Vector calculus. Surface integrals of the first kind

What it is

The integral of a function with respect to surface area:

S \iint f (x, y, z) d S .

Orientation plays no part, $d S > 0$ . With $f \equiv 1$ it gives the area of the surface; in physics, the mass of a shell with surface density $f$ .

Two working formulas

The surface is given explicitly, $z = z (x, y)$ over a domain $D$ :

d S = 1 + z_{x}^{2} + z_{y}^{2} d x d y .

For a plane $A x + B y + C z = D$ this factor is constant and equal to $A^{2} + B^{2} + C^{2} / ∣ C ∣$ — it comes straight out of the integral.

The surface is given parametrically, $r (u, v)$ :

d S = r_{u} \times r_{v} d u d v .

For a sphere of radius $a$ in spherical coordinates $x = a sin θ cos φ$ , $y = a sin θ sin φ$ , $z = a cos θ$ this gives $d S = a^{2} sin θ d θ d φ$ — a formula worth remembering; it saves half the solution.

The steps

Choose how to describe the surface: planes and graphs explicitly, spheres and cylinders parametrically.
Find the projection $D$ (for the explicit form) or the limits of the parameters.
Write out $d S$ .
Substitute the equation of the surface into $f$ — this often simplifies the integrand sharply, because on the surface itself the variables are related.

Common mistakes

Forgetting the factor $d S$ and computing a double integral over the projection. For an inclined plane this understates the answer by exactly the factor $1 + z_{x}^{2} + z_{y}^{2}$ .
Getting the projection wrong. For a piece of a plane in the first octant the projection is the triangle bounded by the plane's trace on $z = 0$ , and its vertices should be written out explicitly.
Getting the limits of the iterated integral wrong. After substituting the upper limit into the inner integral, check the expression at one point (say, $x = 0$ ) — an arithmetic slip shows up at once.

Problems

Problem 1 (A surface integral of the first kind over a piece of a plane). Compute

S \iint (2 x + 3 y + 2 z) d S,

where $S$ is the part of the plane $x + 3 y + z = 3$ lying in the first octant.

Solution.

The surface. The plane meets the axes at the points $(3; 0; 0)$ , $(0; 1; 0)$ , $(0; 0; 3)$ , so $S$ is the triangle with these vertices.

The area element. Write $z = 3 - x - 3 y$ . Then $z_{x} = - 1$ , $z_{y} = - 3$ and

d S = 1 + z_{x}^{2} + z_{y}^{2} d x d y = 1 + 1 + 9 d x d y = 11 d x d y .

For a plane this factor is constant, so it comes straight out of the integral.

The projection. The projection of $S$ onto the $O x y$ plane is the triangle bounded by the trace $x + 3 y = 3$ and the axes, that is

0 ⩽ x ⩽ 3, 0 ⩽ y ⩽ \frac{3 - x}{3} .

Simplifying the integrand. On the surface the variables are related, and substituting $z = 3 - x - 3 y$ usually simplifies the expression sharply: $2 x + 3 y + 2 z = 2 x + 3 y + 2 (3 - x - 3 y) = 6 - 3 y$ .

Integration. The inner integral in $y$ :

\int_{0}^{(3 - x) /3} (6 - 3 y) d y = [6 y - \frac{3 y ^{2}}{2}]_{0}^{(3 - x) /3} = 2 (3 - x) - \frac{( 3 - x ) ^{2}}{6} .

Over a common denominator:

= \frac{12 ( 3 - x ) - ( 3 - x ) ^{2}}{6} = \frac{36 - 12 x - 9 + 6 x - x ^{2}}{6} = \frac{27 - 6 x - x ^{2}}{6} .

It is useful to check a point: at $x = 0$ the expression gives $27/6 = 4.5$ , and direct substitution into $2 \cdot 3 - 9/6$ gives the same. It agrees.

The outer integral:

11 \int_{0}^{3} \frac{27 - 6 x - x ^{2}}{6} d x = \frac{11}{6} [27 x - 3 x^{2} - \frac{x ^{3}}{3}]_{0}^{3} = \frac{11}{6} (81 - 27 - 9) = \frac{45 11}{6} .

= \frac{15 11}{2} .

Answer: $\frac{15 11}{2}$ .

Problem 2 (A surface integral of the first kind over a piece of a plane (variant)). Compute

S \iint (x + z) d S,

where $S$ is the part of the plane $2 x + y + 2 z = 2$ lying in the first octant.

Solution.

The surface. The plane cuts the axes at the points $(1; 0; 0)$ , $(0; 2; 0)$ , $(0; 0; 1)$ .

The area element. From the equation $z = 1 - x - y /2$ , whence $z_{x} = - 1$ , $z_{y} = - 1/2$ and

d S = 1 + 1 + \frac{1}{4} d x d y = \frac{3}{2} d x d y .

The same factor also comes from the coefficients of the plane: $A^{2} + B^{2} + C^{2} /∣ C ∣ = 4 + 1 + 4 /2 = 3/2$ . A handy check.

