01 Vector Calculus.pdf

BREE 319 Summary Package 1 Vector Calculus BREE 319 Vector Calculus Scalar functions A scalar function f (x, y) assigns one real number z to each ordered pair of real numbers (x, y) in some domain in the xy-p...

BREE 319 Summary Package 1 Vector Calculus BREE 319 Vector Calculus Scalar functions A scalar function f (x, y) assigns one real number z to each ordered pair of real numbers (x, y) in some domain in the xy-plane. sin(x) sin(y) graph of f (x, y) = xy level curves and color graph of average temperature (◦ F) near sea level in July The graph z = f (x, y) is a surface in 3-space. You should be familiar with the functions below and their graphs. Similarly, a function f (x, y, z) assigns one real number w to each ordered triple of real numbers (x, y, z) in some domain in 3-space. The graph w = f (x, y, z) is harder to visualize. 1 BREE 319 Vector Calculus Vector fields A vector field F~ (x, y) assigns one vector to each (x, y) in some domain in the xy-plane.   F~ (x, y) = P (x, y), Q(x, y) = P (x, y)ı̂ + Q(x, y)̂ velocity vector field of an ideal fluid around a cylinder velocity vector field F~ (x, y) of ocean currents  2  −y 2 F~ (x, y) = 1 − (xx2 +y 2 )2 ı̂ − (x22xy +y 2 )2 ̂ Similarly, a vector field F~ (x, y, z) assigns one vector to each (x, y, z) in some domain in 3-space.   F~ (x, y, z) = P (x, y, z), Q(x, y, z), R(x, y, z) = P (x, y, z)ı̂ + Q(x, y, z)̂ + R(x, y, z) k̂ velocity vector field F~ (x, y, z) of air around a scooter 2 BREE 319 Vector Calculus Gradient vector Recall that the gradient of a scalar function f is a vector:   ∂f ∂f ∂f ∂f ∇f (x, y) = ∂x , ∂y = ∂x ı̂ + ∂y ̂   ∂f ∂f ∂f ∂f ∂f ∂f ∇f (x, y, z) = ∂x , ∂y , ∂z = ∂x ı̂ + ∂y ̂ + ∂z k̂ The gradient vector points in the direction of steepest increase of f. red curves: level curves of the saddle f (x, y) = x2 − y 2 blue arrows: gradient vector field ∇f (x, y) = 2xı̂ − 2y ̂ 3 BREE 319 Vector Calculus Gradient, curl, and divergence The symbol ∇ is called the del vector operator.   ∂ ∂ ∂ ∂ ∂ ∂ ∇= ∂x , ∂y , ∂z = ∂x ı̂ + ∂y ̂ + ∂z k̂ You can manipulate the del vector operator just like a vector. The gradient of a scalar is a vector. This is just like multiplication of a vector times a scalar.     ∂ ∂ ∂ ∂f ∂f ∂f ∇f = ∂x , ∂y , ∂z f= ∂x , ∂y , ∂z The divergence of a vector is a scalar. This is just like the vector dot product.   ∇ · F~ = ∂ ∂ ∂ ∂x , ∂y , ∂z · (P, Q, R) = ∂P ∂x + ∂Q ∂y + ∂R ∂z The curl of a vector is a vector. This is just like the vector cross-product. ı̂ ̂ k̂     ∇ × F~ = ∂ ∂ ∂ ∂R ∂Q ∂R ∂P ∂Q ∂P
∂x ∂y ∂z = ı̂ ∂y − ∂z − ̂ ∂x − ∂z + k̂ ∂x − ∂y P Q R Laplace operator The divergence of the gradient of f (x, y, z) is a scalar.     ∂2f ∂2f ∂2f ∇ · ∇f = ∂x ∂ ∂ , ∂y ∂ , ∂z · ∂f , ∂f ∂f , ∂x ∂y ∂z = ∂x2 + ∂y 2 + ∂z 2 The operator ∇ · ∇ appears so often in partial differential equations that it has its own name and symbol, the Laplace operator, ∇2. We can apply the Laplace operator to a scalar function, or to a vector field component by component. The Laplacian of a scalar is a scalar, and the Laplacian of a vector is a vector. ∂2f ∂2f ∂2f ∇2 f = ∂x2 + ∂y 2 + ∂z 2 ∇2 F~ = ∇2 P, Q, R = ∇2 P, ∇2 Q, ∇2 R

4 BREE 319 Vector Calculus Line integrals of scalar functions R Given a scalar function f (x, y) and a curve C in the xy-plane, C f (x, y) ds is the line integral of f along C. The line integral represents the net area of the “curtain” between C and the surface z = f (x, y). Suppose the curve C is smooth and is given by the parametrization
~r(t) = x(t), y(t) , a ≤ t ≤ b. We break up the parameter interval [a, b] into n subintervals of equal length ∆t, which divides the curve C into n subarcs. The ith subarc joins the points Pi−1 and Pi , and has a length ∆si. We can approximate this length using a line segment. s 2  2 ∆xi ∆yi q 2 ∆si ≈ ∆xi + ∆yi = 2 + ∆t ∆t ∆t The area of the slice of curtain above the ith subarc is f (x∗i , yi∗ )∆si , where (x∗i , yi∗ ) is any point in the subarc. We add up all these areas and take the limit as n → ∞ to obtain an expression for the line integral. s Z n 2  2 X ∗ ∗ ∆xi ∆yi f (x, y) ds = lim f (xi , yi ) + ∆t C n→∞ i=1 ∆t ∆t If the curve C is smooth and the function f is continuous, this limit always exists and we can evaluate the line integral as follows. Z Z b
