Question

Evaluate $$ \oint \frac{x^{2} y d x-x^{3} d y}{\left(x^{2}+y^{2}\right)^{2}} $$ around the square with vertices \((\pm 1, \pm 1\) ) (note that \(\partial P / \partial y=\partial Q / \partial x\) ).

Short answer

Answer: The value of the given line integral is 0.

Step-by-step solution

Parameterize the curve

We have a square with vertices at \((\pm1, \pm1)\). Let's parameterize each side of the square: 1. Right side: Start at \((1, 1)\) and move to \((1, -1)\). $$ \mathbf{r}_1(t) = (1, 1-t), \quad t \in [0, 2] $$ 2. Bottom side: Start at \((1, -1)\) and move to \((-1, -1)\). $$ \mathbf{r}_2(t) = (1-t, -1), \quad t \in [0, 2] $$ 3. Left side: Start at \((-1, -1)\) and move to \((-1, 1)\). $$ \mathbf{r}_3(t) = (-1, -1+t), \quad t \in [0, 2] $$ 4. Top side: Start at \((-1, 1)\) and move to \((1, 1)\). $$ \mathbf{r}_4(t) = (-1+t, 1), \quad t \in [0, 2] $$ Next, find the derivative of each parameterization, to get the tangent vector \(\mathbf{r'}_i(t)\) for each side.

Compute the tangent vectors

Compute the tangent vectors for each parameterization: $$ \mathbf{r'}_1(t) = (0, -1), \quad \mathbf{r'}_2(t) = (-1, 0), \quad \mathbf{r'}_3(t) = (0, 1), \quad \mathbf{r'}_4(t) = (1, 0) $$

Evaluate the integrand

Evaluate the integrand \(\frac{x^2 y dx - x^3 dy}{(x^2 + y^2)^2}\) along each parameterization using \(dx = x'(t)dt, dy = y'(t)dt\) and \(x(t),y(t)\): 1. Right side: \(\int_0^2 \frac{1^2 (1-t)}{-dt} \left( 1^2 + (1-t)^2 \right )^{-2}dt\). 2. Bottom side: \(\int_0^2 \frac{(1-t)^2 (-1)}{-dt} \left((1-t)^2 + (-1)^2\right)^{-2}dt\). 3. Left side: \(\int_0^2 \frac{(-1)^2 (-1+t)}{dt} \left( (-1)^2 + (-1+t)^2 \right )^{-2}dt\). 4. Top side: \(\int_0^2 \frac{(-1+t)^2 (1)}{dt} \left( (-1+t)^2 + 1^2 \right )^{-2}dt\).

Evaluate the line integral

Add up the integrals from each parameterization: $$ \oint \frac{x^2 y dx - x^3 dy}{(x^2 + y^2)^2} = \sum_{i=1}^4 \int_0^2 \text{Integrand}_i(t) \, dt $$ Since \(\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}\), we know that the curl is zero. Consequently: $$ \oint \frac{x^2 y dx - x^3 dy}{(x^2 + y^2)^2} = 0 $$ Thus, the value of the given line integral around the square with vertices \((\pm 1, \pm 1)\) is 0.

Key concepts

Parameterization

Parameterization is an essential process in analyzing curves and lines in calculus, especially when dealing with line integrals. It involves expressing the path of a curve as a set of parametric equations. In simpler terms, we describe each point on the curve with functions depending on a variable parameter, usually denoted as \( t \). This helps in evaluating integrals over the path.

For instance, in the given exercise, a square is parameterized by considering each of its four sides separately. Each side is described by a unique function \( \mathbf{r}(t) \), converting the geometric path into a mathematical form that is easier to work with. The right side, for example, is parameterized as \( \mathbf{r}_1(t) = (1, 1-t) \) from \( t=0 \) to \( t=2 \).

• This allows us to compactly describe any point along each side of the square as \( t \) varies.
• By breaking down the path into segments, parameterization simplifies the process of introducing these paths into the integral calculations.

Understanding parameterization is crucial as it transforms complex geometries into manageable calculations.

Tangent Vector

The tangent vector plays a vital role in mechanics and calculus when dealing with parameterized curves. Intuitively, it gives us the direction in which the path is proceeding at any point. Mathematically, the tangent vector is obtained by differentiating the parameterized path with respect to its parameter \( t \).

In this exercise, after parameterizing each side of the square, the tangent vector \( \mathbf{r'}_i(t) \) is calculated for each side, representing the direction and rate of change of the path. For example, for the right side of our square, the tangent vector is \( \mathbf{r'}_1(t) = (0, -1) \).

• The tangent vector helps in expressing differentials \( dx \) and \( dy \) as products involving \( dt \), essential in calculating line integrals.
• Each tangent vector shows how the path is oriented, making it easier to understand and visualize the movement along the path.

Essentially, tangent vectors illuminate how curves flow, paving the way for more complex integral evaluations.

Curl of a Vector Field

The curl of a vector field provides insights into the rotational properties of the field. It is a vector operation denoted by \( abla \times \mathbf{F} \), measuring the infinitesimal rotation at a point. For a field \( \mathbf{F} = \langle P, Q \rangle \), the curl is computed as \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \).

In problems involving line integrals and vector fields, noting the conditions where the curl is zero can greatly simplify calculations. In this particular exercise, the equality \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \) ensures that the curl of the vector field is zero.

• A zero curl implies the lack of rotation in the field, which can provide direct solutions in certain integral problems.
• This observation simplifies integral computations and adheres to broader theorems, such as Green's Theorem.

Understanding curl is important as it helps diagnose the behavior of vector fields, its tendency to rotate, and links it to integral evaluations over closed paths.

Green's Theorem

Green's Theorem is a fundamental theorem in vector calculus that provides a connection between a line integral around a simple closed curve and a double integral over the region it encloses. The theorem is formally stated as:

\[ \oint_{C} P \, dx + Q \, dy = \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \] where \( C \) is the closed curve, and \( R \) is the region bounded by \( C \).

In this problem, the use of Green's Theorem is clearly justified because the curl of the vector field, as discussed earlier, is zero (\( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \)). Therefore, the right-hand side of the theorem disappears, simplifying the evaluation of the line integral to zero.

• Green's Theorem leverages simple calculations to evaluate complex line integrals by reducing them to double integrals.
• In situations where the integrand's expressions exhibit certain symmetries or special conditions, the theorem can drastically simplify computations.

Learning Green's Theorem equips one with a formidable tool for converting challenging line integrals into more manageable double integrals over a region.