The projection. The trace of the plane on $z = 0$ is the line $2 x + y = 2$ , so $0 ⩽ x ⩽ 1$ , $0 ⩽ y ⩽ 2 - 2 x$ .

Substitution.

x + z = x + 1 - x - \frac{y}{2} = 1 - \frac{y}{2} .

The integrand has lost its dependence on $x$ — the usual effect of substituting the equation of the surface.

Integration.

\frac{3}{2} \int_{0}^{1} \int_{0}^{2 - 2 x} (1 - \frac{y}{2}) d y d x .

The inner integral:

[y - \frac{y ^{2}}{4}]_{0}^{2 - 2 x} = (2 - 2 x) - \frac{( 2 - 2 x ) ^{2}}{4} = (2 - 2 x) - (1 - x)^{2} .

Write $u = 1 - x$ : the expression is $2 u - u^{2}$ . Then

\frac{3}{2} \int_{0}^{1} (2 u - u^{2}) d x, u = 1 - x .

As $x$ goes $0 \to 1$ , the variable $u$ runs $1 \to 0$ , and with $d x = - d u$

\frac{3}{2} \int_{0}^{1} (2 u - u^{2}) d u = \frac{3}{2} [u^{2} - \frac{u ^{3}}{3}]_{0}^{1} = \frac{3}{2} \cdot \frac{2}{3} = 1.

Answer: $1$ .

Problem 3 (A surface integral over part of a sphere). Compute

S \iint (x + y + z) d S,

where $S$ is the part of the sphere $x^{2} + y^{2} + z^{2} = a^{2}$ ( $a > 0$ ) lying in the first octant.

Solution.

Here the explicit form $z = a^{2} - x^{2} - y^{2}$ would put a root in the denominator and make the integral improper on the boundary of the projection. Spherical coordinates remove this entirely.

Parametrisation. $x = a sin θ cos φ$ , $y = a sin θ sin φ$ , $z = a cos θ$ .

The first octant corresponds to $θ \in [0; π /2]$ (the upper hemisphere) and $φ \in [0; π /2]$ (the first quadrant of the $O x y$ plane).

The area element. For a sphere of radius $a$ : $d S = a^{2} sin θ d θ d φ$ .

The factor $sin θ$ is essential: near the pole the band of area degenerates, and without it the integral is overstated.

Substitution.

x + y + z = a (sin θ (cos φ + sin φ) + cos θ) .

I = a^{3} \int_{0}^{π /2} \int_{0}^{π /2} (sin θ (cos φ + sin φ) + cos θ) sin θ d θ d φ .

Split into two terms. The variables separate, so each double integral falls apart into a product of single ones.

The first term:

a^{3} \int_{0}^{π /2} sin^{2} θ d θ \cdot \int_{0}^{π /2} (cos φ + sin φ) d φ = a^{3} \cdot \frac{π}{4} \cdot 2 = \frac{π a ^{3}}{2} .

The second term:

a^{3} \int_{0}^{π /2} sin θ cos θ d θ \cdot \int_{0}^{π /2} d φ = a^{3} \cdot \frac{1}{2} \cdot \frac{π}{2} = \frac{π a ^{3}}{4} .

Result.

I = \frac{π a ^{3}}{2} + \frac{π a ^{3}}{4} = \frac{3 π a ^{3}}{4} .

A dimension check: the integral of a length over an area must be of third order in $a$ — and it is.

Answer: $\frac{3 π a ^{3}}{4}$ .

Problem 4 (A surface integral over a hemisphere (variant)). Compute

S \iint z d S,

where $S$ is the hemisphere $z = 4 - x^{2} - y^{2}$ .

Solution.

Parametrisation. The hemisphere is the upper half of the sphere of radius $a = 2$ : $x = 2 sin θ cos φ$ , $y = 2 sin θ sin φ$ , $z = 2 cos θ$ , $θ \in [0; π /2]$ , $φ \in [0; 2 π]$ .

Unlike the previous problem, $φ$ runs over the full circle: there is no restriction on the signs of $x$ and $y$ .

The area element. $d S = a^{2} sin θ d θ d φ = 4 sin θ d θ d φ$ .

Integration.

S \iint z d S = \int_{0}^{2 π} \int_{0}^{π /2} 2 cos θ \cdot 4 sin θ d θ d φ = 8 \int_{0}^{2 π} d φ \int_{0}^{π /2} sin θ cos θ d θ .

The inner integral: $\int_{0}^{π /2} sin θ cos θ d θ = [\frac{s i n ^{2} θ}{2}]_{0}^{π /2} = \frac{1}{2}$ .

= 8 \cdot 2 π \cdot \frac{1}{2} = 8 π .

A check through the mean. The area of a hemisphere of radius $2$ is $2 π a^{2} = 8 π$ , so the mean value of $z$ over the hemisphere is $8 π /8 π = 1 = a /2$ . This is a known fact (the projection onto an axis of a point uniformly distributed on a sphere is uniformly distributed), and it confirms the answer.

Answer: $8 π$ .

What it is

Two working formulas

The steps

Common mistakes

Problems

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