You want to compute ∮_C P dx + Q dy around a triangle C. After computing ∂Q/∂x − ∂P/∂y, you find it equals the constant 5. What does Green's theorem reduce this to?
A5 times the area of the triangle
B5 times the perimeter of the triangle
CA sum of three separate line integrals, one per side
DZero, because a closed curve has no net work
Green's theorem converts the line integral to ∬_D (∂Q/∂x − ∂P/∂y) dA = ∬_D 5 dA = 5 · area(D). When the 2D curl is constant, the double integral reduces to that constant times the area — far simpler than computing the line integral around three edges. This is exactly the 'trade' Green's theorem offers: a complicated boundary computation becomes a simple area computation.
Question 2 Multiple Choice
A vector field F has zero divergence everywhere inside a closed curve C. What does the flux form of Green's theorem tell you about the total outward flux across C?
AThe total outward flux is zero
BThe total outward flux equals the area enclosed by C
CThe flux is undefined because divergence is zero
DThe flux depends on the shape of C, not just the divergence
The flux form states ∮_C F · n ds = ∬_D div(F) dA. If div(F) = 0 everywhere in D, the right side is ∬_D 0 dA = 0, so the total outward flux is zero. Physically: if there are no sources or sinks inside D, whatever flows in must flow out. This is the 2D analogue of incompressible fluid flow.
Question 3 True / False
Green's theorem can be used to compute the area of a region by evaluating an integral along the boundary curve alone, without setting up a double integral over the interior.
TTrue
FFalse
Answer: True
This is one of Green's theorem's elegant applications. Because ∬_D 1 dA = ½ ∮_C (x dy − y dx), the area of D is computable from the boundary curve C. Other equivalent formulas are ∮_C x dy and −∮_C y dx. This converts a 2D computation into a 1D one, which is especially useful for regions bounded by parameterizable curves.
Question 4 True / False
Green's theorem primarily helps when the line integral is difficult and the double integral is easy — if the double integral is harder, you cannot apply the theorem in reverse.
TTrue
FFalse
Answer: False
Green's theorem is a two-way trade. You can go either direction: convert a difficult line integral into a double integral, or convert a difficult double integral into a line integral along the boundary. The strategic skill is recognizing which side of the trade is simpler in each problem. A double integral over a complicated region might become tractable as a line integral around a simple boundary.
Question 5 Short Answer
What is the 'trade' that Green's theorem offers, and what strategic judgment is required to apply it well?
Think about your answer, then reveal below.
Model answer: Green's theorem converts a line integral around a closed curve C into a double integral over the enclosed region D, or vice versa. The 'trade' is that whichever side is computationally simpler, you can work from the other side. The strategic judgment is: Is the 2D curl (or divergence) of the field simple over the interior? If so, convert the line integral to a double integral. Is the boundary curve simple to parameterize? If so, convert the double integral to a line integral. Neither direction is universally preferred — the choice depends on what simplifies.
This is the core skill for Green's theorem problems. Students who memorize the formula but don't internalize the 'trade' metaphor tend to apply it in only one direction or fail to recognize when it's the right tool. The theorem is at its most powerful when one side of the equation is obviously simpler — for example, when the 2D curl is a constant, reducing the double integral to a simple area computation.