The actual final tutorial will be on line integrals.
Part 16. Line Integrals
With line integrals, x and y are coordinates on a curve C. First, we must parametrize the curve, i.e [latex]x=g(t)[/latex] and [latex]y=h(t)[/latex]. The parametrized curve will be written as a vector function [latex]r(t)=g(t)i + h(t)j[/latex]. It is also assumed to be smooth. When a curve is smooth, [latex]r(t)[/latex] is continuous and never 0 for any value of t.
When you have a function f(x, y) along a curve C, the line integral is
[latex]\int_C f(x, y) \ ds = \oint f(x, y) \ ds[/latex]
The ds means that the integral is with respect to the arc length, not x or y.
In parametric form,
[latex]ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}dt[/latex]
Rewriting the equation in parametric form gives us:
[latex]\oint f(x, y) \ ds = \int_a^b f(g(t), h(t)) \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}dt[/latex]
Because
[latex]\sqrt{\left(\frac{dx}{dt}\right) ^2 + \left(\frac{dy}{dt}\right) ^2}=\left|\left|\frac{dr}{dt}\right|\right|[/latex]
we obtain the equation for the line integral:
[latex]\oint f(x, y) \ ds = \int_a^b f(g(t), h(t))\left|\left|\frac{dr}{dt}\right|\right|dt[/latex]
For example, we will work out the line integral from 2 to 6 of f(x, y) = x2 + y2, where x = cos(t) and y = sin(t). For this integral,
[latex]\left|\left|\frac{dr}{dt}\right|\right|=\sqrt{1-2sin^2(t)}[/latex]
and
[latex]f(g(t), h(t))=sin^2(t)+cos^2(t)=1[/latex]
The integral now becomes
[latex]\int_2^6 \sqrt{1-sin^2(t)} \ dt[/latex]
The antiderivative of this function is
[latex]\frac{tan(t)}{\sqrt{tan^2(t)+1}}[/latex] Evaluating the antiderivative gives us
[latex]\int_2^6 \sqrt{1-sin^2(t)} \ dt=0.63[/latex]
We can also find line integrals of piecewise smooth curves. A piecewise smooth curve is a smooth curve made up of parts.
The line integral for a piecewise smooth curve is:
[latex]\oint f(x, y) \ ds = \sum \int_C f(x, y) \left|\left|\frac{dr}{dt}\right|\right| dt[/latex]
i.e it is the sum of it's parts.
Part 17. Line Integrals with Respect to x and y
Line integrals can also be taken with respect to x or y.
Because
[latex]dx=\frac{dx}{dt} dt[/latex]
and
[latex]dy=\frac{dy}{dt}dt[/latex]
We can definite the line integral with respect to x or y as
[latex]\oint f(x, y) dx = \int_a^b f(x(t), y(t))\frac{dx][dt}dt[/latex]
and
[latex]\oint f(x, y) dy=\int_c^d f(x(t), y(t)) \frac{dy}{dt}dt[/latex]
For example, we will work out the integral from 6 to 9 of f(x, y) = x^2 + y^2 where x=sin(t) and y=cos(t).
f(x, y) becomes 1, leaving us with
[latex]\int_6^9 cos(t) dt[/latex]
The antiderivative of the function is -sin(t).
Evaluating this gives us
[latex]\int_6^9 cos(t) dt=-0.69[/latex]
That concludes the tutorial on integral calculus.
What will happen to these tutorials?