Question

Find the value of the line integral $$\int_{C} \mathbf{F} \cdot d \mathbf{r}$$ (Hint: If \(\mathbf{F}\) is conservative, the integration may be easier on an alternative path.) \(\int_{C}(2 x-3 y+1) d x-(3 x+y-5) d y\) (a) \(C_{1}:\) line segments from (0,0) to (2,3) to (4,1) to (0,0) (b) \(C_{2}:\) counterclockwise along the semicircle \(x=\sqrt{1-y^{2}}\) from (0,-1) to (0,1) (c) \(C_{3}:\) along the curve \(y=e^{x}\) from (0,1) to \(\left(2, e^{2}\right)\) (d) \(C_{4}:\) the closed path consisting of the semicircle \(x=\sqrt{1-y^{2}}\) from (0,-1) to (0,1) and the line segment from (0,1) to (0,-1)

Short answer

(a) 0, (b) 0, (c) \(6e^2 - e^4 - 4\), (d) 0

Step-by-step solution

Determine if \(\mathbf{F}\) is conservative

To find whether a vector field \(\mathbf{F}\) is conservative, it's needed to calculate the curl of \(\mathbf{F}\) and see if that equals to zero. The curl of \(\mathbf{F}\) in two dimensions is defined as \(\nabla \times \mathbf{F} = \frac{\partial ((2x-3y+1))}{\partial y} - \frac{\partial ((3x+y-5))}{\partial x}\). Calculating this gives: \(3 - 3 = 0\). As \(\nabla \times \mathbf{F} = 0\), \(\mathbf{F}\) is a conservative vector field.

Find the potential function of \(\mathbf{F}\)

The potential function \(f\) of a conservative vector field \(\mathbf{F} = \) can be found by integrating \(M\) with respect to \(x\) and \(N\) with respect to \(y\), and equating those two results: \(\int(2x - 3y +1) dx = \int (3x + y - 5) dy\). By integrating these expressions, we get \(x^2 -3xy + x = 3xy + y^2 - 5y\). Solving for \(f\), we get \(f = x^2 - 3xy + y^2\). This is the potential function of \(\mathbf{F}\), which allows us to compute the line integrals more directly.

Calculate the line integral over \(C_1\)

As \(\mathbf{F}\) is conservative, the line integral over any closed path, such as \(C_1\), equals 0. This is due to the fundamental theorem of line integrals: \(\int_{C} \mathbf{F} \cdot d \mathbf{r} = f(B) - f(A)\), where \(A\) and \(B\) are the endpoints of the path. If the path is closed, \(A = B\), and therefore \(f(A) = f(B)\), leading to a result of 0.

Calculate the line integral over \(C_2\)

The line integral over \(C2\) is \(\int_{C2} \mathbf{F} \cdot d \mathbf{r} = f(B) - f(A)\), where \((A) = (0,-1)\) and \((B) = (0,1)\). Substituting these points into the potential function, we get \(f(B) - f(A) = 1 - 1 = 0\). Thus, the line integral over \(C2\) is also 0.

Calculate the line integral over \(C_3\)

The line integral over \(C3\) is \(\int_{C3} \mathbf{F} \cdot d \mathbf{r} = f(B) - f(A)\), where \(A = (0,1)\) and \(B = (2, e^2)\). Substituting these points into the potential function, we get \(-1 - ((2^2 - 3*2*e^2 + e^4) - (0^2 - 3*0 + 1^2)) = -1 - 4 + 6e^2 - e^4 + 1 = 6e^2 - e^4 - 4\). Thus, the line integral over \(C3\) is \(6e^2 - e^4 - 4\).

Calculate the line integral over \(C_4\)

\(C_4\) is a closed path, which is a combination of \(C_2\) and the vertical line segment, but in the opposite direction. Since \(\mathbf{F}\) is conservative, the line integral over any closed path is 0. So, the line integral over \(C_4\) is also 0.