# Green's Theorem

Physics I · Multivariable Calculus · https://tryals.app/learn/physics-i/greens-theorem

**Green's Theorem** bridges a 1D boundary trip and a 2D interior sweep:

$$\oint_C (P\,dx + Q\,dy) = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)dA$$

Here, $C$ is a positively oriented, simple, closed curve bounding region $D$.

| Feature | Description |
| --- | --- |
| **$C$** | Simple, closed, counterclockwise |
| **$D$** | The region enclosed by $C$ |
| **Integrand** | $z$-component of $\nabla\times\vec{F}$ for $\vec{F} = (P, Q, 0)$ |

Think of it as 2D Stokes' theorem. It converts hard line integrals into easy double integrals, and vice versa.

## Area and Pitfalls

Use line integrals to sweep out 2D area ($A$) using these formulas:

- $A = \oint_C x\,dy$
- $A = -\oint_C y\,dx$
- $A = \frac{1}{2}\oint_C (x\,dy - y\,dx)$

**Common Pitfall:** Green's theorem strictly requires a **closed** curve, positive counterclockwise orientation, and a vector field with continuous partial derivatives throughout the **entire** enclosed region $D$.

If your field has a single interior singularity—like a vortex or point charge at the origin where the denominator hits zero—the equality breaks down completely. Always check your interior domain for holes or undefined points before applying the theorem!

## Practice questions

6 of this lesson's 14 practice questions, with answers. The full set is in the app.

### 1. Which conditions are required for Green's theorem to hold?

A. P and Q have continuous partial derivatives on D
B. F must be conservative
C. C is positively oriented (CCW)
D. C is a simple closed curve

**Answer:** A. P and Q have continuous partial derivatives on D; C. C is positively oriented (CCW); D. C is a simple closed curve

**Why:** Green's theorem needs a closed curve with smooth field; it does not require F to be conservative.

Page: https://tryals.app/practice/physics-i/greens-theorem/which-conditions-are-required-for-greens-theorem-to-hold

### 2. Order the steps to apply Green's theorem ∮_C (P dx + Q dy) = ∬_D ($\partial Q/\partial x$ − $\partial P/\partial y$) dA.

**Answer:**

1. Verify C is a simple, closed, positively oriented curve
2. Identify P and Q from the line integral
3. Compute $\partial Q/\partial x$ − $\partial P/\partial y$
4. Set up ∬_D ($\partial Q/\partial x$ − $\partial P/\partial y$) dA
5. Evaluate the double integral

**Why:** Green's theorem converts a line integral to a double integral.

Page: https://tryals.app/practice/physics-i/greens-theorem/order-the-steps-to-apply-greens-theorem-c-p-dx-q-dy-d

### 3. A surveyor drives a GPS-equipped car once around a lake and computes the lake’s **area** without ever crossing the water. Which theorem makes this possible?

A. Green's theorem: a boundary integral gives the enclosed area
B. Pythagoras’ theorem applied to GPS triangles
C. The fundamental theorem of calculus, areas need antiderivatives
D. It is impossible; the surveyor must grid the interior

**Answer:** A. Green's theorem: a boundary integral gives the enclosed area

**Why:** With the right integrand, the boundary tour computes exactly the enclosed area: $A = \frac{1}{2}\oint(x\,dy - y\,dx)$. Mechanical planimeters, instruments that measure any drawn region by tracing its edge, are Green’s theorem built in brass.

Page: https://tryals.app/practice/physics-i/greens-theorem/a-surveyor-drives-a-gps-equipped-car-once-around-a-lake-and-computes

### 4. Complete the statement of Green’s theorem.

**Answer:** Green’s theorem converts the circulation $\oint_C P\,dx + Q\,dy$ around a closed curve into the double integral over the enclosed region of **Qₓ − Pᵧ (the 2D curl)** — a local rotation density. The curve must be traversed **counterclockwise**, and the theorem breaks if the field has a **singularity** inside the region.

**Why:** Green’s theorem equates boundary circulation with the total of the 2D curl inside. The hypotheses have teeth: one singular point inside $C$ (think of a vortex at the origin) and the two sides genuinely disagree, the basis for winding numbers in complex analysis.

Page: https://tryals.app/practice/physics-i/greens-theorem/complete-the-statement-of-greens-theorem

### 5. Green's theorem cannot be applied when the curve is traversed clockwise.

**Answer:** False

**Why:** False. Reversing orientation simply flips the sign of the line integral, so a clockwise curve is handled by negating the result. The counterclockwise convention fixes which sign is which; it is a bookkeeping rule, not a restriction on when the theorem applies.

Page: https://tryals.app/practice/physics-i/greens-theorem/greens-theorem-cannot-be-applied-when-the-curve-is-traversed

### 6. Green’s theorem converts a line integral around a closed curve into a double integral over the region inside. Why is a CLOSED curve required?

A. Because closed curves are easier to parametrise
B. Because open curves are not orientable
C. Because an open curve does not enclose a region
D. Because the integrand is undefined on open curves

**Answer:** C. Because an open curve does not enclose a region

**Why:** Boundary and interior come as a pair. Stokes and the divergence theorem are the same idea in higher dimensions, and each needs a genuine boundary for the same reason.

Page: https://tryals.app/practice/physics-i/greens-theorem/greens-theorem-converts-a-line-integral-around-a-closed-curve-into-a