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Fundamental Theorem for Line Integrals - Calculus 3

  1. Consider the vector field F(x,y)=(2xy+3y2)i+(x2+6xy−1)j\mathbf{F}(x,y) = (2xy + 3y^2)\mathbf{i} + (x^2 + 6xy - 1)\mathbf{j}. If CC is any smooth curve from point (1,0)(1,0) to point (2,3)(2,3), what property must be verified before applying the Fundamental Theorem for Line Integrals, and what is the value of ∫CF⋅dr\int_C \mathbf{F} \cdot d\mathbf{r}?
    Answer: Verify that F\mathbf{F} is conservative by checking ∂P∂y=∂Q∂x\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}; the integral equals f(2,3)−f(1,0)=21f(2,3) - f(1,0) = 21 where f(x,y)=x2y+3xy2−yf(x,y) = x^2y + 3xy^2 - y To apply the Fundamental Theorem for Line Integrals, we must verify that F is conservative by checking if ∂P/∂y = ∂Q/∂x. Here P = 2xy + 3y² and Q = x² + 6xy - 1, so ∂P/∂y = 2x + 6y and ∂Q/∂x = 2x + 6y, which are equal. To find the potential function f, we integrate: ∂f/∂x = 2xy + 3y² gives f = x²y + 3xy² + g(y), and ∂f/∂y = x² + 6xy + g'(y) = x² + 6xy - 1, so g'(y) = -1 and g(y) = -y. Thus f(x,y) = x²y + 3xy² - y. The integral equals f(2,3) - f(1,0) = (12 + 54 - 3) - (0) = 21. Choice B incorrectly requires a closed curve. Choice C omits the constant term from integration. Choice D incorrectly adds an arbitrary constant and miscalculates.
  2. A vector field G(x,y)=P(x,y)i+Q(x,y)j\mathbf{G}(x,y) = P(x,y)\mathbf{i} + Q(x,y)\mathbf{j} satisfies ∂P∂y=∂Q∂x\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} everywhere in a region RR. However, when computing ∮CG⋅dr\oint_C \mathbf{G} \cdot d\mathbf{r} around a closed curve CC entirely within RR, a student gets a nonzero result. What is the most likely explanation for this apparent contradiction?
    Answer: The region RR is not simply connected, so despite satisfying the curl condition, G\mathbf{G} may not be conservative due to topological obstructions like holes or singularities The condition ∂P/∂y = ∂Q/∂x (zero curl) is necessary for a vector field to be conservative, but it is sufficient only when the domain is simply connected (no holes). If the region R is not simply connected (contains holes, punctures, or has non-trivial topology), then a vector field can satisfy the curl condition locally but still not be conservative globally. Classic examples include fields like (-y/(x²+y²), x/(x²+y²)) around the origin, which has zero curl everywhere it's defined but nonzero circulation around curves enclosing the origin. Choice A incorrectly assumes the field must be conservative. Choice C incorrectly states the theorem's scope. Choice D is incorrect because the curl condition is indeed sufficient for simply connected domains.
  3. A physics student evaluates ∫CF⋅dr\int_C \mathbf{F} \cdot d\mathbf{r} where F(x,y)=(2x−3y)i+(−3x+4y)j\mathbf{F}(x,y) = (2x - 3y)\mathbf{i} + (-3x + 4y)\mathbf{j} along two different paths from (0,0)(0,0) to (1,1)(1,1): Path 1 is y=xy = x, and Path 2 is y=x3y = x^3. The student gets different values and concludes the field is non-conservative. What error did the student make?
    Answer: The computational error occurred in parameterization; when calculated correctly, both paths yield the same result 00 because ∂P∂y=∂Q∂x=−3\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} = -3, confirming conservativeness To determine if F is conservative, we check if ∂P/∂y = ∂Q/∂x. Here P = 2x - 3y and Q = -3x + 4y, so ∂P/∂y = -3 and ∂Q/∂x = -3. Since these are equal, the field is conservative and the line integral should be path-independent. The student must have made a computational error. For a conservative field, we can find the potential function: ∂f/∂x = 2x - 3y gives f = x² - 3xy + g(y), and ∂f/∂y = -3x + g'(y) = -3x + 4y, so g'(y) = 4y and g(y) = 2y². Thus f(x,y) = x² - 3xy + 2y². The integral equals f(1,1) - f(0,0) = (1 - 3 + 2) - 0 = 0. Both paths should give the same result when calculated correctly. Choice B incorrectly states ∂Q/∂x = 4. Choice C incorrectly limits the theorem's scope. Choice D mentions the wrong final value.
  4. Given that F(x,y)=∇f(x,y)\mathbf{F}(x,y) = \nabla f(x,y) where f(x,y)=x3+xy2−2y3f(x,y) = x^3 + xy^2 - 2y^3, consider three different curves from (1,1)(1,1) to (2,0)(2,0): C1C_1 is the straight line, C2C_2 is the parabola y=3−2x+x2−x3y = 3 - 2x + x^2 - x^3, and C3C_3 consists of two line segments via (1,0)(1,0). What can be concluded about ∫C1F⋅dr\int_{C_1} \mathbf{F} \cdot d\mathbf{r}, ∫C2F⋅dr\int_{C_2} \mathbf{F} \cdot d\mathbf{r}, and ∫C3F⋅dr\int_{C_3} \mathbf{F} \cdot d\mathbf{r}?
    Answer: All three integrals equal f(2,0)−f(1,1)=8−(−1)=9f(2,0) - f(1,1) = 8 - (-1) = 9 because path independence holds when the vector field is the gradient of a scalar function Since F = ∇f where f(x,y) = x³ + xy² - 2y³, the vector field F is conservative by definition. The Fundamental Theorem for Line Integrals states that for a conservative vector field, ∫_C F·dr = f(endpoint) - f(startpoint), regardless of the specific path C. We calculate: f(2,0) = 2³ + 2(0)² - 2(0)³ = 8, and f(1,1) = 1³ + 1(1)² - 2(1)³ = 1 + 1 - 2 = -1. Therefore, all three integrals equal f(2,0) - f(1,1) = 8 - (-1) = 9. Choice A incorrectly calculates f(1,1) = 0. Choice B incorrectly suggests the integrals depend on path length. Choice C incorrectly calculates f(1,1) = -2.

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