Surface integrals — scalar and vector (flux): Worked examples

This page is a drill floor. The parent note built the two machines:

scalar: $\iint_{S} f d S = \iint_{D} f ∥ r_{u} \times r_{v} ∥ d u d v$ ,
flux: $\iint_{S} F \cdot d S = \iint_{D} F \cdot (r_{u} \times r_{v}) d u d v$ .

Now we run them through every kind of surface and every trap so you never meet a scenario you haven't seen.

Before we start, two words we will lean on:

Component of a vector means one of its three numbers, e.g. $F = (P, Q, R)$ has components $P, Q, R$ .
Normal means "sticking straight out of the surface, at $9 0^{\circ}$ to it" — like a nail hammered perpendicular into a wall.

The scenario matrix

Every surface-integral problem falls into one of these cells. The examples below are labelled [C#] so you can see we cover them all.

Cell	What makes it different	Watch out for
C1 Graph $z = g (x, y)$ , scalar	flat parameter square, need the square root	slanted patches are bigger than they look
C2 Graph $z = g (x, y)$ , flux	root cancels, use $(- P g_{x} - Q g_{y} + R)$	which way does the normal point?
C3 Explicit parametrization (cylinder / sphere)	tangents from $\partial / \partial u, \partial / \partial v$	plug parametrization into $F$
C4 Orientation sign flip	outward vs inward, upward vs downward	flux changes sign
C5 Degenerate / zero input	flat plate ( $g_{x} = g_{y} = 0$ ), or $F$ purely sideways	flux can be exactly zero
C6 Closed surface + limiting check	glue faces; verify with Divergence Theorem	did you include every face?
C7 Real-world word problem	density → mass, or velocity → flow rate	units must come out right
C8 Exam twist: constant field through slanted disk	$F$ constant, tilted surface	only the perpendicular part counts

C1 — Scalar integral over a slanted plane (mass of a sheet)

Forecast: the flat triangle in the $x y$ -plane has area $\frac{1}{2} \cdot 2 \cdot 2 = 2$ . The tilted sheet is bigger. Guess: is your answer $> 2$ ?

Parametrize as a graph $r (x, y) = (x, y, 2 - x - y)$ . Why this step? The surface is written as $z = g (x, y)$ , so we reuse the graph formula from the parent note — no need to invent a new parametrization.
Slopes: $g_{x} = - 1, g_{y} = - 1$ . Why this step? The stretch factor $1 + g_{x}^{2} + g_{y}^{2}$ needs these slopes; they measure how steeply the plane rises.
Stretch factor $1 + (- 1)^{2} + (- 1)^{2} = 3$ . Why this step? This is the number that says "each flat patch below covers $3$ times as much slanted area." See the figure — the shadow patch and its slanted parent.
Integrate: since the factor is constant, area $= 3 \cdot (area of D) = 3 \cdot 2 = 23$ . Why this step? A constant pulls out of the double integral; the leftover $\iint_{D} d x d y$ is just the base triangle's area.

Figure — Surface integrals — scalar and vector (flux)

Verify: $23 \approx 3.464$ . Sanity: a plane at slope so that it rises $2$ over a horizontal run — the tilt factor $3 \approx 1.73$ is a reasonable "how much longer than the shadow." ✔

C2 — Flux through the same graph (no square root!)

Forecast: $F$ points straight up with strength $z$ . The surface tilts up. Guess: is the flux positive?

Recall the graph-flux formula: upward normal gives $\iint_{D} (- P g_{x} - Q g_{y} + R) d x d y$ with $F = (P, Q, R)$ . Why this step? Flux uses $r_{x} \times r_{y} = (- g_{x}, - g_{y}, 1)$ directly — the norm cancels, so no square root here (contrast C1).
Read off $P = 0, Q = 0, R = z = 2 - x - y$ . So integrand $= - 0 - 0 + (2 - x - y) = 2 - x - y$ . Why this step? We substitute the surface's $z = 2 - x - y$ into $R$ ; everything must become a function of $x, y$ before integrating.
Integrate over the triangle: $\int_{0}^{2} \int_{0}^{2 - x} (2 - x - y) d y d x .$ Why this step? For fixed $x$ , $y$ runs from $0$ up to the line $x + y = 2$ , i.e. $y = 2 - x$ .
Inner: $[(2 - x) y - \frac{y ^{2}}{2}]_{0}^{2 - x} = (2 - x)^{2} - \frac{( 2 - x ) ^{2}}{2} = \frac{1}{2} (2 - x)^{2}$ . Outer: $\int_{0}^{2} \frac{1}{2} (2 - x)^{2} d x = \frac{1}{2} \cdot \frac{( 2 - x ) ^{3}}{- 3}_{0}^{2} \cdot (- 1) = \frac{1}{2} \cdot \frac{8}{3} = \frac{4}{3}$ .