q
2
2 f (x, y) ds = f x(t), y(t) x0 (t) + y 0 (t) dt C a The idea is similar if f (x, y, z) is a continuous function
of three variables and C is a smooth curve in 3-space given by the parametrization ~r(t) = x(t), y(t), z(t) for a ≤ t ≤ b. Z Z b
q
2
2
2 f (x, y, z) ds = f x(t), y(t), z(t) x0 (t) + y 0 (t) + z 0 (t) dt C a 5 BREE 319 Vector Calculus Line integrals along piecewise-smooth curves The above formula are valid if C is a smooth curve. A curve given by the parametrization ~r(t) is called smooth if ~r 0 (t) is continuous and ~r 0 (t) 6= 0. Smooth curves have no sharp corners or cusps. A curve is called piecewise-smooth if it is a union of a finite number of smooth curves. a smooth curve
a piecewise-smooth curve ~r(t) = cos(t), sin(t) , 0 ≤ t ≤ π (a union of 5 smooth curves) If the curve C is piecewise-smooth, then we define the line integral of f along C as the sum of the line integrals of f along each of the smooth pieces of C. In the example above right, Z Z Z Z f (x, y) ds = f (x, y) ds + f (x, y) ds + · · · + f (x, y) ds C C1 C2 C5 Some applications of line integrals Length If f (x, y, z) = 1, then the line integral of f along C is just the length of the curve. Z length = ds C Mass If an object has the shape of a curve C and has mass density (or mass per unit length) of ρ(x, y, z), then the line integral of ρ along C is the mass of the object. Z mass = ρ(x, y, z) ds C 6 BREE 319 Vector Calculus Line integrals of vector fields The work done by a constant force F~ that acts on an object as it moves in a straight line displacement D ~ ~ ~ ~ ~ is the dot product W = F · D = |F ||D| cos(θ). More generally, the work done by a force F~ (x, y, z) = P (x, y, z)ı̂ + Q(x, y, z)̂ + R(x, y, z) k̂ that acts on an object as it moves along a curve C is a line integral. Suppose C is smooth and is given by the parametrization
~r(t) = x(t), y(t), z(t) , a ≤ t ≤ b. We break up the parameter interval [a, b] into n subintervals of equal length ∆t, which divides the curve C into n subarcs. The ith subarc joins the points Pi−1 and Pi. We can approx- imate its length by ∆si , and its direction by T~ (t∗i ), the unit tangent vector to the curve at any point (x∗i , yi∗ , zi∗ ) in the subarc. In other words, the displacement vector from Pi−1 to Pi is approximately T~ (t∗i )∆si. The work done by the force as the object moves along the ith subarc is the dot product F~ (x∗i , yi∗ , zi∗ ) · T~ (t∗i )∆si. We add up all the work done along each subarc and take the limit as n → ∞ to obtain an expression for the line integral. Xn Z work = lim F~ (x∗i , yi∗ , zi∗ ) · T~ (t∗i )∆si = F~ · T~ ds n→∞ C i=1 To compute this line integral, you don’t have to compute the unit tangent vector. Recall that ~r 0 (t) q
2
2
2 T~ (t) = 0 , ds = x0 (t) + y 0 (t) + z 0 (t) dt = |~r 0 (t)| dt. |~r (t)| Therefore, Z Z b
~r 0 (t) 0 Z b F~ · T~ ds = F~ ~r(t) · 0 F~ ~r(t) · ~r 0 (t) dt
|~r (t)| dt = C a |~r (t)| a Different notations for this line integral are given below. Z Z Z Z b F~ · T~ ds = F~ · d~r = F~ ~r(t) · ~r 0 (t) dt
P dx + Qdy + Rdz = C C C a 7 BREE 319 Vector Calculus Fundamental theorem for line integrals Recall the fundamental theorem of calculus: if f (x) is a differentiable function whose derivative f 0 (x) is continuous on the interval [a, b], then Z b f 0 (x) dx = f (b) − f (a) a For line integrals, the analogous theorem is the following: if C is a smooth or piecewise-smooth curve given by the parametrization ~r(t), for a ≤ t ≤ b, and f is a differentiable function whose gradient ∇f is continuous on C, then Z

