Double integrals over rectangles — Fubini's theorem

WHAT is a double integral?

Definition Double integral over a rectangle

Let $R = [a,b]\times[c,d] = \{(x,y): a\le x\le b,\ c\le y\le d\}$ and $f$ bounded on $R$ . Partition $R$ into small sub-rectangles of area $\Delta A = \Delta x\,\Delta y$ , pick a sample point $(x_i^*,y_j^*)$ in each, and form the Riemann sum $S = \sum_{i}\sum_{j} f(x_i^*, y_j^*)\,\Delta A.$ If $S$ approaches a single limit as all sub-rectangles shrink, that limit is $\iint_R f(x,y)\,dA = \lim_{\|P\|\to 0}\sum_i\sum_j f(x_i^*,y_j^*)\,\Delta A.$

WHY this definition? Each term $f(x_i^*,y_j^*)\,\Delta A$ is the volume of a thin rectangular column: base $\Delta A$ , height $f$ . Add all columns → approximate volume under the surface. Shrink the base → exact volume. This is the exact same idea as 1-D area, one dimension up.

HOW do we actually compute it? — Iterated integrals

Computing a 2-D limit directly is brutal. Fubini gives an escape: integrate one variable at a time.

DERIVATION from first principles (the "slicing" argument)

WHY can a volume be sliced? Picture the solid under $z=f(x,y)$ over $R$ .

Step 1 — Slab volume by cross-section. Fix $x = x_0$ . The vertical plane at $x_0$ cuts the solid in a flat shape whose area is $A(x_0) = \int_c^d f(x_0,y)\,dy.$ Why this step? On that plane, $x$ is frozen, so the curve $y\mapsto f(x_0,y)$ bounds a 1-D region; its area is an ordinary single integral.

Step 2 — Sum the slabs. A slab of thickness $dx$ at position $x$ has volume $A(x)\,dx$ . Total volume: $V = \int_a^b A(x)\,dx = \int_a^b\left(\int_c^d f(x,y)\,dy\right)dx.$ Why this step? This is exactly $\int (\text{cross-sectional area})\,dx$ — the volume-by-slicing formula you already know from single-variable calculus.

Step 3 — Slice the other way. Nothing forced us to freeze $x$ first. Freezing $y$ gives slabs of area $\int_a^b f(x,y)\,dx$ , hence $V = \int_c^d\left(\int_a^b f(x,y)\,dx\right)dy.$ Why this step? The same solid has the same volume, so both orders give the same number. That equality IS Fubini's theorem.

Figure — Double integrals over rectangles — Fubini's theorem

A special shortcut: separable integrands

Worked Examples

Worked example Example 1 — basic, check both orders

Compute $\displaystyle\iint_R (x + 2y)\,dA$ , $R=[0,2]\times[1,3]$ .

Order $dy$ first: $\int_0^2\!\!\int_1^3 (x+2y)\,dy\,dx.$ Inner: $\int_1^3 (x+2y)\,dy = \big[xy + y^2\big]_{1}^{3} = (3x+9)-(x+1) = 2x+8.$ Why? $x$ is constant; antiderivative in $y$ of $x$ is $xy$ , of $2y$ is $y^2$ . Outer: $\int_0^2 (2x+8)\,dx = [x^2 + 8x]_0^2 = 4+16 = 20.$

Order $dx$ first (verify): Inner: $\int_0^2 (x+2y)\,dx = [\tfrac{x^2}{2}+2yx]_0^2 = 2 + 4y.$ Outer: $\int_1^3 (2+4y)\,dy = [2y+2y^2]_1^3 = (6+18)-(2+2)=20.$ ✅ Same answer.

Worked example Example 3 — when order matters for

effort $\displaystyle\iint_R y\cos(xy)\,dA$ , $R=[0,\pi]\times[0,1]$ . Try $dx$ first (treat $y$ const): $\int_0^\pi y\cos(xy)\,dx = [\sin(xy)]_0^\pi = \sin(\pi y).$ Why? $\frac{d}{dx}\sin(xy)=y\cos(xy)$ , exactly the integrand. The $y$ out front made this clean. Then $\int_0^1 \sin(\pi y)\,dy = [-\tfrac{1}{\pi}\cos(\pi y)]_0^1 = -\tfrac1\pi(\cos\pi-\cos0)=-\tfrac1\pi(-1-1)=\tfrac{2}{\pi}.$ Doing $dy$ first would force integration by parts — same answer, more work. Fubini lets you choose the easy order.

