Parametric surfaces — tangent planes, surface area: Worked examples

This page is the "throw everything at it" companion to Parametric surfaces — tangent planes, surface area. We first lay out a matrix of every kind of situation these problems can hand you, then work examples until every single cell of that matrix has been touched. Nothing here assumes more than: you can differentiate, you know Cross Product gives a vector perpendicular to two others, and you can do a double integral.

Before we start, the one-line reminder that everything hangs on:

The scenario matrix

Every exam problem on this topic is really one of these cells. The column tells you what is asked; the row tells you what makes it awkward.

Complication ↓ \ Task →	Tangent plane	Surface area
Clean / textbook	(a) paraboloid	(b) cone piece
Sign / orientation matters	(c) which way does $n$ point?	—
Degenerate point ( $r_{u} \times r_{v} = 0$ )	(d) cone tip	(d) tip contributes zero
Graph shortcut $z = f (x, y)$	(e) reuses $⟨ - f_{x}, - f_{y}, 1 ⟩$	(e) $f_{x}^{2} + f_{y}^{2} + 1$
Non-orthogonal grid ( $θ \neq = 9 0^{\circ}$ )	—	(f) sheared plane
Real-world word problem	—	(g) paint on a dome
Exam twist (limiting value)	—	(h) area $\to$ known formula

We now hit each cell. Watch the labels — each example says which cell it fills.

(a) Clean tangent plane — cell "textbook / plane"

Step 1. Compute the tangent vectors. $r_{u} = ⟨ 1, 0, 2 u ⟩ = ⟨ 1, 0, 4 ⟩, r_{v} = ⟨ 0, 1, - 2 v ⟩ = ⟨ 0, 1, - 2 ⟩ .$ Why this step? $r_{u}$ is the velocity if you slide in the $u$ -direction; it lies in the surface, so it must lie in the tangent plane. Same for $r_{v}$ . Two in-plane directions is all a plane needs.

Step 2. Cross them for the normal. $n = r_{u} \times r_{v} = ⟨ (0) (- 2) - (4) (1), (4) (0) - (1) (- 2), (1) (1) - (0) (0) ⟩ = ⟨ - 4, 2, 1 ⟩ .$ Why this step? A plane is pinned down by a point plus a perpendicular direction. The Cross Product is the exact tool that manufactures a vector perpendicular to two given ones.

Step 3. Assemble the plane through $(2, 1, 3)$ . $- 4 (x - 2) + 2 (y - 1) + 1 (z - 3) = 0 ⟹ - 4 x + 2 y + z + 3 = 0 ⟹ z = 4 x - 2 y - 3.$ Why this step? Any vector from $(2, 1, 3)$ to another point $(x, y, z)$ in the plane is $⊥ n$ , so its dot product with $n$ is zero — that dot product is the equation.

Verify: plug the point in: $z = 4 (2) - 2 (1) - 3 = 8 - 2 - 3 = 3$ ✓. And $n \cdot r_{u} = - 4 + 0 + 4 = 0$ , $n \cdot r_{v} = 0 + 2 - 2 = 0$ — normal really is perpendicular to both tangents ✓.

Figure — Parametric surfaces — tangent planes, surface area

(b) Clean surface area — cell "textbook / area"

Step 1. Tangent vectors. $r_{u} = ⟨ cos θ, sin θ, 1 ⟩, r_{θ} = ⟨ - u sin θ, u cos θ, 0 ⟩ .$ Why? $r_{u}$ points outward-and-up along a straight ruling of the cone; $r_{θ}$ points around the circle at height $u$ .

Step 2. Cross product. $r_{u} \times r_{θ} = ⟨ - u cos θ, - u sin θ, u ⟩ .$ Why? We want the area stretch factor $∣ r_{u} \times r_{θ} ∣$ ; the Double Integrals machinery needs this scalar so each tiny $(u, θ)$ rectangle counts its true curved area, not its flat area.

Step 3. Magnitude. $∣ r_{u} \times r_{θ} ∣ = u^{2} cos^{2} θ + u^{2} sin^{2} θ + u^{2} = u^{2} + u^{2} = u 2 .$ Why? $cos^{2} θ + sin^{2} θ = 1$ collapses the $θ$ -dependence, leaving a clean stretch factor.

Step 4. Integrate. $A = \int_{0}^{2 π} \int_{0}^{2} u 2 d u d θ = 2 (2 π) [\frac{u ^{2}}{2}]_{0}^{2} = 2 (2 π) (2) = 42 π .$

Verify: $42 π \approx 17.77 > 4 π \approx 12.57$ — bigger than the flat disk, as forecast ✓. Sanity: a cone of slant $ℓ$ and base radius $R$ has lateral area $π R ℓ$ ; here $R = 2$ , $ℓ = 22$ , giving $π (2) (22) = 42 π$ ✓.

(c) Sign / orientation — cell "which way does the normal point?"

