# 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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