 y e y 2 d x d y = ∫ 0 1 e y 2 [ x ] 0 y d y = ∫ 0 1 y e y 2 d y . I=\int_0^1\!\int_0^y e^{y^2}\,dx\,dy=\int_0^1 e^{y^2}\,[x]_0^y\,dy=\int_0^1 y\,e^{y^2}\,dy. I = ∫ 0 1 ∫ 0 y e y 2 d x d y = ∫ 0 1 e y 2 [ x ] 0 y d y = ∫ 0 1 y e y 2 d y . u = y 2 , d u = 2 y d y u=y^2,\ du=2y\,dy u = y 2 , d u = 2 y d y I = 1 2 ∫ 0 1 e u d u = 1 2 ( e − 1 ) . I=\tfrac12\int_0^1 e^u\,du=\tfrac12(e-1). I = 2 1 ∫ 0 1 e u d u = 2 1 ( e − 1 ) .
So I = e − 1 2 I=\dfrac{e-1}{2} I = 2 e − 1
Evaluate J = ∫ 0 1 ∫ x 1 1 1 + y 4 d y d x \displaystyle J=\int_0^1\!\int_{\sqrt{x}}^{1} \frac{1}{1+y^4}\,dy\,dx J = ∫ 0 1 ∫ x 1 1 + y 4 1 d y d x
Reverse the order in K = ∫ 0 2 ∫ y 2 / 2 y sin ( x ) d x d y \displaystyle K=\int_0^2\!\int_{y^2/2}^{y}\!\sin\!\big(x\big)\,dx\,dy K = ∫ 0 2 ∫ y 2 /2 y sin ( x ) d x d y
∫ 0 4 ∫ y 2 x e x 5 d x d y \displaystyle \int_0^4\!\int_{\sqrt{y}}^{2} x\,e^{x^5}\,dx\,dy ∫ 0 4 ∫ y 2 x e x 5 d x d y
K = ∫ 0 1 ∫ y 1 cos ( x 3 ) ? K=\int_0^1\!\int_{\sqrt{y}}^{1} \cos\!\big(x^3\big)\,? K = ∫ 0 1 ∫ y 1 cos ( x 3 ) ?
Let me give a clean fully worked Example 2 instead:
Worked example Reorder and evaluateK = ∫ 0 1 ∫ x 1 sin ( y 3 ? ) \displaystyle K=\int_0^1\!\int_{\sqrt{x}}^{1}\sin\!\left(\frac{y^3}{?}\right) K = ∫ 0 1 ∫ x 1 sin ( ? y 3 ) K = ∫ 0 1 ∫ x 1 sin y y d y d x K=\displaystyle\int_0^1\!\int_{x}^{1}\frac{\sin y}{y}\,dy\,dx K = ∫ 0 1 ∫ x 1 y sin y d y d x
Step 1 — Read. 0 ≤ x ≤ 1 , x ≤ y ≤ 1 0\le x\le 1,\ x\le y\le 1 0 ≤ x ≤ 1 , x ≤ y ≤ 1 Why swap? ∫ sin y y d y \int \frac{\sin y}{y}\,dy ∫ y s i n y d y
Step 2–4 — Swap to y y y 0 ≤ y ≤ 1 0\le y\le1 0 ≤ y ≤ 1 0 ≤ x ≤ y 0\le x\le y 0 ≤ x ≤ y Why? Identical triangle ⇒ identical reordering as Example 1.
Step 5 — Evaluate.