Verify: flux $= \frac{4}{3} > 0$ , matching the forecast (up-field through up-tilted surface). Notice we never wrote $3$ — that's the beauty of flux. ✔

C3 — Explicit parametrization: side of a cylinder

Forecast: $F$ points radially outward (its $x y$ -part is the position). The wall's normal also points radially out. So the field pierces the wall head-on everywhere. Guess: big positive flux.

Parametrize the wall with angle $θ$ and height $z$ : $r (θ, z) = (2 cos θ, 2 sin θ, z), θ \in [0, 2 π], z \in [0, 5] .$ Why this step? A cylinder is not a graph $z = g (x, y)$ (it's vertical, $z$ can't be a function of $x, y$ ). So we choose the two natural knobs: go around ( $θ$ ) and go up ( $z$ ).
Tangent vectors $r_{θ} = (- 2 sin θ, 2 cos θ, 0)$ , $r_{z} = (0, 0, 1)$ . Why this step? $r_{θ}$ runs along the circular direction, $r_{z}$ straight up; their cross product is normal to both, i.e. sticking out of the wall.
Cross product using Cross product: $r_{θ} \times r_{z} = (2 cos θ, 2 sin θ, 0) .$ Why this step? Its $z$ -component is $0$ and its $x y$ -part points outward (same direction as position) — exactly the outward normal the problem wants. See the figure.
Substitute into $F$ : on the wall $F = (2 cos θ, 2 sin θ, 0)$ . Dot with the normal: $F \cdot (r_{θ} \times r_{z}) = 4 cos^{2} θ + 4 sin^{2} θ = 4.$ Why this step? $cos^{2} + sin^{2} = 1$ collapses it to a constant — the field meets the wall perpendicularly with the same strength everywhere.
Integrate: $\int_{0}^{5} \int_{0}^{2 π} 4 d θ d z = 4 \cdot 2 π \cdot 5 = 40 π .$

Verify: the wall area is $2 π r h = 2 π \cdot 2 \cdot 5 = 20 π$ , and $F \cdot \hat{n} = 2$ (radial strength $= r = 2$ ) constant, so flux $= 2 \cdot 20 π = 40 π$ . ✔ Matches.

C4 — Orientation flip: same cylinder, inward normal

Forecast: we just flipped the normal to point into the cylinder. Guess the answer instantly.

Swap the cross-product order: inward normal $= r_{z} \times r_{θ} = - (r_{θ} \times r_{z})$ . Why this step? The parent note's mistake box: reversing $r_{u} \times r_{v} \to r_{v} \times r_{u}$ flips the sign. Orientation is a choice and it changes the answer's sign.
Every integrand becomes its negative: $- 4$ .
Integrate: $- 40 π$ .

Verify: flux flipped sign, $∣ v a l u e ∣$ unchanged: $- 40 π$ . A negative flux means the field is leaving through the outside, i.e. flowing against the inward normal. ✔ This is the "flux has a sign" lesson made concrete.

C5 — Degenerate / zero cases

Forecast (a): the plate is not tilted at all. Its area should equal its shadow, $1$ . Forecast (b): the field is horizontal, the surface is flat and its normal points straight up — does any fluid go through?

Part (a):

$g (x, y) = 3 \Rightarrow g_{x} = g_{y} = 0$ . Why? A constant has zero slope — no tilt, so no stretching.
Stretch factor $1 + 0 + 0 = 1$ . Area $= \iint_{D} 1 d x d y = 1$ . Why this matters: this is the degenerate end of C1 — the square root becomes $1$ and the sheet equals its shadow. Sanity check on the whole $1 + g_{x}^{2} + g_{y}^{2}$ machinery.

Part (b):

The disk lies in the plane $z = 0$ ; upward normal $\hat{n} = (0, 0, 1)$ . Why? The disk is horizontal, so "straight out" is straight up.
$F \cdot \hat{n} = (1, 0, 0) \cdot (0, 0, 1) = 0$ . Why this step? The dot product picks out the part of $F$ along the normal. A horizontal field has nothing in the vertical direction — it slides across the disk, never through it.
Flux $= \iint_{D} 0 d S = 0$ .

Verify: (a) $= 1$ , (b) $= 0$ . Both match forecasts. (b) is the pure "sideways flow slides along, not through" picture — flux is genuinely zero, not just small. ✔

C6 — Closed surface + Divergence-Theorem limiting check

Forecast: closed surface, divergence $\nabla \cdot F = 1 + 1 + 1 = 3$ . By the Divergence Theorem the total flux should be $3 \times (volume)$ . Volume $= π r^{2} h = π \cdot 4 \cdot 5 = 20 π$ . Guess: $60 π$ .

We sum three faces — you must include every face or you undercount.

