Section A Vector Calculus, Maths II REVISION (VECTORS) 1. Position vector of a point P(x, y, z) is given as + y and its magnitude by 2. The scalar components of a vector are its direction ratios, and represent its projections along the respective axes. 3. The magnitude (r), direction ratios (a, b, c) and direction cosines (l, m, n) of any vector are related as: a b l, m, n r r c r 4. The vector sum of the three sides of a triangle taken in order is 5. The vector sum of two co-initial vectors is given by the diagonal of the parallelogram whose adjacent sides are the given vectors. 6. The multiplication of a given vector by a scalar λ, changes the magnitude of the vector by the multiple λ, and keeps the direction same (or makes it opposite) according as the value of λ is positive (or negative). 7. For a given vector, the vector gives the unit vector in the direction of.
8. The position vector of a point R dividing a line segment joining the points P and Q whose position vectors are respectively, in the ratio m: n (i) internally, is given by na mb m n (ii) externally, is given by mb na m n 9. The scalar product of two given vectors having angle θ between them is defined as Also, when is given, the angle θ between the vectors may be determined by. 10. If θ is the angle between two vectors, then their cross product is given as =, where is a unit vector perpendicular to the plane containing such that form right handed system of coordinate axes. If we have two vectors, given in component form as + and + and λ any scalar, then
+( +( + ; λ = (λ ; = ; and. 11. Triple Scalar Product: The product (u v).w is called triple scalar product of u, v, and w and (u v).w = u v 1 w 1 1 u v w 2 2 2 u v w 3 3 3. Directional Derivatives Discussion: f x gives the rate of change of the function f along the x-axis, or in the direction of i. f gives the rate of change of the function f along the y-axis, or in the direction of j. y What we would like to do is to find the rate of change of the function in an arbitrary direction.
Definition: Let w = f(x, y), and let u = u 1 i + u 2 j be a unit vector. The directional derivative of f at P(x, y) in the direction of u, denoted by D u f(x, y), is D u f(x, y) = lim s 0 f ( x su, y 1 su2 ) s f ( x, y). We also refer to D u f(x, y) as the directional derivative of f at P(x, y) in the direction of a, where a is any vector in the direction of the unit vector u. Note: D i f(x, y) = f x (x, y) and D j f(x, y) = f y (x, y). Exercise: Estimate the directional derivative of f(x, y) = x 2 y ln(xy 2 + 5x + 2y) at P(1, 2) in the direction a = 1.2i + 3.7j using the definition with s = 0.02, 0.01, and 0.005. Theorem: If f is differentiable, then f has a directional derivative in the direction of any unit vector u a, b and Du f ( x, y) fx( x, y) a f y( x, y) b fx( x, y), f y( x, y). a, b Note: If u forms an angle with the positive x-axis in the xy-plane, then u = cos, sin Exercise: Find the directional derivative D u f(x, y) if f(x, y) = x 2 + y 2 + (xy) 2 and
u = 1, 3 2 3. The first vector in the dot product version of the theorem has a special name: it is called the gradient vector of f, and is denoted by gradf or f. Definition: (gradient) The gradient of f(x, y) is the vector function defined by f f ( x, y), f ( x, y). There are many applications of the gradient vector. One is in finding directional derivatives, as we have just seen. Another is given by the following theorem: x y Theorem: The maximum value of the directional derivative D u f(x, y) is f ( x, y) and it occurs when u has the same direction as the gradient vector. The fact that the maximum value of the directional derivative occurs when the direction is along the gradient vector is interesting. It tells us that by simply following the direction of the gradient vector, we can find out where a function changes most quickly. Vector-Valued Functions Definition: A vector-valued function r(t ) is a mapping from its domain D to its range R V 3 (V 3 is the set of all vectors in 3-space), so that for each t in D, r(t ) = v for only one vector v V 3. We can always write a vector-valued function as
r(t ) = f (t )i + g(t )j +h(t )k, (1.1) for some scalar functions f, g and h (called the component functions of r ). We can likewise define a vector-valued function r(t ) in V 2 by r(t ) = f (t )i+g(t )j, for some scalar functions f and g. Examples: (1) Sketch a graph of the curve traced out by the endpoint of the two-dimensional vector-valued function r(t ) = (t+1)i+(t 2-2)j. (2) Sketch a graph of the curve traced out by the endpoint of the vector-valued function r(t) = 4cos t i -3sin t j, t. (3) Plot the curve traced out by the vector-valued function Problem: Find the arc length of the curve traced out by the endpoint of the vector-valued function Solution: First, notice that for 1 t e..
