EXAMPLES OF STOKES’ THEOREM AND GAUSS’ DIVERGENCE THEOREM

1. S TOKES ’ T HEOREM Let S be an oriented surface with positively oriented boundary curve C, and let F be a C 1 vector field defined on S. Then Z ZZ (∇ × F) · dS = F · ds (1.1) S

C

Note: We need to have the correct orientation on the boundary curve. The easiest way to remember this is to use the right-hand rule: With your hand on the boundary curve, point your thumb in the direction of the normal vector to the surface and your palm towards the surface, then your fingers will curl around in the direction of the positive orientation of the boundary curve. Alternately, if you imagine yourself walking around the boundary curve with your head in the direction of the normal vector, then the surface will be on your left. See the diagram on p533 for more details. Note also that the boundary curve C always consists of closed loops, since it is the boundary of the surface S. Questions using Stokes’ Theorem usually fall into three categories: R (1) Use Stokes’ Theorem to compute C F · ds. In these examples it will be easier to compute the surface integral of ∇ × F over some surface S with boundary C instead. (i.e. we use the left-hand side of Stokes’ Theorem to help us compute the right-hand side). RR (2) Use Stokes’ Theorem to Rcompute S (∇ × F) · dS. This time it will be easier to compute the line integral C F·ds, i.e. we want to use the right-hand side of Stokes’ theorem to help us compute the left-hand side. (3) Verify Stokes’ Theorem. In these types of questions you will be given a surface S and a vector field F. The question is asking you to compute both sides of Stokes’ theorem and show that they are the same. 2. E XAMPLES Example 1: Let C be the boundary of the part of the plane 2x + y + 2z = 2 in the first octant, oriented counterclockwise asR viewed from above. Let F(x, y, z) = (x + y 2 , y + z 2 , z + x2 ). Use Stokes’ theorem to compute C F · ds. In this case, computing the line integral would require a lot of effort (the boundary has three smooth components and hence you would need to compute three separate integrals). Instead we note that the surface S enclosed by the boundary curve C is just the plane 2x + y + 2z = 2, which has the parametrisation X : D → R3   1 X(s, t) = s, t, (2 − 2s − t) , 2 Date: May 1, 2009. 1

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EXAMPLES OF STOKES’ THEOREM AND GAUSS’ DIVERGENCE THEOREM

where D = {(s, t) : 0 ≤ t ≤ 2 − 2s, 0 ≤ s ≤ 1} (note that X is the standard parametrisation of the surface S as the graph of a function). A calculation shows that Ts = (1, 0, −1)   1 Tt = 0, 1, − 2   1 Ts × Tt = 1, , 1 . 2 (check that the normal vector is positively oriented with respect to the boundary curve). We can also compute ∇ × F = (−2z, −2x, −2y). Therefore the surface integral becomes   ZZ Z 1 Z 2−2s 1 (∇ × F) · dS = (2s + t − 2, −2s, −2t) · 1, , 1 dt ds 2 S 0 0 Z 1 Z 2−2s = (s − t − 1) dt ds, 0

0

which is then easy to evaluate. Example 2: Let S be the part of the paraboloid z = 9 − x2 − y 2 that lies above the plane z = 5, oriented with normal vector pointing upwards, and let F(x, y, z) = (yz, x2 z, xy). Use Stokes’ Theorem to evaluate ZZ (∇ × F) · dS. S

Using the right-hand rule, we orient the boundary curve C in the anticlockwise direction as viewed from above. It is the circle of radius 2 which lies on the plane z = 5, and is centred at the origin. We can parametrise it by the equation c(t) = (x(t), y(t), z(t)) = (2 cos t, 2 sin t, 5) . Therefore the work integral becomes Z Z 2π F · ds = F(c(t)) · c0 (t) dt C 0 Z 2π
= 10 sin t, 20 cos2 t, 4 sin t cos t · (−2 sin t, 2 cos t, 0) dt 0 Z 2π = −20 sin2 t + 40 cos3 t dt, 0

which can be easily evaluated (don’t forget the double angle formulas!). Example 3: Verify Stokes’ theorem for the case where S is the cylinder {(x, y) : x2 +y 2 = 4, −3 ≤ z ≤ 1} with outwards pointing normal vector, and F = (yz, z, y). Note that to give the boundary curves the correct orientation, the top curve C2 is oriented clockwise as viewed from above, and the curve C1 is oriented anti-clockwise as viewed from above. First we compute the left-hand side of Stokes’ theorem in equation (1.1) (the surface integral). A calculation shows that ∇ × F = (0, y, −z), and the standard parametrisation

