State and prove Green's Theorem.
Understand the Problem
The question asks for the statement and proof of Green's Theorem. Green's Theorem relates a line integral around a simple closed curve C to a double integral over the plane region D bounded by C. It's a fundamental theorem in vector calculus.
Answer
Green's Theorem: Let $C$ be a positively oriented, piecewise-smooth, simple closed curve in the plane, and let $D$ be the region bounded by $C$. If $P$ and $Q$ have continuous partial derivatives on an open region that contains $D$, then $$ \oint_C P \, dx + Q \, dy = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA $$
Answer for screen readers
Green's Theorem:
Let $C$ be a positively oriented, piecewise-smooth, simple closed curve in the plane, and let $D$ be the region bounded by $C$. If $P$ and $Q$ have continuous partial derivatives on an open region that contains $D$, then $$ \oint_C P , dx + Q , dy = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) , dA $$
Proof:
For a region $D$ that is both type I and type II, we have $$ \iint_D -\frac{\partial P}{\partial y} , dA = \oint_C P , dx $$ and $$ \iint_D \frac{\partial Q}{\partial x} , dA = \oint_C Q , dy $$ Adding these two equations, we get $$ \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) , dA = \oint_C P , dx + Q , dy $$
Steps to Solve
- State Green's Theorem
Green's Theorem relates a line integral around a simple closed curve $C$ to a double integral over the region $D$ bounded by $C$. Let $C$ be a positively oriented, piecewise-smooth, simple closed curve in the plane, and let $D$ be the region bounded by $C$. If $P$ and $Q$ have continuous partial derivatives on an open region that contains $D$, then $$ \oint_C P , dx + Q , dy = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) , dA $$
- Outline the Proof
We will prove Green's Theorem for a special region $D$ that is both type I and type II. This means that $D$ can be described both as $ D = {(x, y) \mid a \le x \le b, g_1(x) \le y \le g_2(x) } $ and as $ D = {(x, y) \mid c \le y \le d, h_1(y) \le x \le h_2(y) } $.
- Prove $\iint_D -\frac{\partial P}{\partial y} dA = \oint_C P , dx$
From the description of $D$ as a type I region, we have $$ \iint_D -\frac{\partial P}{\partial y} , dA = \int_a^b \int_{g_1(x)}^{g_2(x)} -\frac{\partial P}{\partial y} , dy , dx $$ $$ = \int_a^b \left[ -P(x, y) \right]_{y=g_1(x)}^{y=g_2(x)} , dx = \int_a^b \left( -P(x, g_2(x)) + P(x, g_1(x)) \right) , dx $$ $$ = \int_a^b P(x, g_1(x)) , dx - \int_a^b P(x, g_2(x)) , dx $$
Now, we parameterize the curve $C$. Let $C_1$ be the part of $C$ along $y=g_1(x)$, oriented from $x=a$ to $x=b$, and let $C_2$ be the part of $C$ along $y=g_2(x)$, oriented from $x=b$ to $x=a$. Then $$ \oint_C P , dx = \int_{C_1} P , dx + \int_{C_2} P , dx = \int_a^b P(x, g_1(x)) , dx + \int_b^a P(x, g_2(x)) , dx $$ $$ = \int_a^b P(x, g_1(x)) , dx - \int_a^b P(x, g_2(x)) , dx $$ Therefore, $\iint_D -\frac{\partial P}{\partial y} , dA = \oint_C P , dx$.
- Prove $\iint_D \frac{\partial Q}{\partial x} dA = \oint_C Q , dy$
From the description of $D$ as a type II region, we have $$ \iint_D \frac{\partial Q}{\partial x} , dA = \int_c^d \int_{h_1(y)}^{h_2(y)} \frac{\partial Q}{\partial x} , dx , dy $$ $$ = \int_c^d \left[ Q(x, y) \right]_{x=h_1(y)}^{x=h_2(y)} , dy = \int_c^d \left( Q(h_2(y), y) - Q(h_1(y), y) \right) , dy $$
Let $C_3$ be the part of $C$ along $x=h_2(y)$, oriented from $y=c$ to $y=d$, and let $C_4$ be the part of $C$ along $x=h_1(y)$, oriented from $y=d$ to $y=c$. Then $$ \oint_C Q , dy = \int_{C_3} Q , dy + \int_{C_4} Q , dy = \int_c^d Q(h_2(y), y) , dy + \int_d^c Q(h_1(y), y) , dy $$ $$ = \int_c^d Q(h_2(y), y) , dy - \int_c^d Q(h_1(y), y) , dy $$ Therefore, $\iint_D \frac{\partial Q}{\partial x} , dA = \oint_C Q , dy$.
- Combine the Results
Adding the two results, we have $$ \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) , dA = \iint_D \frac{\partial Q}{\partial x} , dA - \iint_D \frac{\partial P}{\partial y} , dA = \oint_C Q , dy + \oint_C P , dx = \oint_C P , dx + Q , dy $$
This completes the proof for the special region $D$.
Green's Theorem:
Let $C$ be a positively oriented, piecewise-smooth, simple closed curve in the plane, and let $D$ be the region bounded by $C$. If $P$ and $Q$ have continuous partial derivatives on an open region that contains $D$, then $$ \oint_C P , dx + Q , dy = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) , dA $$
Proof:
For a region $D$ that is both type I and type II, we have $$ \iint_D -\frac{\partial P}{\partial y} , dA = \oint_C P , dx $$ and $$ \iint_D \frac{\partial Q}{\partial x} , dA = \oint_C Q , dy $$ Adding these two equations, we get $$ \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) , dA = \oint_C P , dx + Q , dy $$
More Information
Green's Theorem is a special case of the more general Stokes' Theorem. It simplifies calculations of line integrals and double integrals in certain cases. It is crucial in physics, engineering, and other fields.
Tips
A common mistake is forgetting the positive orientation of the curve $C$. Green's Theorem requires the curve to be traversed counterclockwise. Also, the partial derivatives of $P$ and $Q$ must be continuous on an open region containing $D$. The proof given is only valid for a special type of region.
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