Difficulty: Intermediate | Prerequisites: Sections 16.1 – 16.5, parametric curves, partial derivatives, cross product
Parametric surfaces extend the idea of parametric curves to two dimensions: instead of one parameter tracing a curve, two parameters (u, v) sweep out a surface in 3D. This is the machinery you need before you can compute surface integrals in Section 16.7. The core skills are writing a parametrisation for a given surface, computing the tangent and normal vectors from partial derivatives, and recognising the shape of a surface from its parametric equations. If you can parametrise curves confidently, the jump here is mostly bookkeeping: two parameters instead of one, a cross product instead of a derivative.
A parametric surface r(u, v) = ⟨x(u,v), y(u,v), z(u,v)⟩ maps a rectangle in the uv-plane to a surface in 3D. You find the tangent plane at any point using the cross product r_u × r_v. The magnitude of that cross product is the local area-scaling factor you will need for surface integrals. To identify a surface, fix one parameter and see what curve the other traces.
Parametric surface
A vector-valued function r(u, v) = ⟨x(u,v), y(u,v), z(u,v)⟩ defined on a domain D in the uv-plane that traces out a surface in R³. Think of it as a coordinate system painted onto the surface.
Parameter domain
The region D in the uv-plane over which the parameters range. For a sphere in spherical coordinates, the domain is typically 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π.
Grid curves
The curves you get by holding one parameter fixed and letting the other vary. Fixing u gives one family of curves; fixing v gives another. Together they form a grid on the surface.
Tangent vectors (r_u and r_v)
The partial derivatives of r with respect to u and v. Each is a vector tangent to the surface along one family of grid curves.
Normal vector
The cross product r_u × r_v, which is perpendicular to the surface at each point. Its magnitude |r_u × r_v| gives the local area-scaling factor.
Tangent plane
The plane through a point on the surface, spanned by r_u and r_v at that point (equivalently, the plane with normal r_u × r_v). In simple terms, if you zoomed in on the surface at that point, it would look flat, and that flat plane is the tangent plane.
The homework technique: fix one parameter, see what curve the other traces.
Ex 13 (Graph IV, helix surface): r(u,v) = ⟨u cos v, u sin v, v⟩. Fix u: x/u = cos v, y/u = sin v, giving circles, so the surface wraps around. Fix v: straight lines. The surface is a helicoid.
Ex 14 (Graph VI, saddle-like): Fixing v gives u = x/c², turning z = u² − c² into z = x²/c⁴ − c², a parabola. Fixing u similarly gives a downward parabola. The surface has parabolic cross-sections in both directions.
Ex 15 (Graph I): Fixing v gives x = u³ − u, y = c², z = u², a space curve at constant y. Fixing u gives x = c³ − c, z = c², constant y. The surface is a folded sheet.
Ex 16 (Graph V, pinched surface): r involves (3+cos v) factors, making a surface that narrows toward a point. Substituting u = 0 gives a semicircle; u = 1 gives a single point, confirming the pinch.
Ex 17 (Graph III): Fixing v gives x/cos³(u) = cos³(c), relating x and y through powers of trig functions, producing an astroid-like cross-section.
Ex 18 (Graph II): Fixing v gives x = sin u, y = cos(u)·c, z = c, showing circular cross-sections at each height.
Given r(u,v) = ⟨(2+sin v)cos u, (2+sin v)sin u, u + cos v⟩:
The tangent plane is parallel to the yz-plane when the normal vector r_u × r_v is parallel to ⟨1, 0, 0⟩, meaning the j and k components of the cross product are both zero.
Computing r_u and r_v, taking their cross product, and setting the j and k components to zero yields cos v = 0, so v = π/2 or 3π/2. Then the remaining equation gives sin u = 0, so u = nπ.
The surface points with tangent plane parallel to the yz-plane are at (u, v) = (2kπ, π/2), ((2k+1)π, π/2), (2kπ, 3π/2), and ((2k+1)π, 3π/2), giving the points ⟨3, 0, 2kπ⟩, ⟨−3, 0, (2k+1)π⟩, ⟨1, 0, 2kπ⟩, and ⟨−1, 0, (2k+1)π⟩.
Surface: r(u,v) = ⟨u, v cos u, v sin u⟩ for 0 ≤ u ≤ 6π, 2 ≤ v ≤ 4.
To sketch: fix u = 0, the surface starts as the line segment (0, v, 0) for v in [2, 4]. Fix v = constant, the result is a helix. So the surface is a helical ribbon winding around the x-axis.
For the area, compute r_u × r_v, find its magnitude is √(v² + 1), then integrate:
Area = ∫₂⁴ ∫₀⁶π √(v² + 1) du dv = 6π ∫₂⁴ √(v² + 1) dv
Using the hint f(x) = x√(1+x²) + ln(x + √(1+x²)) with f'(x) = 2√(1+x²), the integral evaluates to 3π[ln(√(v²+1) + v) + v√(v²+1)] from 2 to 4.
Problem 7 (upper half of an ellipsoid): The equation 4x² − 8x + 9y² + 36(z+1)² = 32 completes the square to 4(x−1)² + 9y² + 36(z+1)² = 36. Setting a = 2(x−1), b = 3y, c = 6(z+1) gives a² + b² + c² = 36, a sphere of radius 6 in (a, b, c) space. Parametrise in spherical coordinates and convert back.