K = ∫ 0 1 ∫ 0 y sin y y d x d y = ∫ 0 1 sin y y ⋅ y d y = ∫ 0 1 sin y d y = 1 − cos 1. K=\int_0^1\!\int_0^y \frac{\sin y}{y}\,dx\,dy=\int_0^1 \frac{\sin y}{y}\cdot y\,dy=\int_0^1\sin y\,dy=1-\cos 1. K = ∫ 0 1 ∫ 0 y y s i n y d x d y = ∫ 0 1 y s i n y ⋅ y d y = ∫ 0 1 sin y d y = 1 − cos 1. 1 / y 1/y 1/ y cancels because the inner integral in x x x y y y
Reverse ∫ 0 1 ∫ x 2 x f d y d x \displaystyle \int_0^1\!\int_{x^2}^{x}\, f\,dy\,dx ∫ 0 1 ∫ x 2 x f d y d x
Step 1. 0 ≤ x ≤ 1 0\le x\le1 0 ≤ x ≤ 1 x 2 ≤ y ≤ x x^2\le y\le x x 2 ≤ y ≤ x y = x 2 y=x^2 y = x 2 y = x y=x y = x x 2 ≤ x x^2\le x x 2 ≤ x 0 ≤ x ≤ 1 0\le x\le1 0 ≤ x ≤ 1 Step 2 (sketch). A lens between the line and parabola, corners ( 0 , 0 ) (0,0) ( 0 , 0 ) ( 1 , 1 ) (1,1) ( 1 , 1 ) Step 3. New outer y y y 0 ≤ y ≤ 1 0\le y\le1 0 ≤ y ≤ 1 Step 4. Fix y y y x x x y = x ⇒ x = y y=x\Rightarrow x=y y = x ⇒ x = y y = x 2 ⇒ x = y y=x^2\Rightarrow x=\sqrt y y = x 2 ⇒ x = y 0 < y < 1 0<y<1 0 < y < 1 y < y y<\sqrt y y < y y ≤ x ≤ y y\le x\le\sqrt y y ≤ x ≤ y Why this step? Inside the lens, for fixed height y y y Step 5. ∫ 0 1 ∫ y y f d x d y . \displaystyle \int_0^1\!\int_{y}^{\sqrt y} f\,dx\,dy. ∫ 0 1 ∫ y y f d x d y .
Common mistake Steel-manning the classic blundersBlunder A — leaving a variable in the outer limits. Writing ∫ 0 x ∫ 0 1 d y d x \int_0^x\!\int_0^1 dy\,dx ∫ 0 x ∫ 0 1 d y d x x x x Why it feels right: you copied a curve into the outer slot. Fix: the outermost limits must be constants ; only the inner limits may depend on the outer variable. Always reduce the outer to numbers.
Blunder B — forgetting to re-solve the boundary for the new variable. Keeping y = x y=x y = x x x x Why it feels right: the equation is "the same curve." Fix: you must express the inner variable explicitly — solve y = x y=x y = x x x x x = y x=y x = y y = x 2 y=x^2 y = x 2 x x x x = y x=\sqrt y x = y
Blunder C — swallowing limits without sketching. Mechanically copying min/max of all four numbers. Why it feels right: it's fast. Fix: sketch — many regions need splitting in one orientation and not the other. The sketch tells you when one strip exits onto a different curve.
Blunder D — wrong branch of a square root. Using x = − y x=-\sqrt y x = − y Fix: check which side of the region you're on; pick the branch that lies in R R R
Recall Feynman: explain to a 12-year-oldImagine a tiled patio shaped like a triangle. You want the total dust on all tiles. You can count column by column (go along the bottom, add up each tall stack of tiles) or row by row (go up the side, add up each long row). Either way you count every tile once, so the total dust is the same! Sometimes counting columns is a nightmare and rows are easy — so we just switch how we walk through the tiles. To switch, you draw the patio shape, then describe its left/right edges instead of its top/bottom edges (or vice versa).
Mnemonic Remember the limit rule"Outer numbers, inner curves; flip the strip, re-solve the curve."
Vertical strip ↔ d y d x dy\,dx d y d x d x d y dx\,dy d x d y solve each boundary for the new inner variable .
When can you swap the order of a double integral? When
f f f is integrable (continuous suffices) over
R R R — Fubini's theorem guarantees both iterated integrals equal
∬ R f d A \iint_R f\,dA ∬ R f d A .
In a correctly written iterated integral, what kind of limits must the OUTER integral have? Constant (numeric) limits; only the inner limits may depend on the outer variable.