s Definition: For a vector-valued function the limit of r(t ) as t approaches a is given by provided all of the indicated limits exist. If any of the limits indicated on the right-hand side of (2) fail to exist, then (2) Example: does not exist.
Definition: A vector-valued function is continuous at t = a if and only if all of f, g and h are continuous at t = a. Definition: The derivative r' (t ) of the vector-valued function r(t ) is defined by for any values of t for which the limit exists. When the limit exists for t = a, we say that r is differentiable at t = a. Theorem: Let and suppose that the components f, g and h are all differentiable for some value of t. Then r is also differentiable at that value of t and its derivative is given by Example: Find the derivative of
Theorem: Suppose that r(t ) and s(t ) are differentiable vector-valued functions, f (t ) is a differentiable scalar function and c is any scalar constant. Then (i) (ii) (iii) (iv) (v) Definition: The vector-valued function R(t ) is an antiderivative of the vector-valued function r(t ) whenever R' (t ) = r(t ). Note: Notice that if and f, g and h have antiderivatives F, G and H, respectively, then That is, is an antiderivative of r(t ). In fact,
is also an antiderivative of r(t ), for any choice of constants c 1, c 2 and c 3. This leads us to the following definition. Definition: If R(t ) is any antiderivative of r(t ), the indefinite integral of r(t ) is defined to be where c is an arbitrary constant vector. Example: Evaluate the indefinite integral Solution: where is an arbitrary constant vector. Definition: For the vector-valued function,
we define the definite integral of r(t ) by Note: Suppose that R(t ) is an antiderivative of r(t ) on the interval [ a, b]. Then, Example: Evaluate Solution: Curl and Divergence of Vector Fields Definition:
A vector field in three dimensions is a function F whose domain D is a subset of subset of V 3. If (x,y,z) is in D, then F( x, y, z) M( x, y, z) i N( x, y, z) j P( x, y, z) k, 3 R and whose range is a As a special case of the above if D is in the xy-plane and if the range is a subset of the set of all 2- dimensional vectors, then a vector field F in two dimension is given by F(x, y) = M(x,y) i + N(x, y) j, where M and N are scalar functions.
Definition: (Inverse Square Field) Let r xi y j zk be the position vector for (x,y,z), and let the vector u r r denote the unit vector u r that has the same direction as r. A vector field F is an inverse square field if F( x, y, z) c c 2 3 r r where c is a scalar. Definition: (Conservative Field): A vector field F is conservative if for some scalar function f. F( x, y, z) f ( x, y, z ) Note: Every inverse square vector field is conservative. The Vector Differential Operator: The vector differential operator in three dimensions is i j k x y z. If operates on a scalar function f, it produces the gradient We next use we denote by curl F, or F. f f f grad f f i j k x y z as an operator on a vector field to define another vector field called the curl of F which
Definition: (curl of a vector field) Let F(x,y,z)=M(x,y,z)i+N(x,y,z)j+P(x,y,z)k, where M, N and P have partial derivatives in some region. The curl of F, denoted by curl F, or F is given by curl F= F = P N M P N M i j k y z z x x y Determinant form of curl F: i j k Curl F = F = x y z M N P Definition: (divergence of a vector field) Let F(x,y,z)=M(x,y,z)i+N(x,y,z)j+P(x,y,z)k, where M, N and P have partial derivatives in some region. The divergence of F, denoted by div F or.f, is given by div F =.F = M N P x y z. Physical Interpretation of divergence and curl: 1. Divergence: Imagine that is the velocity of a fluid (say, oil or water) at point (x, y, z) and time t. The variable t plays no role in calculating
