Green's Theorem - Multivariable Calculus

1. Let $R$ be a simply connected region in the plane and $C$ its positively oriented boundary. If $\vec{F} = (P, Q)$ is a vector field such that $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$ throughout $R$, and $\oint_C \vec{F} \cdot d\vec{r} = 15$, what can be concluded?
   Answer: This is impossible since $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$ implies $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0$ By Green's theorem, $\oint_C \vec{F} \cdot d\vec{r} = \oint_C P \, dx + Q \, dy = \iint_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA$. If $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$, then $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0$, which means $\oint_C \vec{F} \cdot d\vec{r} = \iint_R 0 \, dA = 0$. Therefore, it's impossible for the line integral to equal $15$ if the given condition holds. This means there's a fundamental contradiction in the problem statement. Choice A is essentially correct but uses less precise language. Choice B suggests a calculation error rather than recognizing the theoretical impossibility. Choice D incorrectly suggests the result is consistent when it clearly contradicts Green's theorem. Choice C correctly identifies that the given condition $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$ implies the curl is zero, making a nonzero line integral impossible.
2. Consider the vector field $\vec{F}(x,y) = \left(\frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2}\right)$ defined on $\mathbb{R}^2 \setminus \{(0,0)\}$. For which of the following curves can Green's theorem be directly applied to evaluate $\oint_C \vec{F} \cdot d\vec{r}$?
   Answer: Any simple closed curve that does not enclose the origin, since the vector field satisfies the hypotheses of Green's theorem on such domains For Green's theorem to apply, we need $P$ and $Q$ to be continuously differentiable on a simply connected domain containing the curve and its interior. Here $P = \frac{-y}{x^2+y^2}$ and $Q = \frac{x}{x^2+y^2}$. We can verify that $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = \frac{(x^2+y^2) - x(2x)}{(x^2+y^2)^2} - \frac{(x^2+y^2)(-1) - (-y)(2y)}{(x^2+y^2)^2} = \frac{y^2-x^2}{(x^2+y^2)^2} - \frac{-x^2-y^2+2y^2}{(x^2+y^2)^2} = \frac{y^2-x^2}{(x^2+y^2)^2} - \frac{y^2-x^2}{(x^2+y^2)^2} = 0$. So the curl is zero wherever the vector field is defined. If a simple closed curve does not enclose the origin, then the vector field is continuously differentiable throughout the region enclosed by the curve, and Green's theorem applies, giving $\oint_C \vec{F} \cdot d\vec{r} = 0$. However, if the curve encloses the origin, the domain is not simply connected (it has a 'hole' at the origin), so Green's theorem cannot be directly applied. Choice B is wrong because Green's theorem requires the domain to be simply connected. Choice C is wrong because the shape of the curve doesn't matter, only whether it encloses the origin. Choice D is wrong because Green's theorem can be applied when the curve doesn't enclose the origin.
3. A student uses Green's theorem to evaluate $\oint_C xy \, dx + x^2y \, dy$ around a simple closed curve $C$ and gets $\iint_D (2xy - x) \, dA$. However, they realize that $C$ was oriented clockwise instead of counterclockwise. To get the correct value of the original line integral, they should:
   Answer: Multiply their result by $-1$ since clockwise orientation gives the negative of the counterclockwise result When you encounter Green's theorem problems, always pay careful attention to the orientation of your curve, as this directly affects the sign of your result. Green's theorem states that $\oint_C P \, dx + Q \, dy = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA$ when $C$ is traversed counterclockwise. The student correctly calculated the partial derivatives: with $P = xy$ and $Q = x^2y$, we get $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2xy - x$. However, Green's theorem assumes counterclockwise orientation. When a curve is oriented clockwise instead of counterclockwise, the line integral equals the negative of what Green's theorem gives you. This happens because clockwise traversal effectively reverses the direction of integration. Therefore, the student should multiply their double integral result by $-1$ to get the correct value of the original clockwise line integral. Answer choice A is incorrect because the factor of 2 has no basis in Green's theorem orientation rules. Choice B is wrong because Green's theorem is definitely dependent on orientation—the theorem specifically requires counterclockwise orientation. Choice C incorrectly suggests recalculating the curl itself with a negative sign, but the curl calculation $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$ doesn't change based on orientation; only the final result's sign changes. Remember: Green's theorem gives positive results for counterclockwise orientation. For clockwise curves, multiply your Green's theorem result by $-1$.
