Multivariable Calculus Quiz: Greens Theorem
6 questions · exam conditions
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Greens TheoremQuestion 1 of 6

Let F⃗(x,y)=(2xy+cos⁡(x),x2+ysin⁡(y))\vec{F}(x,y) = (2xy + \cos(x), x^2 + y\sin(y)) and let CC be any simple closed curve enclosing a region of area AA. If Green's theorem can be applied, what is ∮CF⃗⋅dr⃗\oint_C \vec{F} \cdot d\vec{r} in terms of AA?

The integral equals 2A2A regardless of the specific curve chosen
The integral equals 00 since the vector field is conservative on any simply connected domain
The integral depends on the specific curve and cannot be expressed solely in terms of AA
The integral equals AA since the curl of F⃗\vec{F} has constant magnitude 11
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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Greens Theorem

Practice Greens Theorem in Multivariable Calculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Greens Theorem, giving you a quick way to practice the rules, question types, and explanations that matter most for Multivariable Calculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Let F⃗(x,y)=(2xy+cos⁡(x),x2+ysin⁡(y))\vec{F}(x,y) = (2xy + \cos(x), x^2 + y\sin(y)) and let CC be any simple closed curve enclosing a region of area AA. If Green's theorem can be applied, what is ∮CF⃗⋅dr⃗\oint_C \vec{F} \cdot d\vec{r} in terms of AA?

  1. The integral equals 2A2A regardless of the specific curve chosen
  2. The integral equals 00 since the vector field is conservative on any simply connected domain (correct answer)
  3. The integral depends on the specific curve and cannot be expressed solely in terms of AA
  4. The integral equals AA since the curl of F⃗\vec{F} has constant magnitude 11
Explanation: For F⃗=(P,Q)=(2xy+cos⁡(x),x2+ysin⁡(y))\vec{F} = (P, Q) = (2xy + \cos(x), x^2 + y\sin(y)), we compute ∂Q∂x=∂∂x(x2+ysin⁡(y))=2x\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(x^2 + y\sin(y)) = 2x and ∂P∂y=∂∂y(2xy+cos⁡(x))=2x\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(2xy + \cos(x)) = 2x. Since ∂Q∂x−∂P∂y=2x−2x=0\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2x - 2x = 0, by Green's theorem: ∮CF⃗⋅dr⃗=∬D0 dA=0\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 00, not 22. Choice C is wrong because the integral is always 00 regardless of the curve. Choice D is wrong because the curl magnitude is 00, not 11.

Question 2

Consider the region DD bounded by y=sin⁡xy = \sin x and y=0y = 0 for 0≤x≤π0 \leq x \leq \pi. If we want to use Green's theorem in the form ∮CP dx+Q dy=∬D(∂Q∂x−∂P∂y)dA\oint_C P \, dx + Q \, dy = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA to evaluate ∬Dx dA\iint_D x \, dA as a line integral, which choice for (P,Q)(P, Q) would work?

  1. (P,Q)=(−xy,x22)(P, Q) = (-xy, \frac{x^2}{2}) since ∂Q∂x−∂P∂y=x−(−x)=2x\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = x - (-x) = 2x
  2. (P,Q)=(−xy2,x22)(P, Q) = (-\frac{xy}{2}, \frac{x^2}{2}) since ∂Q∂x−∂P∂y=x−(−x2)=3x2\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = x - (-\frac{x}{2}) = \frac{3x}{2}
  3. (P,Q)=(x2y2,0)(P, Q) = (\frac{x^2y}{2}, 0) since ∂Q∂x−∂P∂y=0−x22=−x22\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0 - \frac{x^2}{2} = -\frac{x^2}{2}
  4. (P,Q)=(0,x22)(P, Q) = (0, \frac{x^2}{2}) since ∂Q∂x−∂P∂y=x−0=x\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = x - 0 = x (correct answer)
Explanation: 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 PP and QQ. Green's theorem states that ∮CP dx+Q dy=∬D(∂Q∂x−∂P∂y)dA\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 ∬Dx dA\iint_D x \, dA, we need to find functions PP and QQ such that ∂Q∂x−∂P∂y=x\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = x. Let's check option D: (P,Q)=(0,x22)(P, Q) = (0, \frac{x^2}{2}). Here, ∂Q∂x=∂∂x(x22)=x\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}\left(\frac{x^2}{2}\right) = x and ∂P∂y=∂∂y(0)=0\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(0) = 0. Therefore, ∂Q∂x−∂P∂y=x−0=x\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = x - 0 = x, which matches our target exactly. Option A gives ∂Q∂x−∂P∂y=x−(−x)=2x\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = x - (-x) = 2x, which is twice what we need. Option B yields ∂Q∂x−∂P∂y=x−(−x2)=3x2\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = x - (-\frac{x}{2}) = \frac{3x}{2}, also incorrect. Option C produces ∂Q∂x−∂P∂y=0−x22=−x22\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.

Question 3

Let RR be a simply connected region in the plane and CC its positively oriented boundary. If F⃗=(P,Q)\vec{F} = (P, Q) is a vector field such that ∂P∂y=∂Q∂x\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} throughout RR, and ∮CF⃗⋅dr⃗=15\oint_C \vec{F} \cdot d\vec{r} = 15, what can be concluded?