Now imagine a small rectangular box in the fluid, as shown in the figure below: We are interested in measuring the rate per unit volume of fluid flow out of the box across its faces at time t. To do this, look at opposite pairs of faces. First, look at the front and back faces, labeled II and I, parallel to the yz-plane. If x is small, then F on face II is approximately The outer normal to this face is i, and the area is Thus, = (normal component of fluid velocity outward across face II) (area of face II) = (flux of flow out of the box across face II). On face I, F is about F(x, y, z. t) and the outer normal is i, hence, the flux out of the box across face I is approximately or.a similar calculation holds for the other sides. Summing, the total flux out of the box is
The outward flux per unit volume is the outward flux divided by the volume x y z. This is As, and this gives us which is then interpreted as the outward flux per unit volume of the flow at point (x, y, z) and time t. Thus, the divergence of F measures the expansion or "divergence away from the point" of the fluid at the point. If the fluid is incompressible, there can be no gain or loss in the volume element. Hence div V = 0, which is known in Hydrodynamics as the equation of continuity for incompressible fluids. If the flux entering any element of space is the same as that leaving it i.e. div V = 0 everywhere, then such a point function is called a solenoidal vector function. 2. Curl: In general, the curl of any vector point function gives the measure of the angular velocity at any point of the vector field.
Consider curl in connection with a rotating object. Imagine that we have a body rotating with uniform angular speed about a line L, If is the angular speed, the angular velocity vector has magnitude and direction along L in the direction a right-handed screw would move if given the same rotation as the object. We suppose that L goes through the origin of our system and R = xi + yj + zk for any point (x, y, z) on the rotating object. Angular velocity as the curl of the linear velocity Let T be the tangential (linear) velocity at any point. Then, Since T is also in the direction of then, in fact, If then But we also have that because are constants. Thus, we have that which says that the angular velocity of the uniformly rotating body is a scalar multiple of the curl of the linear velocity. This is one motivation for the term curl. In fact, curl
was once commonly known as rotation, and curl F was written rot F. It is also motivation for the term irrotational for a vector field whose curl is the zero vector. In the context of fluids, direction. measures the degree to which a fluid swirls, or rotates, about a given Any motion in which the curl of the velocity vector is zero is said to be irrotational, otherwise rotational. Del and Laplacian Use of the del operator ( ) gives us a compact way of writing divergence, curl, and the identities involving them. It is also used to define the del squared or Laplacian operator: If is a scalar field, then is a vector, and the Laplacian of is defined to be the scalar field This is often denoted and the equation is called Laplace's equation. It arises in heat and wave conduction, and is one of the fundamental equations of physics and engineering. Here are two basic relationships between gradient, divergence, and curl. (1) If is continuous with continuous first and second partial derivatives, then [In words, this says that is the zero vector.] Proof: First, Then,
Each component of is a difference of equal mixed partials, and hence is 0. (2) Let F be a continuous vector field with continuous first and second partials. Then, [In words, div(curl F) is identically zero.] Proof: Let F = F 1 i + F 2 j + F 3 k. Referring to the definition of we have After rearranging terms, we get Exercise: Prove: grad div F = curl curl F + 2 F Line Integrals To develop the notion of an integral of vector fields over curves in space, we need some terminology concerning curves. Usually, a curve C is given in parametric form as x = x(t), y = y(t), z = z(t); a t b.
