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Unformatted text preview: Section 17.4 Green’s Theorem
“Calculating Line Integrals using Double Integrals” In the previous section, we saw an easy way to determine line integrals in the special case when a vector ﬁeld F is conservative. This still leaves the problem of ﬁnding a line integral over a vector ﬁeld which is not conservative. In this section we shall consider a way to determine line integrals of vector ﬁelds which are not necessarily conservative over closed curves. 1. Greens Theorem Green’s Theorem gives us a way to transform a line integral into a double integral. To state Green’s Theorem, we need the following definition. Deﬁnition 1.1. We say a closed curve C has positive orientation if it is traversed counterclockwise. Otherwise we say it has a negative orientation. The following result, called Green’s Theorem, allows us to convert a line integral into a double integral (under certain special conditions). Result 1.2. (Green’s Theorem) Let C be a positively oriented piecewise smooth simple closed curve in the plane and let D be the region bounded by C . If P and Q have continuous partial derivatives on an open region which contains D and F = P i + Qj , then
C F · dr = P dx + Qdy =
C D ∂ Q ∂P dA. − ∂x ∂y Note that we are integrating the curl of the vector ﬁeld F , so we sometimes write Green’s Theorem as:
C F · dr = curl(F )dA.
D Notation: Sometimes we use the notation ∂D to denote C (meaning the boundary of D ). Notation: Sometimes we use the notation over a closed curve. to denote a line integral We illustrate Green’s Theorem with some examples. Example 1.3. Evaluate the line integral y dx − xdy where C is the unit circle centered at the origin oriented counterclockwise both directly and using Green’s Theorem.
1 2 (i ) To evaluate it directly, we ﬁrst note that C is parameterized by r (t) = cos (t)i + sin (t)j with 0 t 2π . Then we have x′ (t) = − sin (t) and y ′(t) = cos (t), so we get
2π 2π y dx − xdy = 0 (sin (t))(− sin (t))dt −
2π 0 (cos (t))(cos (t))dt
0 =− (cos2 (t) + sin2 (t))dt = −2π (ii ) To evaluate using Green’s Theorem, we have ∂Q − ∂P = −1 − ∂x ∂y 1 = −2. Since the curve is oriented counterclockwise, we can apply Green’s Theorem:
2π C 1 0 ydx − xdy = D −2dA = −2 0 rdrdϑ = −4π r2 2 1 0 = −2π Example 1.4. Use Green’s Theorem to evaluate √ √ ( x + y 3)i + (x2 + y )j · r
C where C is the curve y = sin (x) from (0, 0) to (π, 0) and the line segment from (π, 0) to (0, 0). First observe that this curve is oriented in the wrong direction (see illustration). However, as we observed before, we have −C F · dr = − C F · dr, so we need to ﬁnd the line integral and then simply negate the answer. Using Greens Theorem, we have
π −C π sin (x) 0 π F · dr =
sin (x) 3 0 (2x − 3y 2)dydx 2x sin (x) − (sin (x))3 dx 4 3 =
0 (2xy − y )
π dx =
0 0 =
0 2x sin (x) − (sin (x))3 dx = 2π − (integration by parts and trig integrals). 3 Example 1.5. Let F= y x i− 2 j x2 + y 2 x + y2 t 2π , and observe that ∇f = F . If r = cos (t)i + sin (t)j with 0 calculate
C F · dr directly and then explain why you cannot calculate it using either FTC or Green’s Theorem. (i ) Directly we have
2π C ∇f · dr =
2π 0 x2 x y i− 2 j 2 +y x + y2 · (− sin (t)i + cos (t)j )dt
2π =
0 sin (t)i−cos (t) ·(− sin (t)i+cos (t)j )dt = 0 − sin2 (t)−cos2 (t)dt = −2π (ii ) If we were to try to use Greens Theorem, we would have ∂Q ∂P y 2 − x2 x2 − y 2 − =− 2 −2 = 0, ∂x ∂y (x + y 2 )2 (x + y 2 )2 so Green’s Theorem would say
C F · dr = 0dA = 0
D (iii ) If we were to try to use the Fundamental Theorem, since F = ∇(arctan (x)), the fundamental theorem would say
