Final Flashcards
Set up for line integrals with multiple parts (Without Green’s theorem)
Paramaterize the curves using
(1-t)⟨x0,y0⟩ + t⟨x1,y1⟩
same for x1,y1 -> x2,y2
same for x2,y2 -> x0,y0
this will give you r(t)
∫f(r(t))|r’(t)|dt is the integral you use
The bounds for all the integrals now become t=0 to t=1
Then the whole integral for all the line segments, using the parameterization for each
Then you add them together
∫∮⟨⟩√∇·
yeah
If F(x, y) is a conservative vector field then ∮F · dr = 0 for
any simple closed curve C.
True
If F is a vector field defined on R3 whose component functions have continuous second-order partial derivatives, then div(curl(F))= 0.
True
If F is a vector field defined on R3 whose component functions have continuous partial derivatives and curl(F) =/= 0, then F is a conservative vector field.
False
(u(t) x v(t))’ = u’(t) x v’(t)?
True
Surface integral where you are given x, y, and z bounds and a function that can be solved into z = f(x,y)
(∫∫f(x,y,z)dS)
∫∫f(x,y,z)dS = ∫∫f(x,y,f(x,y))√[(dz/dx)² + (dz/dy)² + 1]dA
Line integrals where C is given to you in ∫c f(x,y,z)dr
Parameterize the CURVE such that r(t) = ⟨x(t),y(t),z(t)⟩ 0<t<1 follows the curve. Multiply f(r(t)) that with |r’(t)| and integrate from t = 0 to t = 1
Stoke’s theorem
∫F·dr = ?
∫F·dr = ∬curl(F)·n ds
thus
∬curl(F)·(rx x ry) ds
Distance between two parallel planes
Find an x,y,z that satisfies the first equation
then find the coefficients of the other plane (a,b,c, and d) if you solve it for zero
use |ax + by + cz + d| / √[a² + b² + c²]
Fundamental theorem of line integrals?
What are the conditions?
∫∇f·dr = f(b) - f(a)
dP/dy = dQ/dx
Green’s theorem?
What are the conditions?
∫F·dr or ∮F·nds or ∮Pdx + Qdy =
∫∫(dQ/dx - dP/dy)dA
Simple closed curve (going counter-clockwise)
Reverse Green’s theorem?
What are the substitutions?
P(x,y) = 1, -y, -1/2 y
Q(x,y) = 0, x, 1/2 x
∫∫1dA = 1/2∮xdy + ydx
Surface area of a graph of a function?
(of the form x=x, y=y, z=f(x,y))
∫∫√[1 + (dz/dx)^2 + (dz/dy)^2]dA
Surface integrals over parametric equations?
∫∫f(x,y,z)dS = ∫∫f(r(u,v))|ru x rv|dA