C is the graph that consists of all points (x(t), y(t), z(t)) for a t b. One may also think of the curve C as a pivoted vector R(t) = x(t)i + y(t)j + z(t)k which sweeps through the graph as t varies from a to b. In this sense, the direction of the parameter assigns a direction to C. If t varies from a to b, we go from Circle swept out by R = cos(t)i + sin(t)j + k as t varies from 0 to 2π. (x(a), y(a), z(a)) to (x(b), y(b), z(b)). We call (x(a), y(a), z(a)) the initial point of C and (x(b), y(b), z(b)) the terminal point. If we vary t from b to a, then we reverse direction on C, and the initial and terminal points reverse their roles. Often, we put an arrow on C to indicate direction and we write t: a b to indicate that t varies from a to b or, similarly, t: b a to say that t varies from b to a. Example 1: Let x(t) = cos(t), y(t) = sin(t), z(t) = 1, where 2 0 t. This is the circle x 2 + y 2 = 1 in the plane z = 1 (see figure above). Here, t varies from 0 to 2π; so on C we go from initial point
(x(0), y(0), z(0)) = (1, 0, 1) to terminal point (x(2π), y(2π), z(2π)), which is also (1,0, 1). As t varies from 0 to 2π, (x(t), y(t), z(t)) goes around C counterclockwise if viewed from a point above (1, 0, 1) on the positive z-axis. If t varied from 2π to 0, we would reverse direction on C and go around clockwise. Note that the vector R(t) = cos(t)i + sin(t)j + k sweeps out C as t: 0 2. Definitions: We call C a closed curve if the initial and terminal points are the same. A closed curve C is called a simple closed curve if C does not cross itself. C is continuous if x, y, and z are continuous functions of t on [a, b]. A curve is a smooth curve if x', y', and z' are continuous on [a, b] and not simultaneously zero for < t < b. If C is smooth then C has a tangent vector R'(t) = x'(t)i + y'(t)j + z'(t)k. a Finally, C is piecewise smooth if C consists of a finite number of smooth curves. That is, C is smooth if x', y', and z' are continuous except for finitely many points in a < t < b. We call a curve regular if it is simple and piecewise smooth. The positive direction for the line integral around a simple closed curve C is that direction a point on the curve must move in order to keep the region R bounded by C to the left. If C is not a closed curve, then the positive direction on C is the direction corresponding to the increasing values of the parameter t. Example 2: Show that the following curve C is regular: Solution: Here, R(t) = 2t i - 3t 2 j + t k sweeps out C as t varies from 1 to 2. The initial point is (2, -3, 1), and the terminal point is (4, -12, 2). This curve is smooth, as R'(t) = 2i - 6tj + k is continuous for1 t 2. It is not closed, as the terminal and initia1 points differ. If (2t 1, -3t 2 1, t 1 ) = (2t 2, -3t 2 2, t 2 ), then z = t 1 = t 2, so C is simple as the curve visits each point only once. Thus C is regular. Exercise: Show that the following curve H is regular: Example 3: Let D be given by Since x 2 y 2 = 1, this is a circle of radius 1 about the (0, 0, 1) in the plane z = 1. It is a closed curve, as
the initial point (1, 0, 1) coincides with the terminal point. While the graph of D is the same as that of C in Example 1, there is a considerable difference since D goes around the circle twice, once as t: 0 2, and once again as t: 2 4. Thus, D is not simple (e.g., we hit (0, 1, 1) at t = π/2 and again at t = 5π/2). D is smooth, as the coordinate functions have continuous derivatives on (0, 4π). Note: This example emphasizes that our curves are really more than just graphs. If we imagine a particle moving around the circle, we can see that going around twice, as with D, might expend more energy than going around just once, as with C. Line Integral of a Vector Field over a Curve Let curve C which is given by C is denoted be continuous at least on the graph of a regular Then, the line integral of F over and is defined by Here, R(t) = x(t)i + y(t)j + z(t)k, and Then, In words, replace x, y, and z in F 1, F 2, and F 3 by the parametrized functions of t describing C, take the dot product with the tangent to C, and integrate the resulting function of t from a to b in the usual way. Properties:
1. For vector fields F and G 2. For any scalar α, 3. If we reverse direction on C to get a curve K (same graph, but initial and terminal points reversed), then 4. If C is regular, consisting of smooth pieces C 1, C 2,, C n (with the terminal point of C i, the same as the initial point of C i+1 ), then Example 4: Let and let C be given by x = t, y = 2t, z = -4t; Then R(t) = t i + 2t j - 4t k and Replacing x by t, y by 2t, and z by - 4t in F gives us
Then and we have Example 5: Let and let C be given by Here, Putting x = t 2, y = 3t, and z = 0 into F gives us F = 2 t 2 i - e t j + k. Then, and Example 6: Let F = -i + xyz j y 2 k, and let C be given by x = t, y = t, z = l; 1 t 1. Then, C consists of smooth pieces C 1 and C 2, where
C 1 : x = t, y = -t, z = 1; t: 1 0 and C 2 : x = t, y = t, z = 1; t: 0 1 On C 1 : On C 2 : Hence, Then, Example 7: Evaluate over the quarter-circle in the plane z = 1 from (2, 0, 1) to (0, 2, 1). Here, we may use polar coordinates and write the curve C as On C:
and so Equivalent Notations for Line Integrals: 1. The line integral is also written as Suppose the curve C is given by x = x(t), y = y(t), and z = z(t), for t: a b. Then R = x i + y j + z k and If then 2. There is also a differential notation which is useful in remembering how to evaluate line integrals. Since
we often write In actually evaluating the above integral, one simply substitutes x = x(t), y = y(t), and z = z(t) from C and integrates over the parameter interval. Example 8: Evaluate where and C is given by On C, dx = - dt, dy ( 1/ 2 t ) dt, and dz = 3dt. Then, Example 9: Evaluate
where F = 2xy i + zy j e z k and C is the parabola y = x 2, z = 0, from (0, 0, 0) to (2, 4, 0) in the xy-plane. Solution: On C, considering x as a parameter, we have Thus, dx = dx, dy = 2x dx, dz = 0 dx Alternative solution using y as a parameter: Write C as On C, C: x = y, y = y, z = 0; y: 0 4. Then, Physical Interpretation of Line Integral: A line integral
has a physical interpretation as work done by the force F along C from initial to terminal point. To understand this, recall from elementary physics that F.D is by definition the work done by constant force F along a constant direction given by vector D. Now, if F(x, y, z) is a (not necessarily constant) vector field and C is a regular curve (not necessarily a straight line), then at any point (x(t), y(t), z(t)) on C gives the work done by F at that point in the direction of C (i.e., along its tangent direction). "Summing" over all the points on C is achieved by integrating this gives us as the total work done by F along C. over the parameter interval, and Example 10: Find the work done by (a) F = x i + y j, and along the curve C traced by r(t) = cos t i + sin t j from t = 0 to t = π. Solution: (a) The vector function r(t) gives the parametric equations x = cos t, y = sin t, 0 t, which is a half circle. As seen in the figure, the force field F is perpendicular to C at every point. (radius of a circle is perpendicular to the tangent). Since the tangential components of F are zero, the work done along C is zero. To see this we use the line integral giving the work done by F:
(b) The work done by F is Line Integrals of Scalar Fields The differential notation for a line integral suggests a definition of line integrals of scalar fields which is sometimes useful. If (x, y, z) is a scalar field, and C is a regular curve parametrized by x = x(t), y = y(t), z = z(t); t: a b, then we define three kinds of line integral of : 1. 2.
