Green's Theorem
TLDRGreen's Theorem relates a line integral over a closed curve to a double integral over the plane region it encloses. It equates the line integral of a vector field along a closed, counterclockwise curve C to the double integral over the enclosed domain D of the partial derivatives of the vector field. Green's Theorem makes line integrals easier to evaluate. The direction of C is important - clockwise curves require a sign change. Setting up the bounds of integration properly allows Green's Theorem to be applied to any shape domain D. Overall, Green's Theorem provides a simpler way to evaluate line integrals over closed curves.
Takeaways
- ๐ Green's Theorem relates line integrals over closed curves to double integrals over the enclosed region
- ๐ It applies to counterclockwise oriented closed curves; reverse sign for clockwise curves
- ๐๐ป The line integral equals the double integral of โQ/โx - โP/โy over the enclosed region D
- ๐กQ and P come from the vector field F=Pรฎ+Qฤต being integrated over
- ๐ To use Green's Theorem, express the double integral bounds correctly for the enclosed region D
- ๐งฎ It provides a simpler way to evaluate line integrals, converting them to double integrals
- ๐ Use it when the boundary curve is difficult to integrate directly
- โ ๏ธ Ensure curve orientation and integral signs are handled correctly
- ๐ค Apply similar logic as setting up double integrals to bound the region D
- ๐ง Conceptually, it links line integrals over boundaries to double integrals over interiors
Q & A
What is Green's Theorem used for?
-Green's Theorem allows us to calculate line integrals over closed curves by converting them into double integrals over the enclosed area. This is often simpler.
What is the form of Green's Theorem?
-The line integral of P dx + Q dy around a closed curve C is equal to the double integral over the enclosed area D of (โQ/โx - โP/โy) dA.
How does curve direction affect Green's Theorem?
-Green's Theorem applies directly to counterclockwise oriented closed curves. For clockwise oriented curves, the theorem still applies but with a negative sign on the double integral.
What were the key steps in applying Green's Theorem to the example with the square curve?
-First the double integral was set up over the square domain from 0 to 1 in x and y. Then the partial derivatives โQ/โx and โP/โy were computed and plugged into the integrand. Finally, the double integral was evaluated over the bounds.
Why was the double integral equal to 1/2 in the square curve example?
-When the partial derivatives were computed, the integrand simplified to just x. Integrating this first from 0 to 1 in x gave x^2/2 evaluated from 0 to 1, or 1/2. The y integral then had no y dependence, so integrating the constant 1/2 in y gave another 1/2.
How do you set up the bounds for an irregular domain?
-The same ideas from double integrals apply. You must consider the shape carefully and determine the appropriate order and bounds for x and y to cover the full enclosed area.
What were the bounds for the triangle domain example?
-Since y went from 0 to x, y was integrated first from 0 to x. Then x went from 0 to 1, so the bounds were โซ01 โซ0x (โQ/โx - โP/โy) dy dx.
When would you not use Green's Theorem?
-Green's Theorem only applies to closed curves bounding some area. For open curves, you would have to evaluate the line integral directly.
What fields does Green's Theorem apply to?
-Green's Theorem is valid for any continuously differentiable vector field. The components P and Q can be any well-behaved functions.
How is Green's Theorem applied in practice?
-It is used extensively in physics and engineering for calculating properties like work and fluid flow around boundaries. Any closed loop process can take advantage of Green's Theorem.
Outlines
๐ Understanding Green's Theorem. animosity to what's thither
Paragraph 1 introduces Green's Theorem, which relates a line integral around a closed curve C to a double integral over the enclosed domain D. It states that the line integral of P dx + Q dy around C equals the double integral of โQ/โx - โP/โy over D. Important points are: applies to closed curves, clockwise curves need negative sign, line integrals harder than double integrals so this is useful. An example is shown with a square from (0,0) to (1,0) to (1,1) to (0,1) and vector field F. The double integral ends up being simple compared to direct line integral.
๐ Setting up bounds of integration for Green's Theorem. off the beaten track
Paragraph 2 continues with Green's Theorem. While simple shapes are easy, real boundaries require thought to set up bounds of integration properly. An example curve with 3 line segments forming a triangle is shown. To integrate y first, y runs from 0 to x, and x from 0 to 1. This completes the basics and it's just a matter of setting up the bounds correctly from here.
Mindmap
Keywords
๐กline integral
๐กclosed curve
๐กGreen's Theorem
๐กorientation
๐กbounds of integration
๐กvector field
๐กpartial derivatives
๐กsimplicity
๐กarea
๐กwork
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Transcripts
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