Guides / Integrals

Surface integrals and the divergence theorem: dS = |r_u × r_v| du dv, the normal sets the sign, and close the surface before using div F

A surface integral adds something up over a curved surface, the way a line integral adds it up along a curve. Every one is done the same way: describe the surface with two parameters, get the area element from a cross product, and do an ordinary double integral. Integrating a vector field this way gives its flux, the rate it flows through the surface, and two theorems turn flux into something easier: the divergence theorem for a closed surface and Stokes' theorem for a surface with an edge. Most mistakes are a missing area factor, a normal pointing the wrong way, or a theorem used on a surface it does not apply to.

Why it works

A small rectangle du by dv in the parameter plane maps to a small patch of the surface whose sides are close to the vectors r_u du and r_v dv. The patch is nearly a parallelogram, and the area of a parallelogram is the length of the cross product of its sides. So |r_u × r_v| is the stretch factor from parameter area to surface area, exactly as |r′(t)| is for arc length, and the cross product itself points along the normal.

The divergence of F measures how much the field is spreading out at a point: the net outflow from a tiny box, per unit volume. Adding it up over a solid, the flows across every internal wall cancel, because what leaves one tiny box enters its neighbor, and only the flow through the outer surface survives. That is why the theorem needs the surface to enclose the solid completely. Stokes' theorem is the same cancellation one dimension down: the curl measures circulation around tiny loops, and the internal edges cancel in pairs, leaving the boundary curve.

Example 1: the area of a cone

Find the area of the cone z = √(x² + y²) for z ≤ 2.

1. g_x = x/r, g_y = y/rTreat the cone as a graph over the disk r ≤ 2, with r = √(x² + y²). Each height z equals r.
2. 1 + g_x² + g_y² = 2(x² + y²)/r² = 1, the same at every point.
3. dS = √2 dAEvery bit of the cone is √2 times the area of its shadow on the xy-plane: the side slopes at 45°.
4. area = √2 · 4π ≈ 17.77The shadow is a disk of radius 2, area 4π. Check with the formula for the side of a cone, π · radius · slant = π · 2 · 2√2, which is the same 4√2π.

Example 2: the flux through the side of a cylinder

Find the outward flux of F = (x, y, z) through the side of the cylinder x² + y² = 1, 0 ≤ z ≤ 3, without its top and bottom.

1. r = (cos θ, sin θ, z)Parameters θ from 0 to 2π and z from 0 to 3.
2. r_θ × r_z = (cos θ, sin θ, 0)r_θ = (−sin θ, cos θ, 0) and r_z = (0, 0, 1). The result points away from the axis, which is outward, and has length 1.
3. F · (r_θ × r_z) = 1On the surface F = (cos θ, sin θ, z), and cos² θ + sin² θ = 1; the z part meets a 0.
4. flux = 2π · 3 = 6πThe integrand is constant, so the flux is the parameter area. Here it is also the surface area, since the stretch factor is 1.

Example 3: the divergence theorem on the closed cylinder

Now include the top disk at z = 3 and the bottom disk at z = 0. Find the outward flux of the same F out of the closed cylinder, and check it face by face.

1. div F = 1 + 1 + 1 = 3∂x/∂x + ∂y/∂y + ∂z/∂z.
2. ∭ 3 dV = 3 · 3π = 9πThe solid cylinder has volume π · 1² · 3. The surface is closed now, so the theorem applies.
3. top: F · k = 3, so 3πOn the top the outward normal is (0, 0, 1) and z = 3; the disk has area π.
4. bottom: F · (−k) = −z = 0On the bottom the outward normal points down, and z = 0 there.
5. 6π + 3π + 0 = 9πThe side from Example 2 plus the two disks matches the triple integral.

Example 4: Stokes' theorem on a paraboloid

Let F = (−y, x, z) and let S be the paraboloid z = 4 − x² − y², z ≥ 0, with the upward normal. Find the flux of curl F through S both ways.

1. curl F = (0, 0, 2)(R_y − Q_z, P_z − R_x, Q_x − P_y) = (0, 0, 1 + 1).
2. n dS = (2x, 2y, 1) dAThe upward choice (−g_x, −g_y, 1) for g = 4 − x² − y².
3. ∬ 2 dA = 2 · 4π = 8πThe dot product is 2 everywhere, over the shadow, the disk of radius 2.
4. C: r = (2cos t, 2sin t, 0)The edge is the circle where z = 0. Counterclockwise seen from above, because the normal points up.
5. F · r′ = 4sin² t + 4cos² t = 4F = (−2sin t, 2cos t, 0) and r′ = (−2sin t, 2cos t, 0).
6. ∮ F · dr = 4 · 2π = 8πThe two sides agree. Any surface with this edge, such as the flat disk, gives 8π too.

Common mistakes, and the exact line they happen on

1. Dropping the area factor

The cone of Example 1.

1. shadow: disk r ≤ 2Right so far.
2. area = ∬ dA = 4πThis is the mistake. This is the area of the shadow, about 12.57. The cone is tilted, so each patch is larger than its shadow by √(1 + g_x² + g_y²) = √2. The area is 4√2π ≈ 17.77.

2. The divergence theorem on a surface that is not closed

The side of the cylinder in Example 2.

1. div F = 3Right so far.
2. flux = 3 · 3π = 9πThis is the mistake. The side alone does not enclose a solid, so ∭ div F dV is the flux out of the side, top and bottom together. Subtract the top's 3π and the bottom's 0, as in Example 3: the side carries 6π.

3. The boundary traced the wrong way

The circle in Example 4.

1. ∬ (curl F) · n dS = 8πRight so far, with n pointing up.
2. r = (2cos t, −2sin t, 0)This is the mistake. This goes clockwise seen from above, so ∮ F · dr = −8π, and the two sides of Stokes' theorem seem to disagree. With n up, the edge must be traced counterclockwise from above: walking along it with your head along n, the surface is on your left.

Practice

1. Find the area of the part of the plane z = 2x + 2y + 1 above the unit disk.

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   dS = √(1 + 4 + 4) dA = 3 dA, so the area is 3π.

2. Find the outward flux of F = (x, y, z) through the sphere of radius a, both with the divergence theorem and directly.

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   3 · (4/3)πa³ = 4πa³. Directly, F · n = a on the sphere, times the area 4πa².

3. Find the outward flux of F = (2x, −y, z) out of the unit cube 0 ≤ x, y, z ≤ 1.

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   div F = 2 − 1 + 1 = 2, and the cube has volume 1, so the flux is 2.

4. Find the upward flux of F = (0, 0, 1) through the upper half of the unit sphere.

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   π. Close it with the unit disk: div F = 0, so the flux up through the hemisphere equals the flux up through the disk, which is its area.

5. In Example 4, replace the paraboloid by the flat disk x² + y² ≤ 4 in the plane z = 0, normal up. What is the flux of curl F?

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   8π: curl F · k = 2 over an area of 4π. Same edge, same answer.