∇f · d~r = f ~r(b) − f ~r(a) C A vector field F~ that is the gradient of a scalar function f is called conservative. The line integral of a conservative vector field does not depend on the curve C, only on the initial point ~r(a) and the final point ~r(b). We say the line integral is independent of path. A curve is called closed if its initial point coincides with its final point, that is, ~r(a) = ~r(b). not closed closed I The line integral along a closed curve is denoted F~ · d~r. C The line integral of a conservative vector field along a smooth or piecewise-smooth closed curve is 0. I ∇f · d~r = 0 C 8 BREE 319 Vector Calculus Double integrals over rectangles The double integral ZZ f (x, y) dA R is the net volume between the region R in the xy-plane and the surface z = f (x, y). Suppose the region R in the xy-plane is a rectangle:  R = [a, b] × [c, d] = (x, y) | a ≤ x ≤ b, c ≤ y ≤ d. We break up the rectangle R into an m × n grid of subrectangles, and we label each subrectangle Rij with two indices i and j. Each subrectangle has dimensions ∆x by ∆y, where ∆x = b−am and ∆y = n. d−c Using this grid, we break up the volume into boxes. The area of the base of the box is ∆x∆y, and the height of the box (i, j) is f (x∗ij , yij ∗ ), where (x∗ij , yij ∗ ) is any point in the rectangle Rij. The total volume of all the boxes is m X X n f (x∗ij , yij ∗ )∆x∆y. i=1 j=1 When we use more boxes, the volume of the boxes becomes a better approximation to the volume under the surface. The double integral is the limit as m and n go to infinity of the volume of the boxes. ZZ m X X n f (x, y) dA = lim f (x∗ij , yij ∗ )∆x∆y R m,n→∞ i=1 j=1 Use Fubini’s theorem to evaluate double integrals over rectangles. If f (x, y) is continuous on the rectangle R = [a, b] × [c, d], then ZZ Z b Z d Z d Z b f (x, y) dA = f (x, y) dydx = f (x, y) dxdy R a c c a First compute the inner integral, treating the outer variable as a constant. Then compute the outer integral. 9 BREE 319 Vector Calculus Double integrals over general regions A region D is of type I if  D= (x, y) | a ≤ x ≤ b, g1 (x) ≤ y ≤ g2 (x) You can traverse this region from bottom to top. The function y = g1 (x) is always at the bottom, and the function y = g2 (x) is always on top. For regions D of type I, ZZ Z b Z g2 (x) f (x, y) dA = f (x, y) dydx D a g1 (x) A region D is of type II if  D= (x, y) | c ≤ y ≤ d, h1 (y) ≤ x ≤ h2 (y) You can traverse this region from left to right. The function x = h1 (y) is always on the left, and the function x = h2 (y) is always on the right. For regions D of type II, ZZ Z d Z h2 (y) f (x, y) dA = f (x, y) dxdy D c h1 (y) 10 BREE 319 Vector Calculus Double integrals in polar coordinates Sometimes it is more convenient to describe a region R in the xy-plane using polar coordinates rather than Cartesian coordinates.  The above two regions are examples of polar rectangles: R = (r, θ) | α ≤ θ ≤ β, a ≤ r ≤ b. When we computed a double integral over a rectangle, we divided the rectangle into subrectangles with area ∆x∆y. To compute a double integral over a polar rectangle, we divide the polar rectangle into polar subrectangles of dimensions ∆θ by ∆r, whose area is r∆r∆θ. If R is a polar rectangle, ZZ Z β Z b
f (x, y) dA = f r cos(θ), r sin(θ) r drdθ R α a In other words, substitute x = r cos(θ), y = r sin(θ), and dA = r drdθ. Suppose  D= (r, θ) | α ≤ θ ≤ β, h1 (θ) ≤ r ≤ h2 (θ). You can traverse this region radially outwards, like the spokes of a bicycle wheel. The function r = h1 (θ) is always on the inside, and the function r = h2 (θ) is always on the outside. For this type of region D, ZZ Z β Z h2 (θ)
