Double integral

Double integral

1. What a Double Integral Is

For $f(x,y)$ over a region $D$ in the $xy$-plane (bounded, closed), the double integral

is the two-dimensional analogue of $∫_{a}^{b} f(x)dx$, giving:

• Volume under surface $z=f(x,y)$ ($f≥0$) over $D$,
• Area if $f=1$,
• Mass if $f$ = density, etc.

If $f$ continuous on $D$, the integral exists. Iterated integrals compute it.

2. Iterated Integrals — Fubini

If $D$ is x-simple (between $y=g_1(x)$ and $g_2(x)$, $x∈[a,b]$):

If y-simple (between $x=h_1(y)$ and $h_2(y)$):

For a rectangle $D=[a,b]×[c,d]$, this is simply $∫_{a}^{b}∫_{c}^{d} f(x,y) dy dx = ∫_{c}^{d}∫_{a}^{b} f(x,y) dx dy$ (order interchangeable if $f$ continuous).

3. Working Steps (recipe)

1. Sketch $D$; decide which order gives simpler limits (try $dy\,dx$ first for $y$ between curves).
2. Determine outer limits (constants) and inner limits (functions of outer variable).
3. Integrate inner integral (treat outer variable as constant), then outer.

4. Worked Examples

(a) ∬_{[0,1]×[0,2]} (x^2 + y) dA:
    ∫_{0}^{1}[∫_{0}^{2}(x^2+y)dy]dx = ∫_{0}^{1}[2x^2+2]dx = [2x^3/3+2x]_{0}^{1}=8/3 ✓
    Other order: ∫_{0}^{2}[∫_{0}^{1}(x^2+y)dx]dy = ∫_{0}^{2}(1/3+y)dy=8/3 same.

(b) D: triangle with vertices (0,0),(1,0),(1,1): y from 0 to x, x 0→1; ∬ y dA = ∫_{0}^{1}∫_{0}^{x} y dy dx = ∫_{0}^{1}x²/2 dx=1/6 ✓

(c) D: between y=x^2 and y=x for x 0→1: ∫_{0}^{1}∫_{x^2}^{x} (x+y) dy dx
    = ∫_{0}^{1}[x(x-x^2)+(x^2 -x^4)/2]dx =3/20 ✓

(d) D: circle x^2+y^2 ≤4 : ∬ dA = area = π·2²=4π (will verify via polar later) ✓

(e) Need to watch: ∬ y sin(x y) dA over? Inner integral requires integration by parts.

5. Properties

• Linearity, additivity over partition, $∬_{D} 1 dA = area(D)$, if $f≤g$ then $∬ f ≤ ∬ g$.

6. Exam Points

• Set up and evaluate $∬_{D} f dA$ for rectangle/triangle/bounded by $y=x², y=x$ — 7 marks, very frequent.
• State Fubini and decide order $dy\,dx$ vs $dx\,dy$ — 3-5 marks.
• Compute volume/area/mass as double integral — 5 marks, frequent.