Question

Evaluate each line integral using the given curve \(C\). \(\int_{C} \frac{y}{\sqrt{x^{2}+y^{2}}} d x-\frac{x}{\sqrt{x^{2}+y^{2}}} d y ; C\) is a quarter-circle from (0,4) to (4,0)

Short answer

Question: Compute the line integral of the vector field F(x,y) = (y/√(x^2+y^2), -x/√(x^2+y^2)) along the curve C, a quarter-circle from (0,4) to (4,0). Answer: The line integral of the given vector field along the curve C is π/2.

Step-by-step solution

Parameterize the curve C

The curve C is a quarter-circle from (0,4) to (4,0). We can parameterize this curve using trigonometric functions, as follows: $$ x=4\cos(\theta), \;\; y=4\sin(\theta) $$ Since the curve is going from (0,4) to (4,0), we know that \(\theta\) varies from \(\frac{\pi}{2}\) to 0.

Compute the differentials of x and y

Now, we need to compute the differentials of x and y with respect to the parameter θ: $$ \frac{dx}{d\theta} = -4\sin(\theta), \;\; \frac{dy}{d\theta} = 4\cos(\theta) $$ Thus, the differentials are given by: $$ dx = -4\sin(\theta) d\theta, \;\; dy = 4\cos(\theta) d\theta $$

Compute the line integral

Now we can compute the given line integral along the curve C using the parameterized form and differentials: $$ \int_{C} \frac{y}{\sqrt{x^{2}+y^{2}}} dx - \frac{x}{\sqrt{x^{2}+y^{2}}} dy = \int_{\frac{\pi}{2}}^{0} \frac{4\sin(\theta)}{\sqrt{(4\cos(\theta))^{2}+(4\sin(\theta))^{2}}} (-4\sin(\theta))d\theta - \frac{4\cos(\theta)}{\sqrt{(4\cos(\theta))^{2}+(4\sin(\theta))^{2}}} (4\cos(\theta))d\theta $$ Notice that the terms inside the square root simplify to \(4^{2}(\cos^2(\theta) + \sin^2(\theta))=16\). Taking the square root yields 4. Additionally, the integral can be split into two separate integrals: $$ \int_{\frac{\pi}{2}}^{0} (-\sin^{2}(\theta) - \cos^{2}(\theta)) d\theta $$ The curve is traced from \(\frac{\pi}{2}\) to 0, so limits of our integral will be from \(\frac{\pi}{2}\) to 0. Now, we use the fundamental trigonometric identity \(\sin^{2}(\theta) + \cos^{2}(\theta) = 1\) and simplify the integral: $$ \int_{\frac{\pi}{2}}^{0} (-1) d\theta $$ Now, we can integrate with respect to θ: $$ \int_{\frac{\pi}{2}}^{0} -d\theta = -\Big[\theta\Big]^\frac{\pi}{2}_0 $$ Finally, we compute the definite integral: $$ -\Big(-\frac{\pi}{2}\Big) = \frac{\pi}{2} $$ Hence, the line integral of the given vector field along the curve C is \(\frac{\pi}{2}\).

Key concepts

Parameterization of Curves

Parameterization is an essential concept in calculus, especially when dealing with curves and surfaces. It involves representing a curve or trajectory using a set of equations that express each coordinate as a function of one or more parameters. This is particularly useful for line integrals, where we deal with curves in the xy-plane.

In our exercise, the curve C is a quarter-circle, which we can represent using the trigonometric functions sine and cosine. The parameterization process involves assigning a variable, typically denoted as \(\theta\), that will vary over an interval corresponding to the curve's domain. For our quarter-circle, starting from the point (0,4) and ending at (4,0), \(\theta\) varies from \(\frac{\pi}{2}\) to 0—the standard interval for the first quadrant of a circle.

A practical advantage of parameterization is that it allows us to move from the potentially complicated geometry of a curve to a simple, one-dimensional interval. The power of parameterization shines when we convert difficult geometrical problems into manageable computations, such as when transforming complex line integrals into standard definite integrals.

Trigonometric Functions in Calculus

Trigonometric functions play an indispensable role in calculus, particularly in the study of periodic phenomena and in problems involving right angles and circles, like the parameterization of curves.

In the context of our exercise, the functions cosine and sine are used to define the x and y coordinates of any point on our quarter-circle curve. This particular use of trigonometric functions is powerful because it not only describes the position but also facilitates the computation of differentials — an essential step for evaluating line integrals.

Applying Trigonometric Identities

It's also important to know and use trigonometric identities, such as \(\sin^{2}(\theta) + \cos^{2}(\theta) = 1\), which enable further simplification in calculus problems. These identities are powerful tools that make complex problems solvable, as seen when we simplify the integral in step 3 of our solution. They allow us to evaluate the line integral with ease, demonstrating the significant contribution of trigonometric functions beyond their basic geometric interpretations.

Definite Integrals

Definite integrals are pivotal in calculus as they represent the accumulation of quantities and the total change over an interval. They offer a way to quantify areas under curves, total displacement, or the value of a line integral along a path.

In line integral calculus, when the curve is parameterized, a definite integral is used to calculate the integral along the given path from one endpoint to another. In our exercise, after parameterization and simplification, we calculated a definite integral with the limits from \(\frac{\pi}{2}\) to 0 representing the starting and ending angles of our quarter-circle curve.

Evaluating Definite Integrals

When evaluating definite integrals, it's essential to follow the fundamental theorem of calculus, which involves finding an antiderivative and then using the evaluated bounds to determine the accumulated value. This concept was beautifully displayed in the step-by-step solution of the exercise, demonstrating the quintessential process from antiderivative to evaluation that lies at the heart of definite integrals in calculus.