f (x, y) dA = f r cos(θ), r sin(θ) r drdθ D α h1 (θ) 11 BREE 319 Vector Calculus Some applications of double integrals Volume If f (x, y) ≥ 0, then the double integral of f over a region D in the xy-plane is the volume above D and below the graph of z = f (x, y). ZZ volume = f (x, y) dA D Area If f (x, y) = 1, the solid that lies above the region D in the xy-plane and below the graph of z = f (x, y) = 1 is a cylinder. Its volume is the area of its base D times its height 1. ZZ volume = 1 dA = (area of D) × 1 D Therefore, the integral of f (x, y) = 1 over the region D is the area of D. ZZ area = dA D Mass If an object has the shape of a flat plate that covers a region D in the xy-plane and has mass density (or mass per unit area) of ρ(x, y), then the double integral of ρ over D is the mass of the object. ZZ mass = ρ(x, y) dA D 12 BREE 319 Vector Calculus Green’s theorem Green’s theorem gives a relationship between the line integral of a vector field around a simple closed curve C and a double integral over the plane region D bounded by C. simple not simple simple not simple closed closed not closed not closed We use the convention that a positive orientation is counterclockwise as seen from above, so that D is always to the left as the curve given by ~r(t) is traversed from t = a to t = b. positive orientation negative orientation Green’s theorem states the following. Suppose C is a positively-oriented, piecewise-smooth, simple closed curve in the xy-plane, and D is the region bounded by C. If P and Q have continuous partial derivatives on an open region that contains D, then I ZZ   ∂Q ∂P P dx + Qdy = − dA C D ∂x ∂y 13 BREE 319 Vector Calculus Tangent planes to surfaces A parallelogram has two sides given by vectors ~a and ~b. The vector ~a × ~b is perpendicular to both ~a and ~b, so it is normal to the surface of the parallelogram. The area of the parallelogram is |~a| |~b| sin(θ) = |~a × ~b|. Consider the tangent plane to a surface z = f (x, y) at a point (xo , yo , zo ). The equation of the tangent plane is z − zo = fx (xo , yo )(x − xo ) + fy (xo , yo )(y − yo )
All points on the plane satisfy this equation.
For example, two points on the tangent plane are xo , yo , zo and xo + ∆x, yo , zo + fx (xo , yo )∆x. One vector on the tangent plane is


~a = xo + ∆x, yo , zo + fx (xo , yo )∆x − xo , yo , zo = ∆x, 0, fx (xo , yo )∆x. Similarly, another vector on the tangent plane is ~b = 0, ∆y, fy (xo , yo )∆y.
A normal vector to the tangent plane is ı̂ ̂ k̂ ~a × ~b = ∆x 0 fx (xo , yo )∆x = −fx (xo , yo )∆x∆y ı̂ − fy (xo , yo )∆x∆y ̂ + ∆x∆y k̂. 0 ∆y fy (xo , yo )∆y The area of the parallelogram on the tangent plane with sides ~a and ~b is q |~a × ~b| = 1 + fx2 (xo , yo ) + fy2 (xo , yo ) ∆x∆y A unit normal vector is ~a × ~b −fx (xo , yo )ı̂ − fy (xo , yo )̂ + k̂ n̂ = = q |~a × ~b| 1 + fx2 (xo , yo ) + fy2 (xo , yo ) 14 BREE 319 Vector Calculus Surface integrals of scalar functions A smooth surface S given by z = f (x, y) lies above or below a region D in the xy-plane. Break up D into subrectangles of dimension ∆x by ∆y. Above rectangle Rij , approximate the area of the surface ∆Sij by the area of its tangent plane ∆Tij , like flat tiles on a curved roof. q The tile ∆Tij is a parallelogram. Its area is 1 + fx2 (x∗ij , yij ∗ ) + f 2 (x∗ , y ∗ ) ∆x∆y. y ij ij The surface integral of a function g(x, y, z) over the surface S is ZZ Xm X n q g(x, y, z) dS = lim g(x∗ij , yij ∗ ∗ , zij ∗ ) + f 2 (x∗ , y ∗ )∆x∆y. ) 1 + fx2 (x∗ij , yij y ij ij S m,n→∞ i=1 j=1 This sum is a double integral, which can be evaluated as follows. ZZ ZZ q g(x, y, z) dS = g(x, y, z) 1 + fx2 (x, y) + fy2 (x, y) dA S D Some applications of surface integrals Surface area If g(x, y, z) = 1, then the surface integral of g along S is just the surface area of the surface. ZZ surface area = dS S Mass If an object has the shape of a surface S and has mass density (or mass per unit area) of ρ(x, y, z), then the surface integral of ρ over S is the mass of the object. ZZ mass = ρ(x, y, z) dS S 15 BREE 319 Vector Calculus Orientable surfaces If a surface S is defined by the equation z = f (x, y), then two unit normal vectors to the surface are fx (x, y)ı̂ + fy (x, y)̂ − k̂ n̂ = ± q 1 + fx2 (x, y) + fy2 (x, y) The sign ± determines the orientation of the