6.6 Surface Integrals

Big idea

A curve is a one parameter object, so integrating over it needed one variable and a speed factor. A surface is a two parameter object, so integrating over it needs two variables and an area factor. Everything in this section is the surface version of the line integral section, one dimension up, and the structure matches line for line: a parametrization, a stretching factor that converts parameter area into real area, a scalar integral, an orientation, and a vector integral.

The new piece of machinery is the fundamental cross product. Freezing one parameter and letting the other run gives a curve on the surface, and its velocity vector is tangent to the surface. Two parameters give two such tangent vectors, and a small rectangle in the parameter domain maps to a small patch spanned approximately by them. The area of that patch is the magnitude of their cross product times the area of the rectangle, which is where the factor $\|\mathbf{t}_u \times \mathbf{t}_v\|$ comes from. The cross product also points perpendicular to the surface, so the same object supplies both the area factor and the normal direction.

That second role is what makes flux possible. To ask how much of a field passes through a surface you must first say through which side, and that choice is an orientation: a continuous choice of unit normal. Most surfaces admit exactly two, and switching between them negates every flux integral, just as reversing a curve negated every work integral. Some surfaces admit none, which is a genuine obstruction and not an oversight.

Flux is the quantity the last two theorems of the chapter are stated in, so the mechanics here are worth drilling. In practice most surfaces you meet are graphs, spheres, cylinders or cones, and each has a standard parametrization whose cross product you should be able to write down without rederiving it.

Decoder

The area element of a parametrized surface is the magnitude of the cross product of the two parameter tangent vectors, times the area element of the parameter domain, and that same cross product, normalized, is a unit normal to the surface.

The load is carried by “cross product”. Its magnitude is the area of the parallelogram spanned by the tangents, which is exactly the distortion caused by the parametrization, and its direction is perpendicular to both tangents, which is what perpendicular to the surface means. One computation supplies two different pieces of the theory.

The order of the factors is a real choice, not a formality. Swapping $u$ and $v$ reverses the cross product, hence reverses the normal, hence negates every flux integral. Scalar surface integrals do not notice, because they use only the magnitude.

The usual error is dropping the area factor and integrating $f$ over the parameter rectangle directly. The test case is a sphere, where forgetting $\sin\phi$ weights the poles as heavily as the equator and produces an answer that is not even close.

Definitions and results

Parametric surface. A parametrized surface is a vector function $\mathbf{r}(u,v) = \langle x(u,v),\, y(u,v),\, z(u,v)\rangle$ defined on a region $D$ in the $uv$-plane. The image is the surface $S$.

Tangent vectors and the fundamental cross product. With $$ \mathbf{t}_u = \left\langle \frac{\partial x}{\partial u}, \frac{\partial y}{\partial u}, \frac{\partial z}{\partial u} \right\rangle, \qquad \mathbf{t}_v = \left\langle \frac{\partial x}{\partial v}, \frac{\partial y}{\partial v}, \frac{\partial z}{\partial v} \right\rangle $$ the vector $\mathbf{t}_u \times \mathbf{t}_v$ is normal to $S$. The surface is smooth where this is continuous and never the zero vector.

Surface area. $$ \text{Area}(S) = \iint_D \left\| \mathbf{t}_u \times \mathbf{t}_v \right\| \, dA $$

Scalar surface integral. For a scalar function $f$ defined on $S$, $$ \iint_S f\,dS = \iint_D f\big(\mathbf{r}(u,v)\big)\left\| \mathbf{t}_u \times \mathbf{t}_v \right\| dA $$ Taking $f = 1$ returns the area; taking $f$ to be a surface density returns mass.

Graphs. For $z = g(x,y)$ over a region $D$, parametrize by $x$ and $y$ themselves. Then $\mathbf{t}_x\times\mathbf{t}_y = \langle -g_x, -g_y, 1\rangle$, so $$ dS = \sqrt{1 + g_x^2 + g_y^2}\;dA $$ and the normal written this way always points upward, since its third component is $1$.

Spheres. For the sphere of radius $a$, use $\mathbf{r}(\phi,\theta) = \langle a\sin\phi\cos\theta,\; a\sin\phi\sin\theta,\; a\cos\phi\rangle$ with $0\le\phi\le\pi$ and $0\le\theta\le 2\pi$. Then $dS = a^2\sin\phi\,d\phi\,d\theta$, and the outward unit normal at a point is the position vector divided by $a$.

Orientation. An oriented surface carries a continuous choice of unit normal $\mathbf{n}$. A surface with two sides admits exactly two orientations; a Mobius band admits none and is called nonorientable. For a closed surface, the standard choice is the outward normal.

Choosing the normal direction. Compute $\mathbf{t}_u\times\mathbf{t}_v$, then check one component against the side you want. For an upward orientation demand a positive $\mathbf{k}$ component; for the outward normal on a closed surface, evaluate at a convenient point and see whether the vector points away from the enclosed solid. If the sign is wrong, negate the cross product, or equivalently swap the two parameters.

Flux integral. For a vector field $\mathbf{F}$ continuous on an oriented surface $S$, $$ \iint_S \mathbf{F}\cdot d\mathbf{S} = \iint_S \mathbf{F}\cdot\mathbf{n}\,dS = \iint_D \mathbf{F}\big(\mathbf{r}(u,v)\big)\cdot\left(\mathbf{t}_u\times\mathbf{t}_v\right)dA $$ The magnitudes cancel between the unit normal and the area factor, so the unnormalized cross product is what you actually use. Reversing the orientation negates the result.