Step 1. Tangent vectors at $(\frac{π}{2}, 0)$ . $r_{ϕ} = ⟨ cos ϕ cos θ, cos ϕ sin θ, - sin ϕ ⟩ = ⟨ 0, 0, - 1 ⟩,$ $r_{θ} = ⟨ - sin ϕ sin θ, sin ϕ cos θ, 0 ⟩ = ⟨ 0, 1, 0 ⟩ .$ Why? We evaluate at the specific point because orientation is a local question.

Step 2. Cross in the given order $r_{ϕ} \times r_{θ}$ . $⟨ 0, 0, - 1 ⟩ \times ⟨ 0, 1, 0 ⟩ = ⟨ (0) (0) - (- 1) (1), (- 1) (0) - (0) (0), (0) (1) - (0) (0) ⟩ = ⟨ 1, 0, 0 ⟩ .$ Why? The order of factors in a Cross Product flips the sign; for flux and orientation the sign is the whole point, so we must respect the given order.

Step 3. Compare with outward direction $⟨ 1, 0, 0 ⟩$ : identical → this ordering gives the outward normal.

Verify: $r_{ϕ} \times r_{θ} = ⟨ 1, 0, 0 ⟩$ points radially outward from the origin at $(1, 0, 0)$ ✓. Reversing to $r_{θ} \times r_{ϕ}$ would give $⟨ - 1, 0, 0 ⟩$ — the inward normal, confirming order controls sign.

(d) Degenerate point — cell " $r_{u} \times r_{v} = 0$ "

Step 1. Evaluate the cross product at $u = 0$ . From (b), $r_{u} \times r_{θ} = ⟨ - u cos θ, - u sin θ, u ⟩$ . At $u = 0$ this is $⟨ 0, 0, 0 ⟩$ . Why? The surface's smoothness test is exactly $r_{u} \times r_{v} \neq = 0$ . Here it fails.

Step 2. Interpret. A zero normal means the two tangent vectors are not independent ( $r_{θ} = ⟨ 0, 0, 0 ⟩$ at the tip — the whole circle has collapsed to a point). No 2D plane is spanned, so there is no tangent plane at the apex. Why? A plane needs two independent in-surface directions; at a cusp you don't have them.

Step 3. Does this wreck the area integral of (b)? No — the bad point is a single value $u = 0$ , a set of zero width. The integrand $u 2 \to 0$ there anyway, so it contributes nothing.

Verify: $u \to 0 lim ∣ r_{u} \times r_{θ} ∣ = u \to 0 lim u 2 = 0$ ✓ — the degenerate point is integrable and the area $42 π$ from (b) is unharmed.

(e) Graph shortcut — cell " $z = f (x, y)$ , both tasks at once"

Part (i).

Step 1. $f_{x} = 2 x, f_{y} = 2 y$ . At $(1, 2)$ : $f_{x} = 2, f_{y} = 4$ , point $(1, 2, 5)$ . Why? The graph parametrization pre-computes $r_{x} = ⟨ 1, 0, f_{x} ⟩$ , $r_{y} = ⟨ 0, 1, f_{y} ⟩$ , so the normal is the fixed pattern $⟨ - f_{x}, - f_{y}, 1 ⟩$ .

Step 2. Normal $= ⟨ - 2, - 4, 1 ⟩$ . Plane: $- 2 (x - 1) - 4 (y - 2) + 1 (z - 5) = 0 ⟹ z = 2 x + 4 y - 5.$ Verify (i): at $(1, 2)$ , $z = 2 + 8 - 5 = 5$ ✓.

Part (ii).

Step 3. Stretch factor $f_{x}^{2} + f_{y}^{2} + 1 = 4 x^{2} + 4 y^{2} + 1$ . In polar ( $x^{2} + y^{2} = ρ^{2}$ , $d A = ρ d ρ d θ$ , see Change of Variables and the Jacobian): $A = \int_{0}^{2 π} \int_{0}^{1} 4 ρ^{2} + 1 ρ d ρ d θ .$ Why polar? The stretch factor depends only on $ρ$ , so polar makes the inner integral a clean substitution.

Step 4. Let $w = 4 ρ^{2} + 1$ , $d w = 8 ρ d ρ$ : $\int_{0}^{1} 4 ρ^{2} + 1 ρ d ρ = \frac{1}{8} \int_{1}^{5} w d w = \frac{1}{8} \cdot \frac{2}{3} [w^{3/2}]_{1}^{5} = \frac{1}{12} (5^{3/2} - 1) .$ $A = 2 π \cdot \frac{5 ^{3/2} - 1}{12} = \frac{π}{6} (55 - 1) .$

Verify (ii): $\frac{π}{6} (55 - 1) \approx \frac{π}{6} (11.18 - 1) \approx 5.33 > π \approx 3.14$ — curved surface beats the flat disk, as forecast ✓.

