Preface
This series is aimed at providing tools for an electrical engineer to analyze data and solve problems in design. The focus is on applying calculus to equations or physical systems.
Introduction
This article will cover integrals.
There are many calculus references, the one I like to use is Calculus by Larson, Edwards and Hostetler.
This also assumes you are familiar with Python or can stumble your way through it.
Concepts
Boundary Condition: a spatial boundary constraint.
Initial Condition: a time boundary constraint.
Definition of an Integral
The integral of f on the closed interval [a,b] is given by
lim_{Δ->0} Σ^{n}_{i=1} f(c_{i}) Δx_{i} = ∫_{a}^{b} f(x)dx = F(b)-F(a).
Therefore integrability implies continuity. But continuity does not imply integrability.
Properties of Integrals
General Form | ∫ u dx |
Constant | ∫ c dx = cx + C |
Sum/Difference | ∫ [u +/- v] dx = ∫ u dx +/- ∫v dx + C |
Chain | ∫ f(u) du = F(u) + C |
General Power | ∫ u^{n} du = u^{n+1} / (n+1) + C |
By Parts | ∫(u dv) = uv - ∫(v du) + C |
Vector | For r(t) = f(t)i + g(t)j + h(t)k, then ∫ r(t) = [∫ f(t) dt] i + [∫ g(t) dt] j + [∫ h(t) dt] k |
Series | For f(x)=Σ^{∞}_{n=0} a_{n}(x-c)^{n}, ∫ f(x)dx = C + Σ^{∞}_{n=0} a_{n}[1/(n+1)](x-c)^{n+1} |
Where u and v are a functions, c and n are constants.
Integral Table
Here.
Average Value of a Function
The average value of a function is given by the following formula:
[1/(b-a)] ∫_{a}^{b} f(x)dx
Initial Conditions
An example of this is integrating the acceleration constant to determine velocity. You can substitute known initial position values to calculate the constants. Integrate again to get the position function. Then you have specific solutions for your problem.
Multiple Integration
The rules for integrating more than once are identical to a single integral, you just have to observe two things:
Surface Area: In the Plane
For a in a plane (flat surface) and bound on four sides by equations f(x), g(x), m(x) and n(x), the area is given by:
A = ∫_{f(x)}^{g(x)} ∫_{n(x)}^{m(x)} dy dx
Surface Area: Multiple Integrals
If f(x,y) and its first partial derivatives are continuous on the closed region R, then the area of the surface S over R is:
S = ∫_{S} ∫ f(x,y,z) dS = ∫_{R} ∫ f(x,y,g(x,y)) dS
where
dS = sqrt[1+ f_{x}(x,y)^{2}+f_{y}(x,y)^{2}] dA
If r(u,v) = x(u,v)i + y(u,v)j + z(u,v)k and D be a region in (u,v), then the surface area surface S:
S = ∫_{S}∫ f(x,y,z) dS = ∫_{D}∫ f( x(u,v), y(u,v), z(u,v) ) dS
where
dS = abs(r_{u} cross r_{v}) dA
where
r_{u} = dx/du (partial derivative)
r_{v} = dy/du (partial derivative)
If r(u,v) = xi + yj + f(x,y)k you will derive the same equation as above.
If r(u,v) = ui + f(u)*cos vj + f(u)*sin vk you will derive the same equation as surface of revolution.
S = ∫_{S}∫ F ⋅ N dS = F ⋅ (r_{u} cross r_{v}) dA = ∫R∫ F ⋅ [-g_{x}(x,y)i-g_{y}(x,y)j + k]dA
Surface Area: Surface of Revolution
If a smooth curve C given by y=f(x), then the area S of the surface is given by
S = 2π∫^{b}_{a}f(x) sqrt(1+(f'(x))^{2}) dx (revolved by x axis)
And for x=g(y)
S = 2π∫^{b}_{a}g(y) sqrt(1+(g'(y))^{2}) dy (revolved by y axis)
Volume: Disc Method
Find the volume of a function rotated about an axis and perpendicular to the axis of revolution.
Horizontal Axis of Revolution
V = π ∫^{b}_{a} f(x)^{2}dx
Vertical Axis of Revolution
V = π ∫^{d}_{c} f(y)^{2}dy
Volume: Shell Method
Find the volume of a function rotated about an axis and parallel to the axis of revolution.
Horizontal Axis of Revolution
V = 2π∫^{b}_{a} p(y)h(y)dy
Vertical Axis of Revolution
V = 2π∫^{d}_{c} p(x)h(x)dx
Volume: Multiple Integrals
If f is defined on a closed, bounded region R, then the double integral of f over R is:
∫_{R} ∫ f(x,y) dA, where dA=dx dy.
If f(x,y,z) is continuous over a bounded solid region Q, then the triple integral of f over Q is:
∫ ∫_{Q} ∫ f(x,y,z) dV, where dV=dx dy dz.
Length: Arc Length
Let the function given by y=f(x) be a continuous function on interval [a,b]. The arc length s of f between a and b is:
s = ∫^{d}_{c} sqrt[1+f'(x)^{2}]dx
For a vector valued function r(t)
s = ∫^{b}_{a} sqrt[x'(t)^{2}+y'(t)^{2}+z'(t)^{2}] dt = ∫^{b}_{a} abs(r'(t)) dt
For a linear (straight line through origin) this function reduces to the standard m=(y2-y1)/(x2-x1) function.
Length: Line Integral
This is for weighted functions. Evaluating the line integral for an unweighted function (f(x,y,z)=1) gives you the arc length above.
Let f be continuous in a region containing a smooth curve C. If C is given by r(t) = x(t)i + y(t)j + z(t)k then
s = ∫_{C} f(x,y,z) ds = ∫^{b}_{a} f(x(t), y(t), z(t)) ds
where ds = abs(r'(t)) dt = sqrt[x'(t)^{2}+y'(t)^{2}+z'(t)^{2}] dt
∫_{C}F ⋅ dr = ∫_{C}F ⋅ T ds = ∫^{b}_{a}F(x(t), y(t), z(t)) ⋅ r'(t) dt
∫_{C} 1 ds = ∫^{b}_{a} abs(r'(t)) dt
Special Relationships
Fundamental Theorem of Line Integrals
If F(x,y)=Mi+Nj then
∫_{C}F ⋅ dr = ∫_{C} ∇f ⋅ dr = f(x(b),y(b)) - f(x(a),y(a))
where f is a potential function of F. In other words for conservative functions the line integral is simply the difference between the potential function of the endpoints and is INDEPENDENT OF PATH.
Divergence Theorem
Let Q be a solid region bounded by a closed surface S oriented by a unit normal vector directed outward from Q. If F is a vector field whose component functions have continuous partial derivatives in Q then
∫_{S}∫ F ⋅ N dS = ∫∫_{Q}∫ div F dV
Both integrals describe total flux. The one on the left evaluates flux across the surface and the one on the right describes flux through the volume. N is the unit normal vector.
Stokes's Theorem
Let S be an oriented surface with unit normal vector N, bounded by a piecewise smooth simple closed curve C. If F is a vector field whose component functions have continuous partial derivatives on an open region containing S and C then
∫_{C}F ⋅ dr = ∫_{S}∫ (curl F) ⋅ N dS
This theorem calculates the circulation of the equation about a point.
∫_{C}F ⋅ dr f(x)dx = ∫_{C} ∇f ⋅ dr = f(x(b),y(b))-f(x(a),y(a)) where F(x,y) = ∇ f(x,y)
Relationship between derivatives and integrals
Next Up
Differential Equations
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