What does a vertical strip correspond to in terms of order? Integrating
y y y first then
x x x :
∫ a b ∫ g 1 ( x ) g 2 ( x ) d y d x \int_a^b\int_{g_1(x)}^{g_2(x)} dy\,dx ∫ a b ∫ g 1 ( x ) g 2 ( x ) d y d x (Type I region).
To change order, what algebraic operation do you do to each boundary curve? Solve the curve for the NEW inner variable (e.g.
y = x 2 → x = y y=x^2 \to x=\sqrt y y = x 2 → x = y , correct branch).
Why reverse ∫ 0 1 ∫ x 1 e y 2 d y d x \int_0^1\int_x^1 e^{y^2}\,dy\,dx ∫ 0 1 ∫ x 1 e y 2 d y d x e y 2 e^{y^2} e y 2 has no elementary antiderivative in
y y y ; swapping gives
∫ 0 1 y e y 2 d y \int_0^1 y e^{y^2}dy ∫ 0 1 y e y 2 d y , easy. Answer
= ( e − 1 ) / 2 =(e-1)/2 = ( e − 1 ) /2 .
Reorder ∫ 0 1 ∫ x 1 sin y y d y d x \int_0^1\int_x^1 \frac{\sin y}{y}dy\,dx ∫ 0 1 ∫ x 1 y s i n y d y d x ∫ 0 1 ∫ 0 y sin y y d x d y = ∫ 0 1 sin y d y = 1 − cos 1 \int_0^1\int_0^y \frac{\sin y}{y}dx\,dy=\int_0^1\sin y\,dy=1-\cos 1 ∫ 0 1 ∫ 0 y y s i n y d x d y = ∫ 0 1 sin y d y = 1 − cos 1 .
For the triangle 0 ≤ x ≤ 1 , x ≤ y ≤ 1 0\le x\le1,\ x\le y\le1 0 ≤ x ≤ 1 , x ≤ y ≤ 1 0 ≤ y ≤ 1 , 0 ≤ x ≤ y 0\le y\le1,\ 0\le x\le y 0 ≤ y ≤ 1 , 0 ≤ x ≤ y .
What is the FIRST thing you should do before changing order? Sketch the region
R R R from the original inequalities.
Reorder ∫ 0 1 ∫ x 2 x f d y d x \int_0^1\int_{x^2}^{x} f\,dy\,dx ∫ 0 1 ∫ x 2 x f d y d x ∫ 0 1 ∫ y y f d x d y \int_0^1\int_{y}^{\sqrt y} f\,dx\,dy ∫ 0 1 ∫ y y f d x d y .
Why might one orientation need splitting into multiple integrals? If a strip in that direction enters/exits on different boundary curves over different sub-ranges of the outer variable.
Sum of tiles is commutative
Change order of integration
No elementary antiderivative
5-step recipe: sketch and re-describe R
Intuition Hinglish mein samjhoDekho, double integral ∬ R f d A \iint_R f\,dA ∬ R f d A R R R vertical strips banao (pehle y y y x x x horizontal strips banao (pehle x x x y y y same aayega — kyunki tum same tiles ko hi count kar rahe ho, bas order alag hai. Yahi cheez Fubini's theorem guarantee karta hai.
To phir order change karne ki zaroorat kyun? Do reasons. Pehla: kabhi-kabhi inner integral solve hi nahi hota — jaise ∫ e y 2 d y \int e^{y^2}\,dy ∫ e y 2 d y ∫ y e y 2 d y \int y\,e^{y^2}\,dy ∫ y e y 2 d y u u u
Karne ka tareeka simple hai: pehle region ki sketch banao (yeh sabse important step hai, isko mat chhodo). Sketch se outer variable ka overall min-max nikaalo — yeh constants honge. Phir ek fixed value ke liye strip kahan enter karta hai aur kahan exit, woh curves inner limits banenge. Sabse bada galti yeh hoti hai ki log boundary curve ko same chhod dete hain — nahi! Tumhe curve ko new inner variable ke liye solve karna hota hai: y = x y=x y = x x = y x=y x = y y = x 2 y=x^2 y = x 2 x = y x=\sqrt{y} x = y