EXAMPLES OF STOKES’ THEOREM AND GAUSS’ DIVERGENCE THEOREM

3

of the cylinder is X : D → R3 X(s, t) = (2 cos s, 2 sin s, t), where D = [0, 2π] × [−1, 3]. The normal vector is Ts = (−2 sin s, 2 cos s, 0) Tt = (0, 0, 1) Ts × Tt = (2 cos s, 2 sin s, 0). (check that the normal vector points in the same direction as the orientation of the surface given above). In terms of the parametrisation, we have ∇×F(c(t)) = (0, 2 sin s, −t), and so the surface integral becomes ZZ ZZ (∇ × F) · dS = (0, 2 sin s, −t) · (2 cos s, 2 sin s, 0) ds dt S D Z 1 Z 2π = 4 sin2 s ds st −3

0

= ··· = 16π. Now we compute the right-hand side of equation (1.1). We parametrise the curve C1 by c(t) = (2 cos t, 2 sin t, −3) for 0 ≤ t ≤ 2π (notice that this is oriented anti-clockwise as viewed from above). Then we have c0 (t) = (−2 sin t, 2 cos t, 0) and F(c(t)) = (−6 sin t, −3, 2 sin t), and so the line integral around C1 becomes Z Z 2π F(c(t)) · c0 (t) dt F · ds = C1

0 2π

Z

12 sin2 t − 6 cos t dt

= 0

= ··· = 12π R To compute C2 F · ds we parametrise the curve C2 by c(t) = (2 cos t, −2 sin t, 1) for 0 ≤ t ≤ 2π (notice that this is oriented clockwise as viewed from above). Then we have c0 (t) = (−2 sin t, −2 cos t, 0) and F(c(t)) = (−2 sin t, 1, −2 sin t), therefore the line integral around C2 becomes Z Z 2π F · ds = F(c(t)) · c0 (t) dt C2

0

Z

2π

= 0

= ··· = 4π

4 sin2 t − 2 cos t dt

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EXAMPLES OF STOKES’ THEOREM AND GAUSS’ DIVERGENCE THEOREM

Therefore the integral around the boundary of S becomes Z Z Z F · ds = F · ds + F · ds = 12π + 4π = 16π, C

C1

C2

the same result that we obtained earlier for Stokes’ theorem in this case.

RR S

(∇ × F) · dS. Therefore we have verified

3. G AUSS ’ D IVERGENCE T HEOREM Let B be a solid region in R3 and let S be the surface of B, oriented with outwards pointing normal vector. Gauss’ Divergence Theorem states that for a C 1 vector field F the following equation holds ZZ ZZZ (3.1) F · dS = (∇ · F) dV. S

B

Note that for the theorem to hold, the orientation of the surface must be pointing outwards from the region B. Otherwise we will get a minus sign in equation (3.1). Note also that since S is the boundary of B then it is always a closed surface, i.e. it has no boundary. Questions about Gauss’ Divergence Theorem usually fall into the same three categories as Stokes’ theorem. RR (1) Compute S F · dS. In this case it will be easier to integrate ∇ · F over the region B which isRRRbounded by S. (2) Compute B (∇·F) dV . In this case it will be easier to compute the surface integral of F over S. (3) Verify Gauss’ Divergence Theorem. In these types of questions you will be given a region B and a vector field F. The question is asking you to compute the integrals on both sides of equation (3.1) and show that they are equal. 4. E XAMPLES RR Example 1: Use the divergence theorem to calculate S F · dS, where S is the surface of the box B with vertices (±1, ±2, ±3) with outwards pointing normal vector, and F(x, y, z) = (x2 z 3 , 2xyz 3 , xz 4 ). Note that the surface integral will be difficult to compute, since there are six different components to parametrise (corresponding to the six sides of the box), and so one would have to compute six different integrals. Instead, using Gauss’ theorem, it is easier to compute the integral of ∇ · F over B. First we compute ∇ · F = 2xz 3 + 2xz 3 + 4xz 3 = 8xz 3 . Now we integrate this function over the region B bounded by S: ZZ ZZZ F · dS = (∇ · F) dV S B Z 3Z 2Z 1 = 8xz 3 dxdydz −3