4. Let $C$ be the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ oriented counterclockwise. Using Green's theorem, $\oint_C (x^3 + 2xy) dx + (y^3 + x^2) dy$ equals:
   Answer: $0$ because $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2x - 2x = 0$ With $P = x^3 + 2xy$ and $Q = y^3 + x^2$, we compute: $\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(y^3 + x^2) = 2x$ and $\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(x^3 + 2xy) = 2x$. Therefore, $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2x - 2x = 0$. By Green's theorem: $\oint_C \vec{F} \cdot d\vec{r} = \iint_D 0 \, dA = 0$. Choice A states the same conclusion but uses incorrect notation (it should be $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0$, not equality of the individual partial derivatives). Choice B incorrectly assumes the integrand is $1$. Choice D has an error in computing the partial derivatives and gets $-2$ instead of $0$.
5. Let $\vec{F}(x,y) = (2xy + \cos(x), x^2 + y\sin(y))$ and let $C$ be any simple closed curve enclosing a region of area $A$. If Green's theorem can be applied, what is $\oint_C \vec{F} \cdot d\vec{r}$ in terms of $A$?
   Answer: The integral equals $0$ since the vector field is conservative on any simply connected domain For $\vec{F} = (P, Q) = (2xy + \cos(x), x^2 + y\sin(y))$, we compute $\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(x^2 + y\sin(y)) = 2x$ and $\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(2xy + \cos(x)) = 2x$. Since $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2x - 2x = 0$, by Green's theorem: $\oint_C \vec{F} \cdot d\vec{r} = \iint_D 0 \, dA = 0$. This means the vector field is conservative, and the line integral is zero for any closed curve. Choice A is wrong because the curl is $0$, not $2$. Choice C is wrong because the integral is always $0$ regardless of the curve. Choice D is wrong because the curl magnitude is $0$, not $1$.
6. Consider the region $D$ bounded by $y = \sin x$ and $y = 0$ for $0 \leq x \leq \pi$. If we want to use Green's theorem in the form $\oint_C P \, dx + Q \, dy = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA$ to evaluate $\iint_D x \, dA$ as a line integral, which choice for $(P, Q)$ would work?
   Answer: $(P, Q) = (0, \frac{x^2}{2})$ since $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = x - 0 = x$ When you encounter a problem asking you to use Green's theorem to convert a double integral to a line integral, you need to work backwards from the desired integrand to find appropriate functions $P$ and $Q$. Green's theorem states that $\oint_C P \, dx + Q \, dy = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA$. Since we want to evaluate $\iint_D x \, dA$, we need to find functions $P$ and $Q$ such that $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = x$. Let's check option D: $(P, Q) = (0, \frac{x^2}{2})$. Here, $\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}\left(\frac{x^2}{2}\right) = x$ and $\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(0) = 0$. Therefore, $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = x - 0 = x$, which matches our target exactly. Option A gives $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = x - (-x) = 2x$, which is twice what we need. Option B yields $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = x - (-\frac{x}{2}) = \frac{3x}{2}$, also incorrect. Option C produces $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0 - \frac{x^2}{2} = -\frac{x^2}{2}$, which has the wrong variable and sign. Study tip: When using Green's theorem to convert double integrals to line integrals, always verify that your partial derivatives combine to give exactly the integrand you want—no extra coefficients or wrong variables allowed.

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