  1. There is a contradiction since conservative vector fields must have zero circulation around any closed curve
  2. The calculation must be wrong since Green's theorem would give ∮CF⃗⋅dr⃗=0\oint_C \vec{F} \cdot d\vec{r} = 0
  3. This is impossible since ∂P∂y=∂Q∂x\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} implies ∂Q∂x−∂P∂y=0\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0 (correct answer)
  4. This result is consistent with Green's theorem since we need ∂Q∂x−∂P∂y=0\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0, which is satisfied
Explanation: By Green's theorem, ∮CF⃗⋅dr⃗=∮CP dx+Q dy=∬R(∂Q∂x−∂P∂y)dA\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 ∂P∂y=∂Q∂x\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}, then ∂Q∂x−∂P∂y=0\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0, which means ∮CF⃗⋅dr⃗=∬R0 dA=0\oint_C \vec{F} \cdot d\vec{r} = \iint_R 0 \, dA = 0. Therefore, it's impossible for the line integral to equal 1515 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 ∂P∂y=∂Q∂x\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} implies the curl is zero, making a nonzero line integral impossible.

Question 4

A student uses Green's theorem to evaluate ∮Cxy dx+x2y dy\oint_C xy \, dx + x^2y \, dy around a simple closed curve CC and gets ∬D(2xy−x) dA\iint_D (2xy - x) \, dA. However, they realize that CC was oriented clockwise instead of counterclockwise. To get the correct value of the original line integral, they should:

  1. Multiply their result by 22 since the curl calculation was off by a factor related to orientation
  2. Keep their result unchanged since Green's theorem is independent of orientation
  3. Recalculate using ∬D−(2xy−x) dA\iint_D -(2xy - x) \, dA since the orientation affects the curl calculation
  4. Multiply their result by −1-1 since clockwise orientation gives the negative of the counterclockwise result (correct answer)
Explanation: 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 ∮CP dx+Q dy=∬D(∂Q∂x−∂P∂y)dA\oint_C P \, dx + Q \, dy = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA when CC is traversed counterclockwise. The student correctly calculated the partial derivatives: with P=xyP = xy and Q=x2yQ = x^2y, we get ∂Q∂x−∂P∂y=2xy−x\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 −1-1 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 ∂Q∂x−∂P∂y\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 −1-1.

Question 5

Consider the vector field F⃗(x,y)=(−yx2+y2,xx2+y2)\vec{F}(x,y) = \left(\frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2}\right) defined on R2∖{(0,0)}\mathbb{R}^2 \setminus \{(0,0)\}. For which of the following curves can Green's theorem be directly applied to evaluate ∮CF⃗⋅dr⃗\oint_C \vec{F} \cdot d\vec{r}?

  1. Any simple closed curve that does not enclose the origin, since the vector field satisfies the hypotheses of Green's theorem on such domains (correct answer)
  2. Any simple closed curve, including those that enclose the origin, since the curl of F⃗\vec{F} is zero everywhere it's defined
  3. Only rectangular curves with sides parallel to the coordinate axes, since these avoid the singularity most effectively
  4. No simple closed curves, since the vector field is not defined at the origin and therefore fails the continuity requirements
Explanation: For Green's theorem to apply, we need PP and QQ to be continuously differentiable on a simply connected domain containing the curve and its interior. Here P=−yx2+y2P = \frac{-y}{x^2+y^2} and Q=xx2+y2Q = \frac{x}{x^2+y^2}. We can verify that ∂Q∂x−∂P∂y=(x2+y2)−x(2x)(x2+y2)2−(x2+y2)(−1)−(−y)(2y)(x2+y2)2=y2−x2(x2+y2)2−−x2−y2+2y2(x2+y2)2=y2−x2(x2+y2)2−y2−x2(x2+y2)2=0\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 ∮CF⃗⋅dr⃗=0\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.

Question 6

Let CC be the ellipse x29+y24=1\frac{x^2}{9} + \frac{y^2}{4} = 1 oriented counterclockwise. Using Green's theorem, ∮C(x3+2xy)dx+(y3+x2)dy\oint_C (x^3 + 2xy) dx + (y^3 + x^2) dy equals:

  1. 00 because the partial derivatives satisfy ∂Q∂x=∂P∂y\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}
  2. 6π6\pi because the area of the ellipse is 6π6\pi
  3. 00 because ∂Q∂x−∂P∂y=2x−2x=0\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2x - 2x = 0 (correct answer)
  4. −12π-12\pi because ∂Q∂x−∂P∂y=−2\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = -2 and the area is 6π6\pi
Explanation: With P=x3+2xyP = x^3 + 2xy and Q=y3+x2Q = y^3 + x^2, we compute: ∂Q∂x=∂∂x(y3+x2)=2x\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(y^3 + x^2) = 2x and ∂P∂y=∂∂y(x3+2xy)=2x\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(x^3 + 2xy) = 2x. Therefore, ∂Q∂x−∂P∂y=2x−2x=0\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2x - 2x = 0. By Green's theorem: ∮CF⃗⋅dr⃗=∬D0 dA=0\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 ∂Q∂x−∂P∂y=0\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 11. Choice D has an error in computing the partial derivatives and gets −2-2 instead of 00.