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How to write line integrals in parametric form?
We write the line segment as a vector function: r = ⟨0, 6, − 1⟩ + t⟨4, − 5, 6⟩, 0 ≤ t ≤ 1, or in parametric form x = 4t, y = 6 − 5t, z = − 1 + 6t. Then ∫Cx2zds = ∫1 0(4t)2( − 1 + 6t)√16 + 25 + 36dt = 16√77∫1 0 − t2 + 6t3dt = 56 3 √77. ◻
How is the height of a line integral determined?
Using each line segment as the base of a rectangle, we choose the height to be the height of the surface f above the line segment. If we add up the areas of these rectangles, we get an approximation to the desired area, and in the limit this sum turns into an integral.
How to approximate the area under a curve?
Figure 16.2.1. Approximating the area under a curve. As usual, we start by thinking about how to approximate the area.
How is work done in the vector setting?
If an object moves subject to this (constant) force, in the direction of v, over a distance equal to the length of v, the work done is F ⋅ v | v | | v | = F ⋅ v. Thus, work in the vector setting is still “force times distance”, except that “times” means “dot product”.
What do you need to know about line integrals?
For a scalar line integral, we let C be a smooth curve in a plane or in space and let be a function with a domain that includes C. We chop the curve into small pieces.
How is the parameterization of a line integral changed?
Here is the parameterization of the curve. for 0 ≤ t ≤ 1 0 ≤ t ≤ 1. Remember that we are switching the direction of the curve and this will also change the parameterization so we can make sure that we start/end at the proper point.
How are line integrals evaluated over smooth curves?
Evaluation of line integrals over piecewise smooth curves is a relatively simple thing to do. All we do is evaluate the line integral over each of the pieces and then add them up. The line integral for some function over the above piecewise curve would be,
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