Problem 8 - Calculus: Early Transcendental Functions

Question

Evaluate the line integral. \(\int_{C}(3 x-y) d s,\) where \(C\) is the quarter-circle \(x^{2}+y^{2}=9\) from (0,3) to (3,0)

Step-by-Step Solution

Answer

-9

Step 1: Parameterize the Path C

The curve C is on a quarter circle with radius 3, and ranges from (0,3) to (3,0). We can map this curve using polar coordinates by using the equations \(x = r cos(θ)\) and \(y = r sin(θ)\) and setting \(r = 3\) since the radius is 3. In addition, since the quarter circle goes from the positive y-axis to the positive x-axis, our θ will range from \(pi/2\) to 0. Therefore, the parameterized path is described by the equations \(x = 3 cos(θ)\) and \(y = 3 sin(θ)\) where \(π/2 ≥ θ ≥ 0\).

Step 2: Calculate ds

Next we calculate the differential arc length ds. In polar coordinates, ds is described by the formula \(ds = r dθ\). So by plugging in \(r = 3\) we get \(ds = 3 dθ\).

Step 3: Substitute into the line integral

Now, we substitute the expressions for x, y, and ds into the line integral and integrate from \(\pi/2\) to 0. This gives us the integral \(\int_{\pi/2}^0 {(3(3 cos(θ)) - (3 sin(θ))) * (3 dθ)}\).

Step 4: Evaluate the integral

Finally, we evaluate the integral. We can simplify our integral to \(9 \int_{\pi/2}^0 {(cos(θ) - sin(θ)) dθ}\). This evaluates to -9.

Key Concepts

Polar Coordinates

Polar coordinates offer a way of describing a point's location in a plane using a distance from a reference point and an angle from a reference direction. Unlike the Cartesian system with x and y axes, polar coordinates use:

• Distance called radius (denoted as r).
• Angle (denoted as θ), typically measured from the positive x-axis.

For a quarter-circle of radius 3, we convert the circle's standard equation to polar coordinates. This involves expressing the Cartesian coordinates (x, y) using the formulas:

• \(x = r \cos(θ)\)
• \(y = r \sin(θ)\)

Given that the radius is 3, the equations simplify to

• \(x = 3 \cos(θ)\)
• \(y = 3 \sin(θ)\)

Here, θ varies as the object moves along the circle.

Quarter-Circle Path

A quarter-circle path is a segment of a full circle, specifically 90 degrees or \(\frac{\pi}{2}\) radians. Understanding this path helps us define our bounds for integration.

Our particular segment stretches from point (0, 3) to (3, 0), corresponding to angles starting from \(\frac{\pi}{2}\) to 0 in polar coordinates. This illustrates the quarter-circle's position in the first quadrant.By knowing these parameters, we can now accurately map the path using those angles, ensuring that it captures the correct curve trajectory. This precise understanding of the path is essential when performing further calculations, such as parameterizing and integrating along it.

Parameterization

Parameterization means expressing a path or curve in terms of a third variable, typically a parameter like θ. It offers a simple, powerful way to define curves that are not easily described by standard Cartesian equations.

With our quarter-circle, parameterization is achieved using θ in place of x or y. The path is given by:

• \(x = 3 \cos(θ)\)
• \(y = 3 \sin(θ)\)

Here, θ moves from \(\frac{\pi}{2}\) to 0. This ensures that the calculation captures the switch from point (0, 3) to point (3, 0). The parameterization captures the essence and navigation of the entire path in terms of θ, aiding in further analyses like integration.Through parameterization, the process of evaluating line integrals becomes streamlined, as each point on the curve is easily mapped and understood in its polar form.

Arc Length Differential

The arc length differential, denoted as \(ds\), represents a small segment of the path's length. Calculating this is crucial for integrating over a path, as it weights the differentials of the function as it varies along the curve.

In polar coordinates, \(ds\) is given by the expression \(ds = r dθ\). For our path, with a radius of 3, it simplifies to \(ds = 3 dθ\).
This expression indicates that a small arc segment changes in direct proportion to the change in θ, scaled by the constant radius of 3. Using this differential, the integrand fully accounts for both the path's geometry and the function's variability on the curve.
The concept of arc length differential is vital for evaluating line integrals, ensuring that the length and positioning of each curve segment impact the integral result accurately.