Question

Evaluate the line integral using the Fundamental Theorem of Line Integrals. Use a computer algebra system to verify your results. \(\int_{C} \frac{y d x-x d y}{x^{2}+y^{2}}\) \(C\) : smooth curve from (1,1) to \((2 \sqrt{3}, 2)\)

Short answer

The value of the line integral is -\( \frac{\pi}{12}\).

Step-by-step solution

Identifying the potential function

We are given the integrand \(\frac{y d x-x d y}{x^{2}+y^{2}}\), we can see that this integrand is the derivative of the potential function \(f = \text{arctan}\(\frac{y}{x}\)\) by the way the original function is composed.

Evaluating the function at the end points

Given the curve \(C: \text{smooth curve from}\) (1,1) \(\text{to}\) \((2 \sqrt{3}, 2)\) we can evaluate the potential function at these end points. For point (1,1), \(f(1,1) = \text{arctan}\(\frac{1}{1}\) = \(\frac{\pi}{4}\). For the point \((2 \sqrt{3}, 2)\), \(f(2 \sqrt{3}, 2) = \text{arctan}\(\frac{2}{2\sqrt{3}}\) = \(\frac{\pi}{6}\).

Applying the Fundamental Theorem of Line Integrals

According to the Fundamental Theorem of Line Integrals, the line integral is equal to the potential function evaluated at the endpoint subtracted by the potential function evaluated at the starting point. Substituting the values we get \(\frac{\pi}{6} - \frac{\pi}{4} = -\frac{\pi}{12}\).

Verifying with a computer algebra system

Use a computer algebra system like Mathematica or Wolfram Alpha to confirm the answer. Enter the integral and the curve into the system. The system will also solve the line integral and the answer should match with the result obtained in the previous step, which is -\(\frac{\pi}{12}\).\n\n Although the tutorial does not demonstrate this step, it is straight forward to follow in any popular computer algebra system.

Key concepts

Fundamental Theorem of Line Integrals

Imagine walking along a hilly path; the Fundamental Theorem of Line Integrals is like understanding that you only need to check the height at the beginning and end of your journey to know how much you've ascended or descended, regardless of the path taken. In calculus, this theorem helps us calculate line integrals by simplifying the process significantly. Instead of computing the integral over every tiny segment of the curve, we evaluate the potential function at the endpoints of the curve.

For example, given the line integral \[\begin{equation}\text{\(\int_C \frac{y dx - x dy}{x^2 + y^2}\)},\end{equation}\] the theorem states that we can find the value of this line integral by simply taking the difference of a potential function at the end points, provided that the vector field is conservative. A conservative vector field is one where any line integral along a path dependent only on the endpoints, not the specific path. In our exercise, after identifying the potential function \text{arctan}(\frac{y}{x}), we apply the theorem to find the integral's value by evaluating \(f(2 \sqrt{3}, 2) - f(1,1)\), leading to our answer of \(-\frac{\pi}{12}\).

Potential Function

The potential function in calculus is akin to the concept of potential energy in physics. It's a scalar function which, when differentiated, gives us the components of the vector field. When dealing with line integrals of vector fields, finding a potential function can turn a potentially complex integral into a much simpler problem.

For instance, in our exercise, the integrand \(\frac{y dx - x dy}{x^2 + y^2}\) is recognized as the gradient of the potential function \(f(x, y) = \text{arctan}(\frac{y}{x})\). Identifying this potential function is crucial because it allows us to apply the Fundamental Theorem of Line Integrals. We determine the potential function values at the curve's endpoints, which grants us the power to calculate the integral without directly considering the curve's shape or equation. This is a powerful technique that simplifies calculations and understanding of vector fields.

Computer Algebra System

A Computer Algebra System (CAS) is an invaluable tool for verifying mathematical work, especially when dealing with complex integrals. It can symbolically manipulate mathematical expressions, rather than just performing numerical calculations. Think of it as having a math-savvy robot at your disposal: you give it a formula or equation, and it provides you an exact solution.

In our line integral problem, after manually evaluating the integral through the potential function, we can use a CAS like Mathematica or Wolfram Alpha to cross-check our work. It's as simple as inputting the line integral and the endpoints of the curve. This step is vital because it confirms the accuracy of our manual calculations and provides additional reassurance. It's particularly helpful when dealing with more involved integrals where the potential for computational error is higher.