Worked examples

Area of a piece of a paraboloid

Find the area of the part of $z = x^2 + y^2$ lying above the unit disk.

This is a graph, with $g_x = 2x$ and $g_y = 2y$, so $dS = \sqrt{1 + 4x^2 + 4y^2}\,dA$. Switch to polar coordinates, where the expression under the root is $1 + 4r^2$: $$ \iint_D \sqrt{1+4r^2}\;r\,dr\,d\theta = \int_0^{2\pi}\!\!\int_0^1 r\sqrt{1+4r^2}\,dr\,d\theta $$ Substituting $u = 1+4r^2$, $du = 8r\,dr$, the inner integral is $\frac18\int_1^5 \sqrt{u}\,du = \frac{1}{12}\left(5^{3/2} - 1\right)$. Multiplying by $2\pi$, $$ \text{Area} = \frac{\pi\left(5\sqrt{5} - 1\right)}{6} $$ Sanity check the size: the flat unit disk underneath has area $\pi$, and the bowl is tilted everywhere, so the answer should exceed $\pi$. It is about $5.33$, which is comfortably larger.

A scalar integral over a sphere

Compute $\iint_S z^2\,dS$ over the unit sphere.

With the spherical parametrization, $z = \cos\phi$ and $dS = \sin\phi\,d\phi\,d\theta$, so $$ \iint_S z^2\,dS = \int_0^{2\pi}\!\!\int_0^{\pi} \cos^2\phi \,\sin\phi\;d\phi\,d\theta = 2\pi\left[-\frac{\cos^3\phi}{3}\right]_0^{\pi} = 2\pi\cdot\frac{2}{3} = \frac{4\pi}{3} $$ Check by symmetry. On the unit sphere $x^2+y^2+z^2 = 1$, so the three integrals $\iint x^2\,dS$, $\iint y^2\,dS$ and $\iint z^2\,dS$ add to $\iint 1\,dS = 4\pi$. The sphere treats the three axes alike, so each must be $4\pi/3$. The direct computation agrees.

Flux through the side of a cylinder

Find the outward flux of $\mathbf{F} = \langle x, y, z\rangle$ through the lateral surface of the cylinder $x^2+y^2 = 1$ between $z=0$ and $z=2$.

Parametrize by $\mathbf{r}(\theta,z) = \langle \cos\theta, \sin\theta, z\rangle$ with $0\le\theta\le2\pi$, $0\le z\le 2$. Then $\mathbf{t}_\theta = \langle -\sin\theta, \cos\theta, 0\rangle$ and $\mathbf{t}_z = \langle 0,0,1\rangle$, so $$ \mathbf{t}_\theta \times \mathbf{t}_z = \langle \cos\theta, \; \sin\theta, \; 0\rangle $$ Test the direction: at $\theta = 0$ this is $\langle 1,0,0\rangle$ at the point $(1,0,z)$, pointing away from the axis, so the order chosen already gives the outward normal. On the surface $\mathbf{F} = \langle\cos\theta,\sin\theta,z\rangle$, and the dot product is $\cos^2\theta + \sin^2\theta + 0 = 1$. Therefore $$ \iint_S \mathbf{F}\cdot d\mathbf{S} = \int_0^{2\pi}\!\!\int_0^2 1\,dz\,d\theta = 4\pi $$ which is the lateral area, as it should be for a field of unit outward component along the whole side.

Flux upward through a graph

Find the upward flux of $\mathbf{F} = \langle x, y, z\rangle$ through the part of the paraboloid $z = 1 - x^2 - y^2$ with $z \ge 0$.

Use the graph formula. With $g = 1 - x^2 - y^2$ we get $g_x = -2x$ and $g_y = -2y$, so the upward normal element is $$ \mathbf{n}\,dS = \langle -g_x, -g_y, 1\rangle\,dA = \langle 2x, \; 2y, \; 1\rangle\,dA $$ Its third component is $1 > 0$, confirming the orientation is upward. On the surface, $\mathbf{F}\cdot\langle 2x, 2y, 1\rangle = 2x^2 + 2y^2 + z = 2x^2+2y^2 + \left(1 - x^2 - y^2\right) = x^2+y^2+1$. The shadow is the unit disk, so in polar coordinates $$ \iint_S \mathbf{F}\cdot d\mathbf{S} = \int_0^{2\pi}\!\!\int_0^1 \left(r^2+1\right) r\,dr\,d\theta = 2\pi\left(\frac14 + \frac12\right) = \frac{3\pi}{2} $$ Had the downward side been wanted, the answer would be $-3\pi/2$ with no other change.

Practice

Start with the geometry. These problems ask you to parametrize standard surfaces, compute the fundamental cross product, and find surface area.

Next, scalar integrals over a surface: mass of a thin shell, averages, and integrals where a symmetry argument checks the answer.

Last, flux. Each problem requires you to fix an orientation first and then verify that your cross product points the way you claimed.

Quiz

Six items across parametrizing surfaces, surface area, scalar surface integrals, orientation, and flux.