Question

\(\int_{c}(x \sin x-y) d x+\left(x-y^{2}\right) d y\), where \(C\) is the triangle in the \((x, y)\) plane with vertices \((0,0),(1,1)\), and \((2,0)\)

Short answer

\( \int_{c}(x \sin x-y) dx+(x-y^2) dy = 0 \)

Step-by-step solution

- Identify the vertices of the curve

The vertices of the given triangle in the \(x, y\) plane are \( (0,0) \), \( (1,1) \), and \( (2,0) \).

- Set up the parameterization of the edges

Parameterize the three edges of the triangle: \( (0,0) \) to \( (1,1) \), \( (1,1) \) to \( (2,0) \), and \( (2, 0) \) to \( (0, 0) \).

- Parameterize edge from (0,0) to (1,1)

Let \((x,y) = (t,t) \) where \( t \) ranges from 0 to 1. The differential elements are \( dx = dy = dt \).

- Integrate along the first edge

Substitute \((x, y) = (t, t) \) and \( dx = dy = dt \) into the integral: \[ \int_{0}^1 \left( t \sin t - t \right) dt + \left( t - t^2 \right) dt \]

- Simplify the integral for the first edge

The integral becomes: \[ \int_{0}^1 t\sin t \,dt - \int_{0}^1 t \,dt + \int_{0}^1 t \,dt - \int_{0}^1 t^2 dt \]

- Parameterize edge from (1,1) to (2,0)

Let \(x = 1 + t \) and \(y = 1 - t \) where \( t \) ranges from 0 to 1. Then, \( dx = dt \) and \( dy = -dt \).

- Integrate along the second edge

Substitute \((x, y) = (1+t, 1-t) \), \( dx = dt \), and \( dy = -dt \) into the integral: \[ \int_{0}^1 \left( (1+t)\sin (1+t) - (1-t) \right) dt + \left( (1+t) - (1-t)^2 \right) (-dt) \]

- Simplify the integral for the second edge

Solve the integrals: \[ \int_{0}^1 (1+t)\sin (1+t) dt - \int_{0}^1 (1-t) dt - \int_{0}^1 ((1+t) - (1-2t+t^2)) (-dt) \]

- Parameterize edge from (2,0) to (0,0)

Let \(x = 2 - 2t \) and \(y = 0 \) where \( t \) ranges from 0 to 1. Then, \( dx = -2dt \) and \( dy = 0 \).

- Integrate along the third edge

Substitute \((x, y) = (2-2t, 0) \) and \( dx = -2dt \): \[ \int_{0}^1 \left( (2-2t)\sin (2-2t) - 0 \right) (-2dt) \]

- Simplify the integral for the third edge

Solve the integral: \[ \int_{0}^1 (2-2t)\sin (2-2t) (-2dt) \]

- Sum the integrals

Add up the integrals calculated in steps 4, 7, and 10.

Key concepts

Parameterization

In the context of line integrals, parameterization is a crucial first step. It involves expressing the coordinates of the points on a curve using a parameter. For instance, to parameterize a straight line, you create a relationship where both x and y are dependent on a single variable, like t. This simplifies evaluating integrals along sections of the curve.

In the given problem, parameterizing the three edges of the triangle is done as follows:

• From (0,0) to (1,1), we use \( (x,y) = (t,t) \) where t ranges from 0 to 1.
• From (1,1) to (2,0), we use \( x = 1 + t \) and \( y = 1 - t \), with t going from 0 to 1.
• From (2,0) to (0,0), we use \( x = 2 - 2t \) and \( y =0 \), with t going from 0 to 1.

Through parameterization, it becomes easier to substitute into the integral and work through the calculations.

Integration Along Curves

Integrating along curves involves taking the line integrals of vector fields along a given path or curve in the plane or in space. In this scenario, we integrate functions along the edges of a triangular path. When performing line integrals, ensure you respect the orientation and limits of each edge parameterization.

Formally, for a curve C parameterized by \( \textbf{r}(t) \) where \( t \) ranges from \(a \) to \( b \), the integral of a function \( F(x,y) \) along the curve is usually written as:
\text{\text{ \[ \int_{C} \textbf{F} \bullet d\textbf{r} = \[ \int_{a}^{b} F(\textbf{r}(t)) \bullet \frac{d\textbf{r}}{dt} dt \] \]}
Here, \( d\textbf{r} \) represents small elements in the direction of the curve. Integration along curves allows you to evaluate complex paths by breaking them into manageable linear sections and summing their contributions.

Triangular Path Integration

Handling triangular path integration means dividing the problem into three segments, each corresponding to the sides of the triangle. Each side can be parameterized using a linear relationship. By integrating each segment individually and then summing the results, we capture the entire path.

In this case, the vertices give us directives for parameterizing the triangle's edges.

• First edge: Integrate from (0,0) to (1,1).
• Second edge: Integrate from (1,1) to (2,0).
• Third edge: Integrate from (2,0) to (0,0).

After finding the expressions for each segment, you substitute and evaluate the integrals over their respective ranges. This method ensures that the contributions from all parts of the triangular path are accounted for.

Differential Elements

The concept of differential elements, or incremental changes along a path, is fundamental in calculating line integrals. In rectangular coordinates, differentials such as \( dx \) and \( dy \) represent small changes along the x and y direction, respectively. They are crucial when substituting back into the integral after parameterization.

For example, when parameterizing a segment, you need to replace \( dx \) and \( dy \) according to the parameterization:

• For the edge from (0,0) to (1,1), both \( dx \) and \( dy \) become \( dt \).
• For the edge from (1,1) to (2,0), \( dx = dt \) and \( dy = -dt \).
• For the edge from (2,0) to (0,0), \( dx = -2dt \) and \( dy = 0 \).

Understanding these differential elements helps you set up and solve the integrals correctly by accurately representing the small displacements along each segment of the path.