Answer
Result: \[ \int_C \mathbf{F} \cdot d\mathbf{r} = -28 \]
Work Step by Step
Step 1: Note that the line integral with respect to \(\vec{r}\) of the vector field \(\mathbf{F}(x,y) = f(x,y)\mathbf{i} + g(x,y)\mathbf{j}\) along \(\mathbf{C}\) is expressed as: \[ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_C (f(x,y)\mathbf{i} + g(x,y)\mathbf{j}) \cdot (d\mathbf{xi} + d\mathbf{yj}) = \int_C f(x,y)dx + g(x,y)dy \] Where, \(\vec{r} = x\mathbf{i} + y\mathbf{j}\) Step 2: Now, let's use \((1)\) in order to evaluate the given integral. Hence, \[ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_C (2x\mathbf{j}) \cdot (d\mathbf{xi} + d\mathbf{yj}) \] Therefore \[ \int_C \mathbf{F} \cdot d\mathbf{r} = 2\int_C x dy \] Step 3: Note that this integral is along the line segment \(\mathbf{C}\) from \(P(-2,4)\) to \(Q(-2,11)\). Since we have \(x = -2\) at both points, then this integral is along the vertical line \(x = -2\). Then, let's substitute \(x = -2\) in \((2)\). Hence, \[ \int_C \mathbf{F} \cdot d\mathbf{r} = 2\int_C (-2) dy = -4\int_C dy \] Step 4: We notice that this integral is only with respect to \(y\), then its limits are \(y = 4\) at \(P\) and \(y = 11\) at \(Q\). Thus, \[ \int_C \mathbf{F} \cdot d\mathbf{r} = -4\int_{4}^{11} dy = [-4y]_{4}^{11} = (-4) \cdot 11 - (-4) \cdot 4 \] Result: \[ \int_C \mathbf{F} \cdot d\mathbf{r} = -28 \]
