Problem 24 - Advanced Engineering Mathematics

Question

Evaluate \(\int_{-c} y d x-x d y\), where \(C\) is given by \(x=2 \cos t, \quad y=3 \sin t, \quad 0 \leq t \leq \pi\).

Step-by-Step Solution

Answer

The integral evaluates to \\(-6\pi\\).

Step 1: Identify the curve parameterization

The curve is parameterized by \(x = 2 \cos t\) and \(y = 3 \sin t\) with \(t\) ranging from 0 to \(\pi\). This parameterization describes an ellipse in the xy-plane.

Step 2: Set up the integral using parameterization

The integral \(\int_{C} y \, dx - x \, dy\) needs to be evaluated. Substituting \(x = 2\cos t\) and \(y = 3\sin t\), we express \(dx\) and \(dy\) in terms of \(t\): \(dx = -2\sin t \, dt\) and \(dy = 3\cos t \, dt\).

Step 3: Substitute and simplify

Substitute the expressions for \(dx\) and \(dy\) into the integral: \[\int_{0}^{\pi} (3\sin t)(-2\sin t) \, dt - (2\cos t)(3\cos t) \, dt\]Simplify this to: \[\int_{0}^{\pi} -6\sin^2 t \, dt - 6\cos^2 t \, dt\].

Step 4: Use trigonometric identity to combine terms

Remember \(\sin^2 t + \cos^2 t = 1\). The integral simplifies further to: \[\int_{0}^{\pi} -6(\sin^2 t + \cos^2 t) \, dt = \int_{0}^{\pi} -6 \, dt\].

Step 5: Evaluate the integral

Since \(-6\) is a constant, the integral becomes:\[-6 \int_{0}^{\pi} 1 \, dt = -6[ t ]_{0}^{\pi} = -6(\pi - 0) = -6\pi\].

Key Concepts

Parametric Equations

Parametric equations are a way to represent curves by expressing the coordinates of the points on the curve as functions of a variable, usually denoted as \(t\). This method is especially helpful for curves that cannot be easily described with a single equation. In the given problem, the curve \(C\) is parameterized by:

• \(x = 2\cos t\)
• \(y = 3\sin t\)

Here, \(t\) is a parameter that varies from 0 to \(\pi\). As \(t\) changes within this range, the pair \((x,y)\) traces out the path of the curve. For this particular problem, these parametric equations describe the path of an ellipse in the xy-plane. By analyzing how \(x\) and \(y\) change with \(t\), one can find both the position and the nature of the curve. The derivatives of \(x\) and \(y\) with respect to \(t\) give us \(dx\) and \(dy\), which are crucial for computing line integrals.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for all values of the variable. These identities are incredibly useful for simplifying expressions, particularly in integrations involving trigonometric functions. One fundamental identity is:

• \(\sin^2 t + \cos^2 t = 1\)

This identity states that for any angle \(t\), the square of the sine plus the square of the cosine equals one. In our problem, using this identity allows us to combine and simplify the integrands. The integral initially has terms \(-6\sin^2 t\) and \(-6\cos^2 t\), which can be rewritten using \(\sin^2 t + \cos^2 t = 1\) to help simplify the expression. This approach reduces the complexity of the integration process, ultimately leading to a straightforward calculation. Familiarity with these identities is essential for efficiently handling trigonometric expressions in calculus.

Ellipse Parameterization

Parametrizing an ellipse is a common method to describe the shape and position of an ellipse in the coordinate plane. An ellipse with semi-major axis \(a\) and semi-minor axis \(b\) can be parameterized using:

• \(x = a \cos t\)
• \(y = b \sin t\)

In this exercise, \(a = 2\) and \(b = 3\), revealing that the ellipse reaches a maximum extent of 2 units along the x-axis and 3 units along the y-axis. The parameter \(t\) varies from 0 to \(\pi\), allowing us to traverse half the ellipse. Such parameterization helps in finding the derivatives \(dx\) and \(dy\), necessary for calculating line integrals over the ellipse. By using the parametric equations, one can easily evaluate the integral along the curve, capturing both the geometry and orientation of the ellipse within the range of \(t\).