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Line Integrals via Green's Theorem - Calculus 3

  1. Consider the line integral ∮C(exsin⁡y−2y) dx+(excos⁡y+x2) dy\oint_C (e^x \sin y - 2y) \, dx + (e^x \cos y + x^2) \, dy where CC is the boundary of the triangular region with vertices at (0,0)(0,0), (2,0)(2,0), and (0,3)(0,3), oriented counterclockwise. Which of the following correctly represents the application of Green's Theorem to evaluate this integral?
    Answer: ∬R(excos⁡y+2x−excos⁡y+2) dA=∬R(2x+2) dA\iint_R (e^x \cos y + 2x - e^x \cos y + 2) \, dA = \iint_R (2x + 2) \, dA Green's Theorem states that ∮CP dx+Q dy=∬R(∂Q∂x−∂P∂y) dA\oint_C P \, dx + Q \, dy = \iint_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) \, dA. Here P=exsin⁡y−2yP = e^x \sin y - 2y and Q=excos⁡y+x2Q = e^x \cos y + x^2. We compute: ∂Q∂x=∂∂x(excos⁡y+x2)=excos⁡y+2x\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(e^x \cos y + x^2) = e^x \cos y + 2x and ∂P∂y=∂∂y(exsin⁡y−2y)=excos⁡y−2\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(e^x \sin y - 2y) = e^x \cos y - 2. Therefore: ∂Q∂x−∂P∂y=(excos⁡y+2x)−(excos⁡y−2)=excos⁡y+2x−excos⁡y+2=2x+2\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = (e^x \cos y + 2x) - (e^x \cos y - 2) = e^x \cos y + 2x - e^x \cos y + 2 = 2x + 2. Choice B has the wrong sign for the constant term. Choice C incorrectly adds the partial derivatives instead of subtracting. Choice D uses exsin⁡ye^x \sin y instead of excos⁡ye^x \cos y in the partial derivative calculation.
  2. Let F=(y2+sin⁡x)i+(2xy−ey)j\mathbf{F} = (y^2 + \sin x)\mathbf{i} + (2xy - e^y)\mathbf{j} and let CC be a simple closed curve oriented counterclockwise. If the area enclosed by CC is AA, what is ∮CF⋅dr\oint_C \mathbf{F} \cdot d\mathbf{r} in terms of AA?
    Answer: ∮CF⋅dr=0\oint_C \mathbf{F} \cdot d\mathbf{r} = 0 Using Green's Theorem: ∮CF⋅dr=∮CP dx+Q dy=∬R(∂Q∂x−∂P∂y)dA\oint_C \mathbf{F} \cdot d\mathbf{r} = \oint_C P \, dx + Q \, dy = \iint_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA. Here P=y2+sin⁡xP = y^2 + \sin x and Q=2xy−eyQ = 2xy - e^y. We compute: ∂Q∂x=∂∂x(2xy−ey)=2y\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(2xy - e^y) = 2y and ∂P∂y=∂∂y(y2+sin⁡x)=2y\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(y^2 + \sin x) = 2y. Therefore: ∂Q∂x−∂P∂y=2y−2y=0\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2y - 2y = 0. This means ∮CF⋅dr=∬R0 dA=0\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_R 0 \, dA = 0, regardless of the area AA. The vector field F\mathbf{F} is conservative. Choice A would result if ∂Q∂x−∂P∂y=2\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2. Choice B would result if the curl were 1. Choice D would result if the curl were -1.
  3. A parametric curve CC encloses a region RR with area 12. If Green's Theorem is applied to evaluate ∮C(ax+by) dx+(cx+dy) dy\oint_C (ax + by) \, dx + (cx + dy) \, dy, and the result is 36, what is the value of c−bc - b?
    Answer: c−b=3c - b = 3 Using Green's Theorem: ∮C(ax+by) dx+(cx+dy) dy=∬R(∂Q∂x−∂P∂y)dA\oint_C (ax + by) \, dx + (cx + dy) \, dy = \iint_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA. Here P=ax+byP = ax + by and Q=cx+dyQ = cx + dy. We compute: ∂Q∂x=∂∂x(cx+dy)=c\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(cx + dy) = c and ∂P∂y=∂∂y(ax+by)=b\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(ax + by) = b. Therefore: ∂Q∂x−∂P∂y=c−b\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = c - b. The double integral becomes: ∬R(c−b) dA=(c−b)∬RdA=(c−b)⋅Area(R)=(c−b)⋅12\iint_R (c - b) \, dA = (c - b) \iint_R dA = (c - b) \cdot \text{Area}(R) = (c - b) \cdot 12. Since we're given that this equals 36: (c−b)⋅12=36(c - b) \cdot 12 = 36, which gives us c−b=3c - b = 3. Choice B would result from (c−b)⋅12=18(c-b) \cdot 12 = 18. Choice C would result from (c−b)⋅12=72(c-b) \cdot 12 = 72. Choice D would result from (c−b)⋅12=4(c-b) \cdot 12 = 4.