−2

−1

which is easy to compute. Example 2: Verify the divergence theorem for the case where F(x, y, z) = (x, y, z) and B is the solid sphere of radius R centred at the origin.

EXAMPLES OF STOKES’ THEOREM AND GAUSS’ DIVERGENCE THEOREM

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Firstly we compute the left-hand side of (3.1) (the surface integral). To do this we need to parametrise the surface S, which in this case is the sphere of radius R. The standard parametrisation using spherical co-ordinates is X(s, t) = (R cos t sin s, R sin t sin s, R cos s) . The tangent vectors to the co-ordinate curves are Ts = (R cos t cos s, R sin t cos s, −R sin s) Tt = (−R sin t sin s, R cos t sin s, 0) , and so the normal vector to the surface is
Ts × Tt = R2 sin2 s cos t, R2 sin2 s sin t, R2 sin t cos t . (check that this points outwards from the solid region B) In terms of the parametrisation, the vector field F is given by F(X(s, t)) = (R cos t sin s, R sin t sin s, R cos s) , and so the surface integral becomes Z 2π Z π ZZ
(R cos t sin s, R sin t sin s, R cos s) · R2 sin2 s cos t, R2 sin2 s sin t, R2 sin t cos t ds dt F · dS = 0 0 S Z 2π Z π = R3 sin s ds dt = 4πR

0 3

0

Now we compute the right-hand side of equation (3.1) (the volume integral). First we compute ∇ · F = 1 + 1 + 1 = 3. Then using spherical co-ordinates, the integral becomes ZZZ ZZZ 3 dV (∇ · F) dV = B B Z π Z 2π Z R 3ρ2 sin φ dρdθdφ = 0 0 0 Z π Z 2π = R3 sin φ dθdφ 0 0 Z π 3 = 2πR sin φ dφ 0

= 4πR

3

Therefore we have obtained the same result for the left-hand side and the right-hand side of the divergence theorem, and so we have verified the theorem in this case.

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EXAMPLES OF STOKES’ THEOREM AND GAUSS’ DIVERGENCE THEOREM

5. T HE F UNDAMENTAL T HEOREM OF C ALCULUS Note: This section will not be tested, it is only here to help your understanding. If f (x) is a continuous function with continuous derivative f 0 (x) then the Fundamental Theorem of Calculus (FTOC) states that Z b f 0 (x) dx = f (b) − f (a) a

In other words we can integrate (which involves computing the area under the graph of f by taking the limit of the Riemann sums) by simply anti-differentiating. This is much easier than computing Riemann sums! In the last few weeks we have studied three different higher-dimensional versions of this: The Fundamental Theorem of Calculus for Line Integrals (1-dimensional FTOC) Z ∇f · ds = f (c(b)) − f (c(a)) C

where C is a curve parametrised by c(t) for a ≤ t ≤ b. Stokes’ Theorem (2-dimensional FTOC) Z ZZ F · ds (∇ × F) · dS = C

S

where S is an oriented surface in R that has positively oriented boundary curve C. Gauss’ Divergence Theorem (3-dimensional FTOC) ZZZ ZZ ∇ · F dV = F · dS 3

B

S

where B is a solid region in R with boundary S, oriented with outwards pointing normal vector. All of these theorems can be expressed using the same statement: ”Integrating the derivative of a function over a region is the same as integrating the function over the boundary”. For 1-dimensional curves, the ”derivative” of a function f is ∇f , for 2-dimensional surfaces in R3 the ”derivative” of a vector field F is ∇ × F, and for 3-dimensional solid regions in R3 the ”derivative” of a vector field is ∇ · F. 3