3. If s (arc length) is used to parametrize C, x = x(s), y = y(s), z = z(s), 0 s L,then the line integral of with respect to s is defined by If C is parametrized by t, and we want to evaluate we need not explicitly reparametrize C in terms of s, since Thus, in terms of t, Example 11: Let (x, y, z) = xyz 2 and let C be given by C: x = 2t, y = t, z = t, 1 t 3. On C: dx = 2 dt, dy = dt, and dz = (1/2 t) dt Further,
Then, Finally, in terms of the given parametrization, The value of the above integral cannot be obtained by elementary means, but can be approximated using numerical methods. Independence of the Path: A line integral whose value is the same for every curve connecting two points A and B is said to be independent of the path. Fundamental Theorem: Suppose there exists a function ( x, y) such that d P dx Q dy, that is, P dx + Q dy is an exact differential. Then
depends only on the endpoints A and B of the path C, and Test for path independence Let P and Q be continuous and have continuous first partial derivatives in a simply connected region. Then is independent of the path C if and only if for all (x, y) in the region. Conservative vector fields If is independent of the path C, we know there exists a function ( x, y) such that
where F = P i + Q j is a vector field and In other words, the vector field F is a gradient of the function ( x, y). Since F is called a gradient field and the function ( x, y) is known as a potential function for F. In a gradient force field F, the work done by the force upon a particle moving from position A to position B is the same for all paths between these points. Moreover, the work done by the force along a closed path is zero. For this reason, such a force field is also said to be conservative. In a conservative field F the law of conservation of mechanical energy holds: For a particle moving along a path in a conservative field, Kinetic Energy + Potential Energy = constant. In a simply connected domain (that is, interior of a simple closed curve without holes), a force field F(x, y) = P(x, y)i + Q(x, y)j is a gradient field (that is, conservative) if and only if Example: Show that the vector field is a gradient field. Find a potential function for F. Solution: By identifying P = y 2 + 5 and Q = 2xy - 8, we see that
Hence, F is a gradient field and so there exists a potential function satisfying Integration of the first equation with respect to x and using the second to get the constant of integration, we find Check: Surface Integrals: Let G(x, y, z) be a function of three variables defined over a region of space containing a surface S. Then the surface integral of G over S, denoted by, is defined in similar steps as in multiple integral. Method of Evaluation: Suppose z = f (x, y) is the equation of S, then the differential (for the surface area) is given by If G, f, f x, and f y are continuous throughout the region containing S, we can evaluate the surface integral of G over S by means of the following double integral over R (R is the projection of the surface S on to the xy-plane):
Note: If G(x, y, z) = 1, the above integral gives the surface area of S. Projection of S into other planes If y = g (x, z) is the equation of a surface S that projects onto a region R of the xz-plane, then Similarly, if x = h(y, z) is the equation of a surface that projects onto the yz-plane, then Integrals of vector fields If F(x, y, z) = P(x, y, z)i + Q(x, y, z)j + R(x, y, z)k is the velocity field of fluid, then as shown in the figure, the volume of the fluid flowing through an element of surface area ds per unit time is approximated by (height). (area of base) = (comp n F) ds = (F. n) ds,
where n is a unit normal to the surface. The total volume of a fluid passing through S per unit time is called the flux of F through S and is given by In the case of a closed surface S, if n is the outer (inner) normal, then the above integral gives the volume of fluid flowing out (in) through S per unit time. Example: Let F(x, y, z) = zj + zk represent the flow of a liquid. Find the flux of F through the surface S given by that part of the plane z = 6-3x - 2y in the first octant. Solution: The vector field and the surface are shown in the figure. If the plane is defined by g(x, y, z) = 3x + 2y + z - 6 = 0, then a unit normal is
Hence, or, Remarks:(i) Depending on the nature of the vector field, the above integral can represent electric flux, magnetic flux, flux of heat, and so on. (ii) There are always two normal to a surface: an "upper" and a "lower" normal. We always take the "upper normal"; that is, the normal with a positive k component. For a closed surface we take n to be an outer normal. Integral Theorems 1. Green'sTheorem (a) Green's Theorem in the Plane: Suppose that C is a piecewise smooth simple closed curve bounding a region R. If P, Q, P / y, and Q / x are continuous on R, then Example: Evaluate where C consists of the boundary of the region in the first quadrant that is bounded by the graphs of y = x 2 and y = x 3.
