--- title: "State and prove Green's theorem" subject: "Mathematics III" chapter: 2 key: "semester_3/mathematics_iii:state-and-prove-greens-theorem" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/state-and-prove-greens-theorem site: BCT Engineering --- # State and prove Green's theorem Mathematics III · Chapter 2: Line, Surface and Volume Integrals - Asked in: 2081 Bhadra, 1+4 marks; 2079 Baishakh, 5 marks; 2075 Chaitra, 1+4 marks; 2074 Chaitra, 5 marks; 2074 Asoj, 5 marks; 2073 Chaitra, 5 marks; 2073 Shrawan, 5 marks; 2071 Chaitra, 5 marks; 2070 Asar, 5 marks; 2068 Chaitra, 5 marks; 2068 Baishakh, 5 marks (asked 11 times) - Video: [State and Proof Green's Theorem || Maths Analysis | Vector Analysis](https://www.youtube.com/watch?v=c3zv1Wp4rrk) — Maths Analysis - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/state-and-prove-greens-theorem - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2 ## Question State and prove Green's theorem in plane. ## Short answer Green’s theorem: for a counter-clockwise, piecewise-smooth simple closed curve C enclosing region R, ∮_C (P dx + Q dy) = ∬_R (∂Q/∂x − ∂P/∂y) dx dy, assuming continuous first partials of P and Q. ## Answer #### Statement **Green's theorem:** Let C be a piecewise-smooth, simple closed curve in the xy-plane, traversed counter-clockwise, and let R be the region it encloses. If P(x, y) and Q(x, y) have continuous first partial derivatives in R and on C, then ```text ∮_C (P dx + Q dy) = ∬_R (∂Q/∂x − ∂P/∂y) dx dy ``` It converts a line integral around the boundary into a double integral over the region. #### Proof (for a region simple in both directions) Take R so that every line parallel to an axis meets C in at most two points. ```text y | y = f₂(x) (upper curve C₂) | ______________ | / \ | ( region R ) | \______________/ | y = f₁(x) (lower curve C₁) | +---a-----------------b------> x ``` **Part 1: ∮ P dx = − ∬ (∂P/∂y) dx dy.** Describe R as a ≤ x ≤ b, f₁(x) ≤ y ≤ f₂(x). Then ```text ∬_R ∂P/∂y dx dy = ∫ₐᵇ [ P(x, y) ] from y=f₁(x) to y=f₂(x) dx = ∫ₐᵇ P(x, f₂(x)) dx − ∫ₐᵇ P(x, f₁(x)) dx ``` Now evaluate ∮ P dx along C counter-clockwise. The lower curve C₁ is traced from x = a to x = b; the upper curve C₂ is traced from x = b to x = a; any vertical parts have dx = 0 and contribute nothing: ```text ∮_C P dx = ∫ₐᵇ P(x, f₁(x)) dx + ∫_b^a P(x, f₂(x)) dx = ∫ₐᵇ P(x, f₁(x)) dx − ∫ₐᵇ P(x, f₂(x)) dx = − ∬_R ∂P/∂y dx dy ... (1) ``` **Part 2: ∮ Q dy = ∬ (∂Q/∂x) dx dy.** Describe R also as c ≤ y ≤ d, g₁(y) ≤ x ≤ g₂(y), where x = g₁(y) is the left curve and x = g₂(y) is the right curve. ```text ∬_R ∂Q/∂x dx dy = ∫_c^d [ Q(g₂(y), y) − Q(g₁(y), y) ] dy ``` Counter-clockwise, the right curve is traced upward (y from c to d) and the left curve downward; horizontal parts have dy = 0: ```text ∮_C Q dy = ∫_c^d Q(g₂(y), y) dy + ∫_d^c Q(g₁(y), y) dy = ∫_c^d Q(g₂(y), y) dy − ∫_c^d Q(g₁(y), y) dy = ∬_R ∂Q/∂x dx dy ... (2) ``` **Adding (1) and (2):** ```text ∮_C (P dx + Q dy) = ∬_R (∂Q/∂x − ∂P/∂y) dx dy ∎ ``` #### Extension A region that is not simple in this sense is cut into finitely many simple sub-regions by lines parallel to the axes. Applying the theorem to each and adding, the line integrals along the common cuts cancel (they are traversed in opposite directions), so the theorem holds for the whole region. The same argument works for a region with holes, taking the outer boundary counter-clockwise and the inner boundaries clockwise. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Green's theorem equates ∮(P dx + Q dy) with: A. ∬(∂P/∂x − ∂Q/∂y) dx dy B. ∬(∂Q/∂x − ∂P/∂y) dx dy C. ∬(∂Q/∂y − ∂P/∂x) dx dy D. ∬(P + Q) dx dy Answer: B. Line integral around boundary equals double integral of ∂Q/∂x − ∂P/∂y. (Key point: Statement) 2. The curve C must be traversed: A. Counter-clockwise B. Clockwise C. Either way D. Along the x-axis only Answer: A. Positive orientation keeps the region on the left. (Key point: Orientation) 3. In Part 1, why does ∮P dx equal −∬ ∂P/∂y dx dy? A. P is negative B. y decreases always C. Vertical parts contribute D. The upper curve is traced from b to a, reversing its sign Answer: D. Counter-clockwise, the upper curve runs from x=b to x=a. (Key point: Why the minus sign) 4. For non-simple regions the theorem extends because: A. The region is rotated B. P is set to zero C. Common cuts are traversed twice in opposite directions and cancel D. The cuts contribute double Answer: C. Interior line integrals cancel. 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