C F · dr = arctan (0) − arctan (0) = 0 Note that the second two answers do not match up with the direct calculation, and so something must have gone wrong? Notice that arctan (x/y ) is not deﬁned at the points (1, 0) and (−1, 0), and F is not deﬁned at (0, 0), In particular, in order to apply Green’s Theorem, F must be deﬁned inside and on C , and in order to apply FTC, f (x, y ) must deﬁned on C  therefore, in this case we cannot apply Green’s theorem or FTC, and the only way to calculate this integral is directly. We can also go in the other direction  Green’s Theorem can be used to turn a double integral into a line integral which sometimes may be more useful. One particular application is in ﬁnding the area of a region. Speciﬁcally, we have the following useful result: Result 1.6. If D is a closed simple region, then the area of D is given by 1 A= xdy = − ydx = xdy − ydx. 2 C C 4 We illustrate with an example. Example 1.7. Find the area under one arch of the cycloid x = t − sin (t), y = 1 − cos (t). One arc of the centroid (oriented clockwise) occurs for 0 t 2π , so we need to calculate 1dA where D is the region bounded by this D parametric equation and below by y = 0. Using Greens Theorem, it suﬃces to calculate xdy
C where C is the curve bounded by the cycloid and y = 0. Let C1 denote the part of the cycloid and C2 the horizontal component along the axis. Since C1 is oriented clockwise, we have
C xdy = − xdy +
C1 C2 xdy t 2π , so we A parameterization of C2 is r(t) = (2π − t)i with 0 have
2π xdy =
C2 0 (2π − t)0dt = 0. Then for C1 , we have
2π xdy =
C1 0 (t − sin (t)) sin (t)dt = −3π. It follows that the area will be 3π . Green’s Theorem can actually be modiﬁed to integrate over regions which contain holes by simply breaking a region up into smaller pieces so each of them are simply connected (observe that the integrals along the curves we cut along cancel out).
Cut Cut Green’s Theorem can also be used indirectly to calculate line integrals of a vector ﬁeld F over curves which are not necessarily closed. Speciﬁcally, if we have a line integral over the complicated curve C1 illustrated below, then we can convert it into a sum of a double integral and a line 5 integral over a much easier curve C2 illustrated below using Green’s Theorem:
C1 C2 Speciﬁcally, we have F dr +
C1 C2 F dr =
D curl(F )dA or equivalently F dr =
C1 D curl(F )dA − F dr.
C2 We illustrate with an example. Example 1.8. Find the line integral
C ((x − y )i + xj ) · dr ϑ π /2 where C is the segment of the circle x2 + y 2 = 9 with 0 oriented counterclockwise. Notice that if we parametrize this portion of the circle and evaluate this integral, we get a very messy trig integral. Therefore, we shall try to use Green’s Theorem indirectly. Let C1 be the line segment from the origin to the edge of the circle along the xaxis, and let C2 be the line from the end of the segment back to the origin (see illustration below). Then using Green’s Theorem, we have
C F · dr + C1 F · dr + C1 F · dr = curl(F )dA,
D 6 and we can calculate each of the integrals individually. Speciﬁcally, 9π 9π curl(F )dA = 1 − (−1)dA = 2dA = 2 × = 8 4 D D D (since we are integrating a constant, it will simply be the area of the region times the constant). For C1 , a parametrization will be r (t) = ti for 0 t 3. Thus,
3 C1 3 F · dr = 0 (ti + tj ) · (i)dt = 0 t2 2tdt = 2 3 0 9 =. 2 For C2 , a parameterization will be r (t) = (3 − t)i +(3 − t)j for 0 t 3, so 3 3 3 9 9 t2 F ·dr = (0i+(3−t)j )·(−i−j )dt = (t−3)dt = −3t = −9 = − . 2 2 2 C1 0 0 0 Therefore, 9π 9 9 9π F · dr = −+= . 4 22 4 C 2. Calculating Line Integrals: A Summary We have seen a lot of diﬀerent ways to calculate line integrals, so it would be a good idea to pause and try to reﬂect on the diﬀerent methods we have considered, and which method to use under which circumstances. Ultimately, this choice comes down to two simple questions: Is C closed? and Is F conservative? Ideally, we never want to calculate a line integral directly because it requires a lot of work, so answering these questions is crucial in making the calculations as easy as possible. As a general case, the decision tree on the following page is a good indicator of which method should be used under which circumstances. Is nC closed? UU
n nnn vnnn
No nnnn nn UUUU UUUU Yes UUUU UUUU U* Is i F iiii Noiiiii ii iiii iii tii Can Green’s be applied indirectly?
No conservative?
Yes Is F conservative? UU
No UUUU UUUUYes UUUU UUUU UU*
C Use FTC Can Green’s Theorem be applied?
No F dr = 0 by FTC Direct Calculation Direct Calculation 7 ...
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This note was uploaded on 10/28/2010 for the course MA 261 taught by Professor Stefanov during the Fall '08 term at Purdue UniversityWest Lafayette.
 Fall '08
 Stefanov
 Calculus, Integrals

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