Solution: If P(x, y) = x 2 - y 2 and Q(x, y) = 2y - x, then P / y = -2y, and Q / x = -1. We have Note that the above line integral could have been evaluated in a straightforward manner using the variable x as a parameter. However, in the next example, ponder the problem of evaluating the given line integral in the usual manner. Example: Evaluate where C is the circle (x - 1) 2 + (y - 5) 2 = 4. Solution: Identifying P(x, y) = x 5 + 3y and Q(x, y) = 2x - 2 y e, we have Hence,
Since the double integral gives the area of the region R bounded by the circle of radius 2, we have Vector form of Green's Theorem If F(x, y) = P(x, y)i + Q(x, y)j is a two-dimensional vector field, then Clearly, Green's Theorem can be written in vector notation as That is, the line integral of the tangential component of F is the double integral of the normal component of curl F. (b) Green's Theorem in 3-space
The vector form of Green's Theorem generalizes from a simple closed curve in the plane to a simple closed curve in 3-space. Suppose z = f(x, y) is a continuous function having continuous first partial derivatives whose graph is a surface S over a region R on the xy-plane. Let C be a simple closed curve forming the boundary of S and let its projection C xy onto the xy-plane form the boundary of R. The positive direction on C is defined by the positive direction on C xy. Furthermore, let T be a unit tangent to C and let n be a unit upper normal to S (see figure). The three dimensional form of Green's Theorem, is called Stokes' Theorem. 2. Stokes' Theorem Let F(x, y, z) = P(x, y, z)i + Q(x, y, z)j + R(x, y, z)k be a vector field for which P, Q, and R are continuous and have continuous first partial derivatives in a region containing a surface S. If C is the boundary of S traversed in the positive direction, then The above is not the full statement of Stokes' Theorem. It is limited to surfaces on which we can take n as an upper normal. The complete statement and proof for two sided surfaces, or surfaces with an orientation, can be found in most advanced calculus texts.
Example: Let S be the part of the cylinder Verify Stokes' Theorem if F = xyi + yzj + xzk. Solution: The surface S, the curve C, which is composed of the union of C 1, C 2, C 3, and C 4, and the region R are shown in the figure below. The Surface Integral: From F = xyi + yzj + xzk, we find Now, if defines the cylinder, the upper normal is
Therefore, Using we get The Line Integral: We write
On C 1 : x = 1, z = 0, dx = 0, dz = 0, so Hence, which agrees with the value of the surface integral. Example: Evaluate
where C is the trace of the cylinder x 2 + y 2 = 1 in the plane y + z = 2. Solution If F = zi + xj + yk, then Moreover, if g(x, y, z) = y + z - 2 = 0 defines the plane, then the upper normal is Hence, from we get
3. Divergence Theorem Let F = Pi + Qj and let T = (dx/ds)i + (dy/ds)j be a unit tangent to a simple closed plane curve C. In vector form of Green s Theorem we saw that can be evaluated by a double integral involving curl F. Similarly, if n = (dy/ds)i - (dx/ds)j is a unit normal to C (check T. n), then can be expressed in terms of a double integral of div F. From Green's Theorem, That is,
The above result is a special case of the Divergence Theorem. The following is a generalization of the above to 3-space. Divergence Theorem (also known as Gauss' Theorem) Let F(x, y, z) = P(x, y, z)i + Q(x, y, z)j + R(x, y, z)k be a vector field for which P, Q, and R are continuous and have continuous first partial derivatives in a region that contains a surface S, which joins the boundary of a closed and bounded region D in space. Then where n is an outer normal to S. Example: Let D be the region bounded by the hemisphere Verify the divergence theorem if F = xi + yj + (z - l)k. Solution The closed region D is shown in the figure below.
The Triple Integral: Since F = xi + yj + (z - l)k, we see div F = 3. Hence, In the last calculation, we used the fact that gives the volume of the hemisphere. The Surface Integral: We write where S 1 is the hemisphere and S2 is the plane z = 1. If S 1 is a level surface of g(x, y, z) = x 2 + y 2 + (z -1) 2, then a unit outer normal is Now,
and so from the formula for the surface integral we have On S 2, we take But, since z = 1, Hence, we see that agrees with the value obtained earlier. Example: Solution: Instead of evaluating six surface integrals for each of the six plane faces of the given cube, we apply the Divergence Theorem to obtain the surface integral. Since
we have,