Difficulty: Intermediate to Advanced | Prerequisites: §15.1–15.3 (double integrals in Cartesian and polar), partial derivatives, basic geometry of cones and cylinders.
Once you can evaluate double integrals, the next question is: what can you compute with them? This section covers two major applications. First, surface area, where you integrate the "stretching factor" of a surface over its projection in the xy-plane. Second, physical applications like total population from a density function, or the volume of a solid under a surface. These problems tie together integration, geometry, and the ability to describe three-dimensional surfaces in two-dimensional terms. If you are comfortable setting up double integrals over general regions, you have the mechanical skill; these problems add the modelling layer.
Surface area of z = f(x, y) over a region D equals ∫∫_D √(1 + f_x² + f_y²) dA. Total population (or mass) of a flat region equals ∫∫_D ρ(x, y) dA, where ρ is the density. Both reduce to setting up the right double integral over the right region.
Surface area (of a graph z = f(x, y))
The area of the surface that sits above (or below) a region D in the xy-plane. Computed by integrating the magnitude of the cross product of the tangent vectors, which simplifies to √(1 + f_x² + f_y²). Think of it as: you tile the surface with tiny parallelograms and add up their areas.
Density function ρ(x, y)
A function giving the "amount per unit area" at each point, whether that is mass per m², people per km², or paint per in². Integrating ρ over a region D gives the total amount.
Lateral surface area
The area of the side of a three-dimensional shape (cone, cylinder), excluding the top and bottom. For a cone frustum with radii r₁ and r₂ and slant height s, the lateral area is π(r₁ + r₂)s.
Completing the square (in surface equations)
A technique for rewriting x² + y² + z² = 4z as x² + y² + (z − 2)² = 4, revealing a sphere of radius 2 centred at (0, 0, 2). Used in Problem 5 to identify the geometry before computing surface area.
The surface is z = 1 + 2(x − 3)² + y², bounded by y = 3, y = x, and the first octant (x ≥ 0, y ≥ 0, z ≥ 0).
The region of integration in the xy-plane is the triangle where 0 ≤ x ≤ y and 0 ≤ y ≤ 3 (since y = x and y = 3 bound the region, and y ≥ x in this triangle).
The volume is ∫∫_D [1 + 2(x − 3)² + y²] dA. Set up as an iterated integral with y from 0 to 3 and x from 0 to y, then expand and integrate term by term.
The funnel has two parts:
A cylinder of radius 1, height 2: lateral surface area = 2π(1)(2) = 4π in².
A cone frustum with smaller radius 1, larger radius 5, and height 5.
For the cone frustum, the slant height is s = √(5² + (5 − 1)²) = √(25 + 16) = √41.
Lateral surface area of the frustum = π(1 + 5)√41 = 6π√41.
Total paint wasted = 4π + 6π√41 in².
The sphere x² + y² + z² = 4z completes the square to x² + y² + (z − 2)² = 4, a sphere of radius 2 centred at (0, 0, 2).
The paraboloid 3z = x² + y² intersects the sphere where substitution gives the boundary circle. Setting r² = x² + y² and solving for z from both equations simultaneously yields the projection region D in the xy-plane.
From the sphere: z = 2 + √(4 − r²) (upper hemisphere) or z = 2 − √(4 − r²) (lower). From the paraboloid: z = r²/3. Setting these equal determines r at intersection.
The surface area integral uses the formula ∫∫_D √(1 + z_x² + z_y²) dA, where z is solved from the sphere equation. Polar coordinates simplify the computation.
The region D = {(x, y) : 5|x| ≤ y ≤ 5} is a triangle (vertex at the origin, base along y = 5).
Population = ∫∫_D 10^(3−y) dA.
By symmetry in x, this equals 2∫₀⁵ ∫₀^{y/5} 10^(3−y) dx dy = 2∫₀⁵ (y/5) · 10^(3−y) dy.
The y-integral requires integration by parts. The answer should be an integer, which serves as a useful check.
Surface area of z = f(x, y) over D
A = ∫∫_D √(1 + (∂f/∂x)² + (∂f/∂y)²) dA
Total mass or population
M = ∫∫_D ρ(x, y) dA
Lateral surface area of a cone frustum
A = π(r₁ + r₂) · s, where s = √(h² + (r₂ − r₁)²) is the slant height
Lateral surface area of a cylinder
A = 2πr h
Volume under z = f(x, y) over D
V = ∫∫_D f(x, y) dA
Surface area integrals are how engineers compute the material needed to coat, paint, or plate a curved surface. The funnel problem (Problem 4) is a practical instance: knowing the inner surface area tells you how much paint is wasted as coating. Population density integrals (Problem 6) are used in urban planning, epidemiology, and resource allocation.
Students forget to include the √(1 + f_x² + f_y²) factor and just compute ∫∫_D dA, which gives the area of the flat projection, not the surface.
When a surface is given implicitly (like x² + y² + z² = 4z), students sometimes differentiate incorrectly. Solve for z first, then take partial derivatives.
For the cone frustum, students confuse the vertical height h with the slant height s. They are related by s = √(h² + Δr²), but they are not the same.
Students often set up the region D incorrectly when |x| appears in the bounds. Remember that |x| splits the integral into x ≥ 0 and x < 0 halves, which are symmetric here.
⚠️ The surface area formula ∫∫ √(1 + f_x² + f_y²) dA is heavily tested. You must be able to compute partial derivatives quickly.
⚠️ Problems that mix geometric shapes (cylinder + cone) test whether you can decompose a solid into standard pieces.
⚠️ Population/mass-from-density problems almost always involve identifying the correct region D and setting up the iterated integral in the right order.
True or false: The surface area of z = f(x, y) over D is always greater than or equal to the area of D.
True. The factor √(1 + f_x² + f_y²) ≥ 1, with equality only when the surface is perfectly flat.
Fill in the blank: The slant height of a cone frustum with height 5 and radii 1 and 5 is ___.
√(25 + 16) = √41
True or false: To find the total population from a density function ρ(x, y), you integrate ρ over the boundary of the region.
False. You integrate ρ over the interior of the region D.
Q: Set up (but do not evaluate) the surface area integral for z = x² + y² over the disk x² + y² ≤ 4.
A: f_x = 2x, f_y = 2y, so 1 + f_x² + f_y² = 1 + 4x² + 4y² = 1 + 4r². In polar: A = ∫₀^{2π} ∫₀² √(1 + 4r²) · r dr dθ.
Q: A region in the xy-plane is bounded by y = |x| and y = 3. If the density is ρ(x, y) = e^y, set up the integral for total mass.
A: By symmetry, M = 2∫₀³ ∫₀^y e^y dx dy = 2∫₀³ y e^y dy. Evaluate with integration by parts.
Q: Why does completing the square matter in Problem 5?
A: The equation x² + y² + z² = 4z hides the geometry. Completing the square reveals a sphere of radius 2 centred at (0, 0, 2), which lets you solve for z and compute partial derivatives correctly.
Surface area connects directly to surface integrals in Chapter 16, where you integrate a function over a surface rather than just computing the surface's area. The density/population problems here are two-dimensional versions of the mass and centre-of-mass calculations in §15.6, which extend to triple integrals. The completing-the-square technique reappears whenever you work with spheres in spherical coordinates (§15.8).
surface area double integral, surface area formula multivariable, √(1 + fx² + fy²), cone frustum lateral area, population density integral, density function double integral, completing the square sphere, MATH 215, Calculus III, §15.4, §15.5, volume under surface, paint surface area funnel