Side wall. From C3-style work with $F = (x, y, z)$ : $F \cdot (r_{θ} \times r_{z}) = (2 cos θ, 2 sin θ, z) \cdot (2 cos θ, 2 sin θ, 0) = 4$ . Flux $_{side} = \int_{0}^{5} \int_{0}^{2 π} 4 d θ d z = 40 π$ . Why: the $z$ -part of $F$ dots with the wall's zero $z$ -normal → drops out, same $4$ as before.
Top disk $z = 5$ , outward normal $= (0, 0, 1)$ . $F \cdot \hat{n} = z = 5$ . Area $= π \cdot 2^{2} = 4 π$ . Flux $_{top} = 5 \cdot 4 π = 20 π$ . Why the normal is $(0, 0, 1)$ : "outward" at the top means upward.
Bottom disk $z = 0$ , outward normal $= (0, 0, - 1)$ (outward means downward here!). $F \cdot \hat{n} = (x, y, 0) \cdot (0, 0, - 1) = 0$ . Flux $_{bottom} = 0$ . Why the normal is $(0, 0, - 1)$ : the bottom face's outside is below it — a classic sign trap (C4 flavour).
Total $= 40 π + 20 π + 0 = 60 π$ .

Verify (limiting / theorem check): Divergence Theorem gives $∭ 3 d V = 3 \cdot 20 π = 60 π$ . ✔ Exactly matches, and it forced us to remember the bottom face flips its normal.

C7 — Real-world word problem: rainfall through a roof

Forecast: rain comes down, roof faces up. We want how much passes through per second — a flow rate in m³/s. Guess the sign convention: use the upward normal, then the physically-caught rate is $-$ (flux) since rain moves down.

Graph flux formula, $g = 4 - \frac{1}{2} x$ , so $g_{x} = - \frac{1}{2}, g_{y} = 0$ , upward normal $(- g_{x}, - g_{y}, 1) = (\frac{1}{2}, 0, 1)$ . Why: the roof is $z = g (x, y)$ , straightforward graph case.
$F = (0, 0, - 3) = (P, Q, R)$ , integrand $= - P g_{x} - Q g_{y} + R = 0 - 0 + (- 3) = - 3$ . Why this step: only $R = - 3$ survives because $P = Q = 0$ (rain has no horizontal motion). The dot product keeps only the vertical part.
Integrate over the base rectangle (area $2 \times 3 = 6$ ): flux $= - 3 \cdot 6 = - 18$ m³/s. Why the base area, not roof area: the graph-flux formula already integrates over the flat $x y$ -region $D$ ; the tilt is baked into the $(- g_{x}, - g_{y}, 1)$ normal (no separate ).
Water caught per second $= - (upward flux) = 18$ m³/s. Why negate: upward flux is negative because rain flows opposite to the upward normal; the amount actually collected is that magnitude.

Verify (units & geometry): rain speed $3$ m/s $\times$ horizontal footprint $6$ m² $= 18$ m³/s. The horizontal footprint is what catches vertical rain — the roof's tilt does not increase catch for vertical rain, which is why the $\frac{1}{2}$ slope cancelled out. Units: (m/s)(m²) $=$ m³/s. ✔

C8 — Exam twist: constant field through a tilted disk

Forecast: flux of a constant field through a flat surface $= (F \cdot \hat{n}) \times (area)$ — no integral needed because nothing varies. Guess: positive but less than $6 π$ because the plate is tilted away from the wind.

Dot product $F \cdot \hat{n} = (0, 0, 6) \cdot \frac{1}{3} (1, 1, 1) = \frac{6}{3} = 23$ . Why: only the component of $F$ along $\hat{n}$ goes through; the dot product extracts it.
Area of plate $= π r^{2} = π$ . Why: the field and normal are both constant, so $\iint (F \cdot \hat{n}) d S = (F \cdot \hat{n}) \iint d S = (F \cdot \hat{n}) \cdot Area$ .
Flux $= 23 \cdot π = 23 π$ .

Verify: $23 π \approx 10.88$ . Compare a plate facing the wind directly ( $\hat{n} = (0, 0, 1)$ ): flux would be $6 π \approx 18.85$ . Our tilted plate catches the fraction $\frac{2 3}{6} = \frac{1}{3} \approx 0.577$ of that — the tilt "cosine" — which matches: only $57.7%$ of the wind is perpendicular. ✔ Positive and less than $6 π$ , as forecast.

Recall One-line recall for each cell

C1 slanted-plane area is bigger than its shadow by ::: $1 + g_{x}^{2} + g_{y}^{2}$ C2 graph flux integrand (upward) is ::: $- P g_{x} - Q g_{y} + R$ , no square root C3 cylinder-wall outward normal is ::: $r_{θ} \times r_{z} = (2 cos θ, 2 sin θ, 0)$ C4 reversing the normal does what to flux? ::: flips its sign C5 flux of a horizontal field through a horizontal disk is ::: exactly $0$ C6 to check a closed-surface flux use ::: the Divergence Theorem, $∭ \nabla \cdot F d V$ C7 vertical rain caught depends on the ::: horizontal footprint area, not the tilt C8 constant field, flat plate: flux $=$ ::: $(F \cdot \hat{n}) \times Area$