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SECTION 13.8 STOKES ’ THEOREM
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P2P213.8 STOKES ’ VS. GREEN ’ S THEOREM Stokes ’ Theorem can be regarded as a higher- dimensional version of Green ’ s Theorem. Green ’ s Theorem relates a double integral over a plane region D to a line integral around its plane boundary curve. Stokes ’ Theorem relates a surface integral over a surface S to a line integral around the boundary curve of S (a space curve).
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P3P313.8 INTRODUCTION Figure 1 shows an oriented surface with unit normal vector n. The orientation of S induces the positive orientation of the boundary curve C shown in the figure.
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P4P413.8 INTRODUCTION This means that: If you walk in the positive direction around C with your head pointing in the direction of n, the surface will always be on your left.
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P5P513.8 STOKES ’ THEOREM Let S be an oriented piecewise-smooth surface bounded by a simple, closed, piecewise-smooth boundary curve C with positive orientation. Let F be a vector field whose components have continuous partial derivatives on an open region in that contains S. Then,
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P6P613.8 STOKES ’ THEOREM The theorem is named after the Irish mathematical physicist Sir George Stokes (1819 – 1903). What we call Stokes ’ Theorem was actually discovered by the Scottish physicist Sir William Thomson (1824 – 1907, known as Lord Kelvin). Stokes learned of it in a letter from Thomson in 1850.
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P7P713.8 STOKES ’ THEOREM Thus, Stokes ’ Theorem says: The line integral around the boundary curve of S of the tangential component of F is equal to the surface integral of the normal component of the curl of F.
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P8P813.8 STOKES ’ THEOREM The positively oriented boundary curve of the oriented surface S is often written as ∂S. So, the theorem can be expressed as:
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P9P913.8 STOKES ’ THEOREM, GREEN ’ S THEOREM, & FTC There is an analogy among Stokes ’ Theorem, Green ’ s Theorem, and the Fundamental Theorem of Calculus (FTC). As before, there is an integral involving derivatives on the left side of Equation 1 (recall that curl F is a sort of derivative of F). The right side involves the values of F only on the boundary of S.
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P1013.8 STOKES ’ THEOREM, GREEN ’ S THEOREM, & FTC In fact, consider the special case where the surface S: Is flat. Lies in the xy-plane with upward orientation.
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P1113.8 STOKES ’ THEOREM, GREEN ’ S THEOREM, & FTC Then, The unit normal is k. The surface integral becomes a double integral. Stokes ’ Theorem becomes:
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P1213.8 STOKES ’ THEOREM, GREEN ’ S THEOREM, & FTC This is precisely the vector form of Green ’ s Theorem given in Equation 12 in Section 13.5 Thus, we see that Green ’ s Theorem is really a special case of Stokes ’ Theorem.
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P1313.8 STOKES ’ THEOREM Stokes ’ Theorem is too difficult for us to prove in its full generality. Still, we can give a proof when: S is a graph. F, S, and C are well behaved.
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P1413.8 PROOF OF A SPECIAL CASE OF STOKES ’ THEOREM We assume that the equation of S is: z = g(x, y), (x, y) D where: g has continuous second-order partial derivatives. D is a simple plane region whose boundary curve C 1 corresponds to C.
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P1513.8 PROOF OF A SPECIAL CASE OF STOKES ’ THEOREM If the orientation of S is upward, the positive orientation of C corresponds to the positive orientation of C 1.
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P1613.8 PROOF OF A SPECIAL CASE OF STOKES ’ THEOREM We are also given that: F = P i + Q j + R k where the partial derivatives of P, Q, and R are continuous. S is a graph of a function. Thus, we can apply Formula 10 in Section 13.7 with F replaced by curl F.
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P1713.8 PROOF OF A SPECIAL CASE OF STOKES ’ THEOREM The result is: where the partial derivatives of P, Q, and R are evaluated at (x, y, g(x, y)).
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P1813.8 PROOF OF A SPECIAL CASE OF STOKES ’ THEOREM Suppose x = x(t) y = y(t) a ≤ t ≤ b is a parametric representation of C 1. Then, a parametric representation of C is: x = x(t) y = y(t) z = g(x(t), y(t)) a ≤ t ≤ b
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P1913.8 PROOF OF A SPECIAL CASE OF STOKES ’ THEOREM This allows us, with the aid of the Chain Rule, to evaluate the line integral as follows:
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P2013.8 PROOF OF A SPECIAL CASE OF STOKES ’ THEOREM We have used Green ’ s Theorem in the last step.
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P2113.8 PROOF OF A SPECIAL CASE OF STOKES ’ THEOREM Next, we use the Chain Rule again, remembering that: P, Q, and R are functions of x, y, and z. z is itself a function of x and y.