Common Mistakes (Steel-manned)

Recall Feynman: explain to a 12-year-old

Imagine a loaf of bread shaped weirdly on top. You want its total volume. You could slice it into thin vertical slices, measure the area of each slice's face, and add them up. Or you could slice it the other direction. Either way you get the same loaf, so you get the same volume. Fubini just says: "slice whichever way is easier, you'll get the same answer." A double integral is just adding up tiny towers; iterated integrals are a tidy way to add them row-by-row then column-by-column.

Connections

Riemann sums — double integrals are 2-D Riemann limits.
Volume by slicing (single-variable) — Fubini's derivation IS this idea.
Double integrals over general regions — next step: non-constant inner limits.
Change of order of integration — uses Fubini to swap when one order is hard.
Fubini–Tonelli theorem — the measure-theoretic generalization.
Triple integrals — slice in three directions.

Flashcards

What does a double integral

\iint_R f\,dA

geometrically represent?

The signed volume between the surface

z=f(x,y)

and the

xy

-plane over

R

(volume above minus volume below).

State Fubini's theorem for a rectangle

R=[a,b]\times[c,d]

f

is continuous on

R

, then

\iint_R f\,dA = \int_a^b\int_c^d f\,dy\,dx = \int_c^d\int_a^b f\,dx\,dy

In the inner integral of

\int_a^b\int_c^d f(x,y)\,dy\,dx

, how is

x

treated?

As a constant.

Why are both iteration orders equal on a rectangle?

They compute the same solid's volume by slicing in two different directions; same solid ⇒ same volume.

Shortcut for

\iint_R g(x)h(y)\,dA

\left(\int_a^b g\,dx\right)\left(\int_c^d h\,dy\right)

, valid because limits are constant and variables separate.

What is

A(x)=\int_c^d f(x,y)\,dy

in the slicing derivation?

The cross-sectional area of the solid at fixed

x

When can Fubini fail?

When

f

is not integrable (unbounded/non-integrable on

R

); the two orders may then differ.

\iint_{[0,2]\times[1,3]} (x+2y)\,dA = ?

20

Mnemonic for the procedure?

FREEZE the outer variable, FILL the inner 1-D integral, FILE into the outer integral.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, double integral ka matlab simple hai: $z=f(x,y)$ ek surface hai, aur ek rectangle $R$ ke upar uske niche jo volume banta hai, wahi double integral $\iint_R f\,dA$ hai. Hum us volume ko chhote-chhote columns (tiny towers) mein todte hain — har column ka base $\Delta x\,\Delta y$ aur height $f$ — phir sabko add kar dete hain. Limit lene par exact volume mil jaata hai.

Ab seedha 2-D limit nikalna mushkil hai, isliye Fubini's theorem kaam aata hai. Yeh kehta hai: ek variable ko freeze karo, doosre ko integrate karo (ek 1-D integral), phir bacha hua variable integrate karo. Aur sabse mast baat — order matter nahi karta. Pehle $dy$ karo ya pehle $dx$ , answer same aayega. Reason intuitive hai: bread ko jis bhi direction se slice karo, total volume same rehta hai.

Practical tip: inner integral mein doosra variable ek constant ki tarah treat karo (jaise woh number $7$ ho). Agar function separable ho, yaani $f=g(x)h(y)$ , to do alag integrals ka product le lo — bahut time bachta hai. Aur jab ek order mushkil lage (jaise integration by parts aa raha ho), to doosra order try karo — Fubini permission deta hai easy raasta chunne ki.

Bas yaad rakho rectangle pe limits hamesha constant hote hain, isliye order swap karna safe hai. Lekin non-rectangular region pe limits variable ho jaate hain — wahan blindly swap mat karna, region dobara dekhna padega. FREEZE, FILL, FILE — yahi mantra hai.

Double integrals over rectangles — Fubini's theorem

WHAT is a double integral?

HOW do we actually compute it? — Iterated integrals

DERIVATION from first principles (the "slicing" argument)

A special shortcut: separable integrands

Worked Examples

Common Mistakes (Steel-manned)

Connections

Flashcards

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Multivariable Calculus

Connections