Double Integrals

Study Sheet

Double Integrals

Volume by slicing twice

Iterated Integration

Tip
Integrate one variable at a time

∬Rf dA\iint_R f\,dA over a rectangle is two ordinary integrals nested: inner in yy (holding xx), then outer in xx — and Fubini says the order does not matter for continuous ff. Over non-rectangles, the INNER limits depend on the outer variable: ∫01∫0x⋯ dy dx\int_0^1\int_0^x \cdots\,dy\,dx covers the triangle under y=xy = x.

Concept
Polar coordinates carry an extra r

On disks and rings, switch to x=rcos⁡θx = r\cos\theta, y=rsin⁡θy = r\sin\theta — but the area element becomes dA=r dr dθdA = r\,dr\,d\theta. That extra rr (a small polar rectangle is r Δθr\,\Delta\theta wide, not Δθ\Delta\theta) is the single most-forgotten factor in the course. It is also the magic that evaluates ∫−∞∞e−x2dx=π\int_{-\infty}^{\infty}e^{-x^2}dx = \sqrt{\pi}.

Example
A separable rectangle

∬[0,2]×[0,3]xy dA=(∫02x dx)(∫03y dy)=2⋅92=9\iint_{[0,2]\times[0,3]} xy\,dA = \left(\int_0^2 x\,dx\right)\left(\int_0^3 y\,dy\right) = 2 \cdot \tfrac92 = 9 — when the integrand factors and the region is a rectangle, the double integral is a product of singles.

Side note
Significance: averages, mass, and probability

A double integral is how you total anything spread over a region: mass from density, rainfall from intensity maps, probability from joint densities. The average value of ff over a region is ∬f dA\iint f\,dA divided by the area — centroids and centers of mass are just coordinate-weighted versions of that same computation.

Try it
Try it: an average the fast way

Find the average of f(x,y)=xf(x,y) = x over the rectangle [0,4]×[0,2][0,4]\times[0,2]. Work: ∬x dA=(∫04x dx)(2)=8⋅2=16\iint x\,dA = \left(\int_0^4x\,dx\right)(2) = 8\cdot2 = 16; divide by area 88: average 22. Sanity: xx ranges 00 to 44 uniformly, so its average must be the midpoint 22 ✓.

Proofs & Why It Matters

Tip
Proof: dA = r dr dθ in polar coordinates

A small polar cell between radii rr and r+Δrr + \Delta r and angles θ\theta and θ+Δθ\theta + \Delta\theta is nearly a rectangle: its radial side has length Δr\Delta r, and its circular side is an arc of radius rr subtending Δθ\Delta\theta, of length r Δθr\,\Delta\theta.

So its area is ≈r Δr Δθ\approx r\,\Delta r\,\Delta\theta (exactly 12[(r+Δr)2−r2]Δθ=r Δr Δθ+12Δr2Δθ\tfrac12\left[(r+\Delta r)^2 - r^2\right]\Delta\theta = r\,\Delta r\,\Delta\theta + \tfrac12\Delta r^2\Delta\theta, and the second term vanishes in the limit). Summing cells gives dA=r dr dθdA = r\,dr\,d\theta. ■\blacksquare

Tip
Proof: the Gaussian integral equals √π

Let I=∫−∞∞e−x2dxI = \int_{-\infty}^\infty e^{-x^2}dx. Then I2=∬e−(x2+y2) dAI^2 = \iint e^{-(x^2+y^2)}\,dA over the plane — switch to polar: ∫02π ⁣ ⁣∫0∞e−r2 r dr dθ\int_0^{2\pi}\!\!\int_0^\infty e^{-r^2}\,r\,dr\,d\theta. The extra rr makes the inner integral elementary (u=r2u = r^2): it equals 12\tfrac12, so I2=2π⋅12=πI^2 = 2\pi\cdot\tfrac12 = \pi and I=πI = \sqrt\pi. No one-variable method ever evaluates this integral — the plane does. ■\blacksquare

Going Deeper: Explanations & Worked Problems

Concept
Setting up limits: the one skill that decides everything

A double integral over a non-rectangle succeeds or fails at the limits. The procedure: (1) sketch the region; (2) choose an order, say inner yy, outer xx; (3) for the OUTER variable record the full interval of xx values the region occupies; (4) for the INNER variable, freeze xx, draw the vertical line through the region, and record where it enters and exits — those (usually xx-dependent) heights are the inner limits. For the triangle with vertices (0,0),(1,0),(1,1)(0,0), (1,0), (1,1) (below the line y=xy = x): xx runs 00 to 11; the vertical line at xx enters at y=0y = 0 and exits at y=xy = x, so ∫01 ⁣∫0x⋯dy dx\int_0^1\!\int_0^x \cdots dy\,dx. Reversing the order re-derives, never guesses: a horizontal line at height yy enters at x=yx = y and exits at x=1x = 1, so the same integral is ∫01 ⁣∫y1⋯dx dy\int_0^1\!\int_y^1 \cdots dx\,dy. Fubini guarantees the two agree — and when one order is hard (or impossible, as with ∫ey2\int e^{y^2}), swapping is the standard rescue.

Example
Worked: a triangle integral both ways

Compute ∬T2y dA\iint_T 2y\,dA over the triangle below y=xy = x on [0,1][0,1].

Order 1 (inner yy):

Step 1 — inner: ∫0x2y dy=[y2]0x=x2\int_0^x 2y\,dy = \left[y^2\right]_0^x = x^2.

Step 2 — outer: ∫01x2dx=13\int_0^1 x^2dx = \tfrac13.

Order 2 (inner xx):

Step 1 — inner: ∫y12y dx=2y(1−y)\int_y^1 2y\,dx = 2y(1 - y) (the integrand is constant in xx, so multiply by the interval length).

Step 2 — outer: ∫01(2y−2y2)dy=1−23=13\int_0^1 (2y - 2y^2)dy = 1 - \tfrac23 = \tfrac13. Both orders give 13\tfrac13 — Fubini in action, and a complete template: inner antiderivative, evaluate at the xx-dependent limits, then an ordinary outer integral.

Example
Worked: a polar integral with the extra r, start to finish

Compute ∬D(x2+y2) dA\iint_D (x^2 + y^2)\,dA over the disk of radius 22.

Step 1 — translate to polar: x2+y2=r2x^2 + y^2 = r^2 and dA=r dr dθdA = r\,dr\,d\theta, so the integrand becomes r2⋅r=r3r^2\cdot r = r^3.

Step 2 — limits: the full disk is 0≤r≤20 \le r \le 2, 0≤θ≤2π0 \le \theta \le 2\pi.

Step 3 — inner integral: ∫02r3dr=[r44]02=4\int_0^2 r^3dr = \left[\tfrac{r^4}{4}\right]_0^2 = 4.

Step 4 — outer: ∫02π4 dθ=8π\int_0^{2\pi}4\,d\theta = 8\pi.

Step 5 — the cautionary rerun: forgetting the extra rr gives ∫02π ⁣∫02r2dr dθ=16π3≈16.8\int_0^{2\pi}\!\int_0^2 r^2dr\,d\theta = \tfrac{16\pi}{3} \approx 16.8 instead of 8π≈25.18\pi \approx 25.1 — a wrong answer with no warning sign attached. Writing "dA=r dr dθdA = r\,dr\,d\theta" before computing anything is the habit that prevents it.