  4. Consider the vector field F=(x3−3xy2)i+(y3−3x2y)j\mathbf{F} = (x^3 - 3xy^2)\mathbf{i} + (y^3 - 3x^2y)\mathbf{j}. If C1C_1 is the square with vertices at (±1,±1)(\pm 1, \pm 1) and C2C_2 is the circle x2+y2=2x^2 + y^2 = 2, both oriented counterclockwise, which statement about ∮C1F⋅dr\oint_{C_1} \mathbf{F} \cdot d\mathbf{r} and ∮C2F⋅dr\oint_{C_2} \mathbf{F} \cdot d\mathbf{r} is correct?
    Answer: Both integrals are zero because ∂Q∂x=∂P∂y\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y} everywhere in both regions Let's check if the vector field is conservative by computing the curl. We have P=x3−3xy2P = x^3 - 3xy^2 and Q=y3−3x2yQ = y^3 - 3x^2y. Computing the partial derivatives: ∂P∂y=∂∂y(x3−3xy2)=−6xy\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(x^3 - 3xy^2) = -6xy and ∂Q∂x=∂∂x(y3−3x2y)=−6xy\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(y^3 - 3x^2y) = -6xy. Since ∂Q∂x=∂P∂y=−6xy\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y} = -6xy, the curl is: ∂Q∂x−∂P∂y=−6xy−(−6xy)=0\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = -6xy - (-6xy) = 0. By Green's Theorem: ∮CF⋅dr=∬R0 dA=0\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_R 0 \, dA = 0 for any simple closed curve CC. This means both ∮C1F⋅dr=0\oint_{C_1} \mathbf{F} \cdot d\mathbf{r} = 0 and ∮C2F⋅dr=0\oint_{C_2} \mathbf{F} \cdot d\mathbf{r} = 0. Choice A is correct in conclusion but uses imprecise language ('conservative throughout the plane' vs. zero curl). Choice B is incorrect because the integrals are zero, not nonzero. Choice C misunderstands that line integrals of conservative fields are path-independent. Choice D correctly identifies the key condition ∂Q∂x=∂P∂y\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y} and the resulting zero curl.
  5. A region RR is bounded by the curves y=x3y = x^3, y=8y = 8, and x=0x = 0. Using Green's Theorem to evaluate ∮∂Rx2 dx+2xy dy\oint_{\partial R} x^2 \, dx + 2xy \, dy (where ∂R\partial R is oriented counterclockwise), the double integral that must be computed is ∬Rh(x,y) dA\iint_R h(x,y) \, dA. What is h(x,y)h(x,y), and what is the value of this double integral?
    Answer: h(x,y)=2yh(x,y) = 2y and the integral equals 7687\frac{768}{7} Using Green's Theorem with P=x2P = x^2 and Q=2xyQ = 2xy: ∂Q∂x=∂∂x(2xy)=2y\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(2xy) = 2y and ∂P∂y=∂∂y(x2)=0\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(x^2) = 0. Therefore: h(x,y)=∂Q∂x−∂P∂y=2y−0=2yh(x,y) = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2y - 0 = 2y. The region is bounded by y=x3y = x^3, y=8y = 8, and x=0x = 0, giving {(x,y):0≤x≤2,x3≤y≤8}\{(x,y) : 0 \leq x \leq 2, x^3 \leq y \leq 8\}. The integral becomes: ∬R2y dA=∫02∫x382y dy dx=∫02[y2]x38dx=∫02(64−x6) dx=[64x−x77]02=128−1287=896−1287=7687\iint_R 2y \, dA = \int_0^2 \int_{x^3}^8 2y \, dy \, dx = \int_0^2 \left[y^2\right]_{x^3}^8 dx = \int_0^2 (64 - x^6) \, dx = \left[64x - \frac{x^7}{7}\right]_0^2 = 128 - \frac{128}{7} = \frac{896 - 128}{7} = \frac{768}{7}.
  6. Consider the line integral ∮C(3x2+y3) dx+(3xy2+x3) dy\oint_C (3x^2 + y^3) \, dx + (3xy^2 + x^3) \, dy around the ellipse x24+y29=1\frac{x^2}{4} + \frac{y^2}{9} = 1 oriented counterclockwise. To apply Green's Theorem most efficiently, which approach should be used?
    Answer: Apply Green's Theorem since ∂Q∂x−∂P∂y=3y2+3x2−3y2=3x2\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 3y^2 + 3x^2 - 3y^2 = 3x^2, then integrate ∬R3x2 dA\iint_R 3x^2 \, dA over the elliptical region Let's check if the vector field is conservative by computing the partial derivatives. We have P=3x2+y3P = 3x^2 + y^3 and Q=3xy2+x3Q = 3xy^2 + x^3. Computing: ∂P∂y=∂∂y(3x2+y3)=3y2\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(3x^2 + y^3) = 3y^2 and ∂Q∂x=∂∂x(3xy2+x3)=3y2+3x2\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(3xy^2 + x^3) = 3y^2 + 3x^2. Since ∂Q∂x≠∂P∂y\frac{\partial Q}{\partial x} \neq \frac{\partial P}{\partial y}, the field is not conservative. The curl is: ∂Q∂x−∂P∂y=(3y2+3x2)−3y2=3x2\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = (3y^2 + 3x^2) - 3y^2 = 3x^2. Therefore, Green's Theorem gives ∮CF⋅dr=∬R3x2 dA\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_R 3x^2 \, dA. Choice A suggests avoiding Green's Theorem, which is inefficient. Choice C incorrectly claims the field is conservative. Choice D has the wrong expression for the curl.

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