Question

Workdone by a particle along the equare formed by the linen \(y=\pm 1\) and \(x=\pm 1\) under the force $$ \mathbf{F}=\left(x^{2}+x y\right) I+\left(x^{2}+y^{2}\right) J \text { is. } $$

Short answer

The work done is 0.

Step-by-step solution

Understand the Problem

We need to find the work done by the vector field \( \mathbf{F} = (x^2 + xy)\hat{i} + (x^2 + y^2)\hat{j} \) as a particle moves along the perimeter of a square.The square is defined by the lines \( y = \pm 1 \) and \( x = \pm 1 \). Remember, work done over a closed path for a non-conservative force is obtained using the line integral of the force along the path.

Determine Path Segments and Direction

The particle moves along a square path composed of four segments:1. From \((-1, -1)\) to \((1, -1)\), along \(y = -1\).2. From \((1, -1)\) to \((1, 1)\), along \(x = 1\).3. From \((1, 1)\) to \((-1, 1)\), along \(y = 1\).4. From \((-1, 1)\) to \((-1, -1)\), along \(x = -1\).We will calculate the line integral over each segment.

Integrate over the First Segment

For the segment from \((-1, -1)\) to \((1, -1)\), parameterize with \(x = t\), \(y = -1\) where \(t\) varies from \(-1\) to \(1\).The differential \(dx = dt\) and \(dy = 0\).Plug into the integral: \[ W_1 = \int_{-1}^{1} ((t^2 + t(-1))dt + (t^2 -1)(0)) = \int_{-1}^{1} (t^2 - t)dt. \]Calculate this integral.

Integrate over the Second Segment

For the segment from \((1, -1)\) to \((1, 1)\), parameterize with \(x = 1\), \(y = t\) where \(t\) varies from \(-1\) to \(1\).The differential \(dx = 0\) and \(dy = dt\).Plug into the integral:\[ W_2 = \int_{-1}^{1} ((1^2 + 1t)0 + (1^2 + t^2)(dt)) = \int_{-1}^{1} (1 + t^2)dt. \]Calculate this integral.

Integrate over the Third Segment

For the segment from \((1, 1)\) to \((-1, 1)\), parameterize with \(x = t\), \(y = 1\) where \(t\) varies from \(1\) to \(-1\).The differential \(dx = dt\) and \(dy = 0\).Plug into the integral:\[ W_3 = \int_{1}^{-1} ((t^2 + t(1))dt + (t^2 + 1)(0)) = \int_{1}^{-1} (t^2 + t)dt. \]Calculate this integral.

Integrate over the Fourth Segment

For the segment from \((-1, 1)\) to \((-1, -1)\), parameterize with \(x = -1\), \(y = t\) where \(t\) varies from \(1\) to \(-1\).The differential \(dx = 0\) and \(dy = dt\).Plug into the integral:\[ W_4 = \int_{1}^{-1} ((-1^2 - 1t)0 + (-1^2 + t^2)(dt)) = \int_{1}^{-1} (1 + t^2)dt. \]Calculate this integral.

Sum Work for All Segments

Approve that all the required line integrals for calculating the work have been established in Steps 3 to 6.Evaluate each integral:1. For the first segment, calculate: \[ W_1 = \int_{-1}^{1} t^2 - t \, dt = [\frac{1}{3}t^3 - \frac{1}{2}t^2]_{-1}^{1} = \left(\frac{1}{3} - \frac{1}{2}\right) - \left(-\frac{1}{3} + \frac{1}{2}\right) = 0. \]2. For the second segment: \[ W_2 = \int_{-1}^{1} 1 + t^2 \, dt = [t + \frac{1}{3}t^3]_{-1}^{1} = (1 + \frac{1}{3}) - (-1 - \frac{1}{3}) = \frac{8}{3}. \]3. For the third segment: \[ W_3 = \int_{1}^{-1} t^2 + t \, dt = [\frac{1}{3}t^3 + \frac{1}{2}t^2]_{1}^{-1} = 0. \]4. For the fourth segment: \[ W_4 = \int_{1}^{-1} 1 + t^2 \, dt = [t + \frac{1}{3}t^3]_{1}^{-1} = -\frac{8}{3}. \]The total work done: \[ W = W_1 + W_2 + W_3 + W_4 = 0 + \frac{8}{3} + 0 - \frac{8}{3} = 0. \]

Conclusion: Final Result

The work done by the force \( \mathbf{F} \) over the closed square path is zero, indicating that the force field is conservative over this closed path or the contributions cancel each other out.

Key concepts

Work Done in Vector Fields

When we talk about work done in vector fields, we are referring to the process of determining how much force is used while moving an object along a path in a field of forces. In our case, this field of forces is represented by the vector field \( \mathbf{F} = (x^2 + xy)\hat{i} + (x^2 + y^2)\hat{j} \).

The work done, \( W \), over a path \( C \) is calculated using a line integral of the vector field along the path. We express the work as:\[W = \int_C \mathbf{F} \cdot d\mathbf{r} = \int (F_x \, dx + F_y \, dy)\]Here, \( \, dx \) and \( \, dy \) represent the differentials of path segments in the \( x \) and \( y \) directions, respectively.

To compute \( W \), break the path \( C \) into simpler segments and evaluate the line integral over each segment. Each segment of the path can often be parameterized, which simplifies the integration process. The sum of these integrals gives the total work done along the path.

Conservative Vector Fields

Conservative vector fields are an important concept when dealing with line integrals. A vector field \( \mathbf{F} \) is called conservative if the line integral of \( \mathbf{F} \) around any closed path is zero.

Mathematically, for a conservative vector field, the work done around any closed loop \( C \) is:\[\oint_C \mathbf{F} \cdot d\mathbf{r} = 0\]This property is crucial because it means that the work done by a conservative field depends only on the starting and ending points of the path, not on the specific path taken between these points.

In the exercise provided, the result of the total line integral being zero indicates that \( \mathbf{F} \) behaves as a conservative vector field over the closed square path. This is why the work done by \( \mathbf{F} \) over the entire loop is zero despite the individual contributions around the path.

Parametric Equations

When evaluating line integrals, in many cases, it is useful to express the path in parametric form. Parametrizing involves representing the path as a set of equations in terms of a third variable, usually \( t \), which is called the parameter.

For example, a straight line from point \((-1, -1)\) to \((1, -1)\) on the path can be parameterized as:

• \( x = t \)
• \( y = -1 \)

where \( t \) varies from \(-1\) to \(1\). By doing this, the line integral becomes a simple one-dimensional integral with respect to \( t \), which makes calculations easier.

Parametric equations help simplify complex integrals by transforming the path into manageable segments. This systematic approach allows the computation of work done over each segment, helping in concluding the total work done along the complete path.