Green’s Theorem

Example: Use Green's Theorem to Evaluate I = ∫ y² dx + xy dy

C

around the closed curve, C, bounding the region, R, where R is the ellipse

defined by (x/3)² + (y/2)² = 1 .

Strategy: Apply the standard form of Green’s Theorem to evaluate the line integral

I = ∫ P dx + Q dy = ∫ ∫ [∂Q/∂x - ∂P/∂y] dA

C R

Here P(x,y) = y² , Q(x,y) = xy and the closed curve, C, is the

Now ∂Q/∂x = y and ∂P/∂y = 2y so ∂Q/∂x - ∂P/∂y = - y

y = 2 x = 3√[1-(y/2)²]

Thus the area integral becomes ∫ ∫ -y dA = -2 ∫ ∫ y dx dy

R y = -2 x = 0

y = 2 x = 3√[1-(y/2)²] y = 2

I = -2 ∫ y x| dy = - 6 ∫ y √[1-(y/2)²] dy

y = -2 x = 0 y = -2

let u = 1- (y/2)² , du = - (1/2) y dy, y dy = - 2 du

2

I = 12 ∫ √u du = 8 [1-(y/2)² ]^3/2 | = 0 (result)

-2

Note this result makes sense since ∫ ∫ -y dA should equal zero when encompassing

R

the region, R, by going completing around the ellipse (the areas on both sides of

the y-axis cancel out).

Example: Evaluate the line integral: ∫ (7y – e^sinx)dx + (15x – sin(y³ + 8y) dy

C

where C is a circle of radius 3 centered at (5, -7). Note that this is a very difficult

integral to evaluate. So try simplifying the calculation using the RHS of Green’s Theorem.

∂Q/∂x = 15, ∂P/∂y = 7 ; ∫ ∫[∂Q/∂x - ∂P/∂y] dA = ∫ ∫ 8 dA = 8 (π 3² ) = 72π (result)

R R

Note the simplification that results in using Green’s Theorem.

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