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.) $$ \int_{C} y^{2} d x+2 x y d y $$ (a) \(C_{1}:\) line segments from (0,0) to (3,4) to (4,4) (b) \(C_{2}:\) clockwise along the semicircle \(y=\sqrt{1-x^{2}}\) from (-1,0) to (1,0) (c) \(C_{3}:\) line segments from (-1,-1) to (-1,1) to (1,1) to (1,-1) to (-1,-1) (d) \(C_{4}:\) the closed path consisting of the semicircle \(y=\sqrt{1-x^{2}}\) from (-1,0) to (1,0) and the line segment from (1,0) to (-1,0)

Short answer

The values of the line integrals are: (a) 72, (b) 0, (c) 0, and (d) 0

Step-by-step solution

Determine If the Field is Conservative

Calculate the curl of the field to check if \(\mathbf{F}\) is conservative (or have zero curl). The 2-dimensional curl of a vector field \(\mathbf{F} = \langle P,Q \rangle\) is given by \(∇ × \mathbf{F} =\left( \frac{∂Q}{∂x} - \frac{∂P}{∂y} \right)\) Here, \(P = y^{2}, Q = 2xy\). Thus, \(\frac{∂Q}{∂x} = 2y\) and \(\frac{∂P}{∂y} = 2y\). By subtracting the two, you get the curl \(∇ × \mathbf{F}= 0\). Therefore, the field \(\mathbf{F}\) is conservative.

Find a Potential Function

\(\mathbf{F}\) is conservative, so we need to determine the potential function, \(f\). The function \(f\) can be determined by integrating the components of \(\mathbf{F}\). \(f(x,y) = \int P dx = \int y^{2} dx = xy^{2} + g(y)\) We don’t have enough information to determine \(g(y)\) so we then take the partial derivative of \(f\) with respect to \(y\). \(f_{y}' = xy + g'(y)\) The \(y\) component of \(\mathbf{F}\), however, was given as \(Q = 2xy\). Setting these equal to each other gives \(g'(y) = xy\). Integrating this with respect to \(y\), we have \(g(y) = \frac{y^{2}}{2}\). Our potential function is \(f(x,y) = xy^{2} + \frac{y^{2}}{2}\)

Integrate Over the Given Paths

Now, we can calculate the line integral over the given contours using the potential function \(f\). For \(C_1\), \(f_b - f_a = f(4,4) - f(0,0) = 72\). For \(C_2\), \(f_b - f_a = f(1,0) - f(-1,0) = 0\). For \(C_3\), \(f_{b}-f_{a} = f(1, -1) - f(-1, -1) = 0\). And for \(C_4\), \(f_{b} - f_{a} = f(-1,0) - f(-1,0) = 0\). This simplifies the computation and makes it easy to handle even complex paths.

Key concepts

Vector Fields

Imagine a field, not of grass, but of arrows pointing in various directions. This is a vector field, and it associates a vector to every point in a space. In calculus, vector fields are used to represent physical quantities that have both magnitude and direction, such as velocity or force at different points in space.

Mathematically, in a two-dimensional space, a vector field is represented by \( \textbf{F}(x, y) = \text{P}(x, y) \textbf{i} + \text{Q}(x, y) \textbf{j} \), where \( P \) and \( Q \) are functions that describe the horizontal and vertical components of the vector field, respectively. In our exercise, the vector field \( \textbf{F} \) had the components \( y^2 \) and \( 2xy \).

Conservative Fields

Certain vector fields, known as conservative fields, have a special property: the work done moving along a path within the field only depends on the start and end points, not the path taken. This is akin to walking on a hilly terrain where the effort depends solely on the height you start and end at, not the route you take.

Conservative vector fields are very important in physics because they can represent fields where energy is conserved, such as gravitational and electrostatic fields. In the context of the given exercise, knowing whether the vector field \( \textbf{F} \) is conservative simplifies the calculation of the line integral, as you can choose any convenient path between two points.

Potential Function

A potential function for a conservative vector field is a function \( f(x, y) \) whose gradient is equal to the vector field. This means that \( \textbf{F} = abla f \), where \( abla f \) is the vector of partial derivatives also known as the gradient of \( f \). The potential function is very much like the altitude on a map, where it determines the value at every point, and the gradient (or slope) at any point matches the vector field.

In our exercise, once determined, the potential function \( f(x, y) = xy^2 + \frac{y^2}{2} \) allowed for straightforward calculation of the line integral over various paths, simply by evaluating the difference in function values at the endpoints of the path—this is a powerful technique when dealing with conservative fields.

Curl of a Vector Field

The curl of a vector field helps us determine if a field is conservative or not. Generally speaking, the curl measures the rotation of the vector field at a point. A nonzero curl indicates the presence of 'circulation' or 'rotation' at that point, like a small whirlpool in a fluid. If a vector field has zero curl everywhere, it's a strong indication that the field is conservative.

In our problem, calculating the curl of \( \textbf{F} \) involved determining the differences between the derivatives of \( P \) and \( Q \)—the components of \( \textbf{F}\). When these differences equaled zero, the curl was determined to be zero, which confirmed that \( \textbf{F} \) is a conservative field. This understanding is essential, as it justifies why we can use the potential function to evaluate line integrals in this field.