surface. surface with surface with upwards orientation downwards orientation closed surface with closed surface with outwards orientation inwards orientation a Möbius strip is not orientable 16 BREE 319 Vector Calculus Surface integrals of vector fields Given a vector field ~ G(x, y, z) = P (x, y, z)ı̂ + Q(x, y, z)̂ + R(x, y, z) k̂ and an orientable surface S with a unit normal vector n̂, the flux of F~ through S is a surface integral. ZZ flux = ~ · n̂ dS G S The flux determines how much of the vector field passes through S. Some physical applications include the flux of fluids, electric flux, magnetic flux, heat flux, etc. To compute this line integral, you don’t have to compute the unit normal vector. Recall that fx (x, y)ı̂ + fy (x, y)̂ − k̂ q n̂ = ± q , dS = 1 + fx2 (x, y) + fy2 (x, y) dA. 1 + fx2 (x, y) + fy2 (x, y) Therefore, ZZ ZZ ~ · n̂ dS = ± ~ x, y, f (x, y) · fx (x, y)ı̂ + fy (x, y)̂ − k̂ dA

G G S D If the surface S is defined piecewise, then we define the flux through S as the sum of the fluxes through each piece of S. In the example on the right, the flux out of the closed surface is ZZ ZZ ZZ G · n̂ dS = G · n̂ dS + G · n̂ dS S S1 S2 where S1 is oriented downwards and S2 is oriented upwards. 17 BREE 319 Vector Calculus Stokes’ theorem Stokes’ theorem is a higher-dimensional version of Green’s theorem. Recall that Green’s theorem gives a relationship between a double integral over a region D and a line integral of a vector field around a curve C that bounds D. Similarly, Stokes’ theorem gives a relationship between a flux through a surface S and a line integral of a vector field around a curve C that bounds S. For Green’s theorem, we use the convention that a positive orientation is counterclockwise as seen from above. If you walk around C, the region D is always on your left. Similarly, for Stoke’s theorem, the orientation of S induces the positive orientation of C. If you walk around C with your head pointing in the direction of n̂, the surface is always on your left. Green’s theorem Stokes’ theorem positive orientation positive orientation Stokes’ theorem states the following. Suppose S is an oriented piecewise-smooth surface that is bounded by a simple, closed, piecewise-smooth boundary curve C with positive orientation. Suppose F~ is a vector field whose components have continuous partial derivatives on S. Then I ZZ F~ · d~r = ∇ × F~ · n̂ dS C S 18 BREE 319 Vector Calculus Triple integrals Consider the box B in 3-space:  B = [a, b] × [c, d] × [r, s] = (x, y, z) | a ≤ x ≤ b, c ≤ y ≤ d, r ≤ z ≤ s }. Analogously to a double integral over a rectangle, we can use Fubini’s theorem to evaluate a triple integral over a box. ZZZ Z bZ dZ s f (x, y, z) dV = f (x, y, z) dzdydx B a c r More generally, suppose E is a region in 3-space defined by  E = (x, y, z) | (x, y) ∈ D, u1 (x, y) ≤ z ≤ u2 (x, y) }, where D is a region in the xy-plane. Then ZZZ Z Z "Z u2 (x,y) # f (x, y, z) dV = f (x, y, z)dz dA E D u1 (x,y) In the above, if D is both of type I and of type II, then E is called a simple solid region. The combination of polar coordinates in the xy-plane, along with a z coordinate, is called the cylindrical coordinate system. 19 BREE 319 Vector Calculus Divergence theorem The divergence theorem relates the triple integral of the divergence of a vector field F~ over a solid E to the flux of F~ out of the boundary surface of E. The divergence theorem states the following. Suppose that E is a simple solid region and S is the boundary surface of E, with outwards orientation. Suppose that F~ is a vector field whose components have continuous partial derivatives on E. Then ZZ ZZZ F~ · n̂ dS = ∇ · F~ dV S E 20 BREE 319 Vector Calculus Summary Notice the similarities between all the theorems in the table below. They each relate the integral of a derivative over a region to the integral of the original function or vector field over the boundary of the region. 21