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P2213.8 PROOF OF A SPECIAL CASE OF STOKES ’ THEOREM Thus, we get:
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P2313.8 PROOF OF A SPECIAL CASE OF STOKES ’ THEOREM Four terms in that double integral cancel. The remaining six can be arranged to coincide with the right side of Equation 2. Hence,
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P2413.8 Example 1 Evaluate where: F(x, y, z) = – y 2 i + x j + z 2 k C is the curve of intersection of the plane y + z = 2 and the cylinder x 2 + y 2 = 1. (Orient C to be counterclockwise when viewed from above.)
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P2513.8 Example 1 SOLUTION The curve C (an ellipse) is shown in Figure 3. could be evaluated directly. However, it ’ s easier to use Stokes ’ Theorem.
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P2613.8 Example 1 SOLUTION We first compute:
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P2713.8 Example 1 SOLUTION There are many surfaces with boundary C. The most convenient choice, though, is the elliptical region S in the plane y + z = 2 that is bounded by C. If we orient S upward, C has the induced positive orientation.
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P2813.8 Example 1 SOLUTION The projection D of S on the xy-plane is the disk x 2 + y 2 ≤ 1. So, using Equation 10 in Section 13.7 with z = g(x, y) = 2 – y, we have the following result.
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P2913.8 Example 1 SOLUTION
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P3013.8 Example 2 Use Stokes ’ Theorem to compute where: F(x, y, z) = xz i + yz j + xy k S is the part of the sphere x 2 + y 2 + z 2 = 4 that lies inside the cylinder x 2 + y 2 =1 and above the xy-plane.
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P3113.8 Example 2 SOLUTION To find the boundary curve C, we solve: x 2 + y 2 + z 2 = 4 and x 2 + y 2 = 1 Subtracting, we get z 2 = 3. So, (since z > 0). So, C is the circle given by: x 2 + y 2 = 1,
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P3213.8 Example 2 SOLUTION A vector equation of C is: r(t) = cos t i + sin t j + k 0 ≤ t ≤ 2 Therefore, r ’ (t) = – sin t i + cos t j Also, we have:
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P3313.8 Example 2 SOLUTION Thus, by Stokes ’ Theorem,
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P3413.8 STOKES ’ THEOREM Note that, in Example 2, we computed a surface integral simply by knowing the values of F on the boundary curve C. This means that: If we have another oriented surface with the same boundary curve C, we get exactly the same value for the surface integral!
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P3513.8 STOKES ’ THEOREM In general, if S 1 and S 2 are oriented surfaces with the same oriented boundary curve C and both satisfy the hypotheses of Stokes ’ Theorem, then This fact is useful when it is difficult to integrate over one surface but easy to integrate over the other.
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P3613.8 CURL VECTOR We now use Stokes ’ Theorem to throw some light on the meaning of the curl vector. Suppose that C is an oriented closed curve and v represents the velocity field in fluid flow.
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P3713.8 CURL VECTOR Consider the line integral and recall that v ‧ T is the component of v in the direction of the unit tangent vector T. This means that the closer the direction of v is to the direction of T, the larger the value of v ‧ T.
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P3813.8 CIRCULATION Thus, is a measure of the tendency of the fluid to move around C. It is called the circulation of v around C. See Figure 5.
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P3913.8 CURL VECTOR Now, let P 0 (x 0, y 0, z 0 ) be a point in the fluid. S a be a small disk with radius a and center P 0. Then, (curl F)(P) ≈ (curl F)(P 0 ) for all points P on S a because curl F is continuous.
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P4013.8 CURL VECTOR Thus, by Stokes ’ Theorem, we get the following approximation to the circulation around the boundary circle C a :
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P4113.8 CURL VECTOR The approximation becomes better as a → 0. Thus, we have:
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P4213.8 CURL & CIRCULATION Equation 4 gives the relationship between the curl and the circulation. It shows that curl v ‧ n is a measure of the rotating effect of the fluid about the axis n. The curling effect is greatest about the axis parallel to curl v.
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P4313.8 CURL & CIRCULATION The paddle wheel rotates fastest when its axis is parallel to curl v.
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P4413.8 CLOSED CURVES Finally, we mention that Stokes ’ Theorem can be used to prove Theorem 4 in Section 13.5: If curl F = 0 on all of, then F is conservative.
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P4513.8 CLOSED CURVES From Theorems 3 and 4 in Section 13.3, we know that F is conservative if for every closed path C. Given C, suppose we can find an orientable surface S whose boundary is C. This can be done, but the proof requires advanced techniques.
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P4613.8 CLOSED CURVES Then, Stokes ’ Theorem gives: A curve that is not simple can be broken into a number of simple curves. The integrals around these curves are all 0.
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P4713.8 CLOSED CURVES Adding these integrals, we obtain: for any closed curve C.
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