Problem 8 (sphere inside cones): For a sphere of radius 4 centred at the origin, the constraint "inside the cone z = √(3(x² + y²))" translates in spherical coordinates to a bound on the polar angle φ. The cone z = √3·r (where r = √(x²+y²)) gives ρ cos φ = √3 · ρ sin φ, so tan φ = 1/√3, meaning φ = π/6. The parametrisation is then r(θ, φ) = ⟨4 sin φ cos θ, 4 sin φ sin θ, 4 cos φ⟩ with 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ π/6.
For different cones (x = √(⅓(y²+z²)) or y = √(x²+z²)), you swap the roles of x, y, z in the spherical coordinate setup, and the angle bound changes accordingly.
\mathbf{r}(u,v) = \langle x(u,v),\; y(u,v),\; z(u,v) \rangle\mathbf{r}_u = \left\langle \frac{\partial x}{\partial u},\; \frac{\partial y}{\partial u},\; \frac{\partial z}{\partial u} \right\rangle, \quad \mathbf{r}_v = \left\langle \frac{\partial x}{\partial v},\; \frac{\partial y}{\partial v},\; \frac{\partial z}{\partial v} \right\rangle\text{Normal vector: } \mathbf{n} = \mathbf{r}_u \times \mathbf{r}_v\text{Surface area} = \iint_D |\mathbf{r}_u \times \mathbf{r}_v|\, dASphere of radius ρ₀ centred at (x₀, y₀, z₀):
\mathbf{r}(\phi, \theta) = \langle \rho_0 \sin\phi\cos\theta + x_0,\; \rho_0 \sin\phi\sin\theta + y_0,\; \rho_0 \cos\phi + z_0 \rangleFor a sphere of radius ρ₀, |r_φ × r_θ| = ρ₀² sin φ.
Students often mix up the roles of φ and θ in spherical coordinates. In the Stewart convention, φ is the angle from the positive z-axis (polar angle) and θ is the angle in the xy-plane (azimuthal angle). Some physics textbooks swap them.
A frequent mistake is computing r_u × r_v in the wrong order. Reversing the order flips the sign of the normal. This matters when orientation is important (Section 16.7).
Students sometimes think they need to parametrise every surface from scratch. Many standard surfaces (sphere, cylinder, cone, torus) have known parametrisations worth memorising.
When a cone is written as z = √(3(x² + y²)), students sometimes forget to convert to spherical coordinates and instead try to integrate in Cartesian, making the problem far harder.
⚠️ "Fix one parameter and describe the resulting curve" is a very common exam question format. Practise this technique until it is automatic.
⚠️ Computing r_u × r_v and its magnitude is the single most-tested skill from this section. It feeds directly into surface area and surface integral calculations.
⚠️ Parametrising spheres inside cones (Problem 8) requires translating a Cartesian inequality into a bound on φ. This type of problem appears frequently on finals.
⚠️ Know how to complete the square and change variables to recognise an ellipsoid or shifted sphere (Problem 7).
True or false: for a parametric surface r(u,v), the vector r_u × r_v is tangent to the surface. False – it is normal (perpendicular) to the surface.
Fill in the blank: the surface area element for a parametric surface is dS = ___ dA. |r_u × r_v|
True or false: a sphere of radius 4 in spherical coordinates has |r_φ × r_θ| = 4 sin φ. False – it is 16 sin φ (that is 4² sin φ).
Fill in the blank: the cone z = √(3(x² + y²)) corresponds to the polar angle φ = ___. π/6
True or false: grid curves on a parametric surface are obtained by fixing both parameters. False – you fix one and vary the other.
Q: Given r(u,v) = ⟨u, v cos u, v sin u⟩, find the tangent vectors r_u and r_v.
A: r_u = ⟨1, −v sin u, v cos u⟩ and r_v = ⟨0, cos u, sin u⟩.
Q: For the surface r(u,v) = ⟨u cos v, u sin v, v⟩, describe the grid curves you see when u is fixed.
A: With u fixed at u = c, the curve is r = ⟨c cos v, c sin v, v⟩, a helix of radius c winding upward.
Q: Parametrise the part of the sphere x² + y² + z² = 16 that lies inside the cone z = √(3(x² + y²)).
A: r(θ, φ) = ⟨4 sin φ cos θ, 4 sin φ sin θ, 4 cos φ⟩ with 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ π/6.
Q: How do you determine when a tangent plane to a parametric surface is parallel to the yz-plane?
A: The normal vector r_u × r_v must be parallel to ⟨1, 0, 0⟩, so the j and k components of the cross product must both be zero.
Q: What does the equation 4(x−1)² + 9y² + 36(z+1)² = 36 represent?
A: An ellipsoid centred at (1, 0, −1) with semi-axes a = 3, b = 2, c = 1 (along the x, y, z directions respectively).
Parametric surfaces are the foundation for surface integrals (Section 16.7). The cross product r_u × r_v that you compute here becomes the dS in every surface integral you will evaluate.
The tangent plane construction connects back to linearisation from Calculus II: the tangent plane is the best linear approximation to the surface at a point.
The spherical coordinate parametrisations here are the same coordinates you used for triple integrals in Chapter 15. The difference is that here ρ is fixed (you are on a surface), while previously ρ varied.
parametric surface, parametrisation, parameter domain, grid curves, tangent plane, normal vector, cross product, surface area, spherical coordinates, helicoid, ellipsoid, completing the square, cone inside sphere, Stewart 16.6, Calculus III, multivariable calculus, MATH 215, University of Michigan, Homework 9