(f) Non-orthogonal grid — cell " $θ \neq = 9 0^{\circ}$ , dot vs cross matters"

Step 1. Tangents: $r_{u} = ⟨ 1, 0, 0 ⟩$ , $r_{v} = ⟨ 1, 1, 0 ⟩$ . Why? Simple differentiation; note they point in different, non-perpendicular directions.

Step 2. The wrong naive factor $∣ r_{u} ∣∣ r_{v} ∣ = 1 \cdot 2 = 2$ . Why show the wrong one? Students assume the stretch factor is length-times-length. That's only right when the grid is orthogonal.

Step 3. The correct cross product: $r_{u} \times r_{v} = ⟨ 0, 0, 1 ⟩, ∣ r_{u} \times r_{v} ∣ = 1.$ Why cross, not the length product? $∣ r_{u} \times r_{v} ∣ = ∣ r_{u} ∣∣ r_{v} ∣ sin θ$ . Here the angle $θ$ between the tangents satisfies $cos θ = \frac{r _{u} \cdot r _{v}}{∣ r _{u} ∣∣ r _{v} ∣} = \frac{1}{2}$ , so $θ = 4 5^{\circ}$ and $sin θ = \frac{1}{2}$ . Then $2 \cdot \frac{1}{2} = 1$ .

Step 4. Area $= \int_{0}^{1} \int_{0}^{1} 1 d u d v = 1$ .

Verify: cross-product area $= 1$ ; the naive product gave $2 \approx 1.414$ — an over-count by the factor $1/ sin 4 5^{\circ}$ . Forecast of $1$ confirmed ✓.

(g) Real-world word problem — cell "paint on a dome"

Step 1. Parametrize the top half: $ϕ \in [0, \frac{π}{2}]$ , $θ \in [0, 2 π]$ , $R = 3$ . Why? $ϕ = 0$ is the top; $ϕ = \frac{π}{2}$ is the equator — exactly the upper half.

Step 2. Reuse the sphere stretch factor $∣ r_{ϕ} \times r_{θ} ∣ = R^{2} sin ϕ = 9 sin ϕ$ (derived in the parent note via $sin^{2} + cos^{2} = 1$ ).

Step 3. Area: $A = \int_{0}^{2 π} \int_{0}^{π /2} 9 sin ϕ d ϕ d θ = 9 (2 π) [- cos ϕ]_{0}^{π /2} = 9 (2 π) (1) = 18 π m^{2} .$ Why the limit $π /2$ ? Only the top half is painted, so $ϕ$ stops at the equator.

Step 4. Litres $= \frac{18 π}{8} = \frac{9 π}{4} \approx 7.07$ litres.

Verify: $18 π \approx 56.5$ m² matches $2 π R^{2}$ ✓; $9 π /4 \approx 7.07$ litres, matching the forecast. Units: $m^{2} \div (m^{2} / litre) = litres$ ✓.

(h) Exam twist — limiting value — cell "area $\to$ known formula"

Step 1. For $z = c$ : $f_{x} = 0, f_{y} = 0$ , so $f_{x}^{2} + f_{y}^{2} + 1 = 1 = 1$ . Why? This is the sanity check every formula must pass — a flat surface must not be stretched.

Step 2. $A = \iint_{D} 1 d A = area (D)$ . ✓ The formula degenerates correctly.

Step 3. For $z = k x$ over the unit square $[0, 1]^{2}$ : $f_{x} = k, f_{y} = 0$ , factor $= 1 + k^{2}$ , constant, so $A = 1 + k^{2} \cdot area = 1 + k^{2} .$ Why? Tilting a unit square by slope $k$ stretches its length along $x$ by exactly $1 + k^{2}$ (the hypotenuse of a rise- $k$ , run- $1$ triangle) — pure Arc Length logic promoted to a surface.

Step 4. Limits: $k \to 0 \Rightarrow A \to 1$ (flat); $k \to \infty \Rightarrow A \to \infty$ (vertical wall, infinite as projected onto $D$ ).

Verify: at $k = 0$ , $A = 1$ ✓; at $k = 1$ , $A = 2 \approx 1.414$ ✓; monotone increasing in $∣ k ∣$ ✓.

Recall check

Recall Which product for which job?

Area needs which product of $r_{u}, r_{v}$ , and why not the other? ::: The cross product's magnitude $∣ r_{u} ∣∣ r_{v} ∣ sin θ$ — it measures spread and is zero for parallel edges; the dot product measures alignment and is zero for perpendicular edges (opposite of what area wants).

Recall Degenerate points

What does $r_{u} \times r_{v} = 0$ tell you? ::: The tangents are dependent (a cusp/tip); no tangent plane exists there, but a single such point contributes zero to an area integral.

Recall Non-orthogonal grid trap

When does $∣ r_{u} ∣∣ r_{v} ∣$ equal the true area factor? ::: Only when the grid is orthogonal ( $θ = 9 0^{\circ}$ , $sin θ = 1$ ); otherwise it overestimates by $1/ sin θ$ .