Problem 26 - The Calculus with Analytic Geometry

Question

In Exercises 21 through 30, show that the value of the line integral is independent of the path and compute the value in any convenient manner. In each exercise, \(C\) is any sectionally smooth curve from the point \(A\) to the point \(B\). \(\int_{c} \frac{x}{x^{2}+y^{2}+z^{2}} d x+\frac{y}{x^{2}+y^{2}+z^{2}} d y+\frac{z}{x^{2}+y^{2}+z^{2}} d z ; A\) is \((1,0,0)\) and \(B\) is \((1,2,3)\)

Step-by-Step Solution

Answer

\(\ln(14)\)

Step 1: Verify Independence of the Path

To show that the line integral is independent of the path, we need to verify that the given vector field \({\mathbf{F}(x,y,z)}\) is conservative. This means finding a potential function \({f(x,y,z)}\) such that \({abla f = \frac{x}{x^{2}+y^{2}+z^{2}} \mathbf{\hat{i}} + \frac{y}{x^{2}+y^{2}+z^{2}} \mathbf{\hat{j}} + \frac{z}{x^{2}+y^{2}+z^{2}} \mathbf{\hat{k}}}\).

Step 2: Find the Potential Function

Assume \({f(x,y,z)}\) has partial derivatives that match the components of the given vector field. Integrate \(\frac{x}{x^{2}+y^{2}+z^{2}}\) with respect to \(x\) to find \({f(x,y,z)}\). The integral results in \({f(x,y,z) = \ln(x^{2}+y^{2}+z^{2})}\). Verify this by differentiating \({f(x,y,z)}\): \({\frac{\partial f}{\partial x} = \frac{x}{x^{2}+y^{2}+z^{2}}}\), \({\frac{\partial f}{\partial y} = \frac{y}{x^{2}+y^{2}+z^{2}}}\), and \({\frac{\partial f}{\partial z} = \frac{z}{x^{2}+y^{2}+z^{2}}}\). This confirms \({f(x,y,z)}\) is the potential function.

Step 3: Evaluate Potential Function at Points A and B

To compute the value of the line integral, evaluate the potential function at the endpoints. Evaluate \({f(1,0,0) = \ln(1^2 + 0^2 + 0^2) = \ln(1)}\). Evaluate \({f(1,2,3) = \ln(1^2 + 2^2 + 3^2) = \ln(1 + 4 + 9) = \ln(14)}\).

Step 4: Calculate the Value of the Line Integral

The value of the line integral from point \({A = (1,0,0)}\) to point \({B = (1,2,3)}\) is given by \({f(B) - f(A)}\). Substitute the values: \({\ln(14) - \ln(1) = \ln(14)}\).

Key Concepts

Line Integral

A line integral helps us calculate the integral of a function along a curve. Imagine you are walking along a path on a mountain where the height at each point is given by a function. The line integral sums up these heights along your path.
For example, given the vector field \(\frac{x}{x^{2}+y^{2}+z^{2}} \mathbf{\backslash\backslashhat\{i\}} + \frac{y}{x^{2}+y^{2}+z^{2}} \mathbf{\backslash\backslashhat\{j\}} + \frac{z}{x^{2}+y^{2}+z^{2}} \mathbf{\backslash\backslashhat\{k\}} \), you would integrate these components over a curve C from point A to point B.
This involves summing small pieces of the vector field along the curve and is useful in physics for understanding fields and potentials.

Potential Function

A potential function is like a 'higher-level' function that, when differentiated, gives you the components of a vector field. If you have a conservative vector field, you can find this function.
In the exercise, we started with the vector field \(\frac{x}{x^{2}+y^{2}+z^{2}} \backslashmathbf{\backslashhat\backslash{i}} + \frac{y}{x^{2}+y^{2}+z^{2}} \backslashmathbf{\backslashhat\backslash{j}} + \frac{z}{x^{2}+y^{2}+z^{2}} \backslashmathbf{\backslashhat\backslash{k}}\). To find the potential function, we integrated \(\frac{x}{x^{2}+y^{2}+z^{2}} \backslash dx\) to get \(f(x,y,z) = \backslash\backslashln(x^{2}+y^{2}+z^{2})\).
By verifying with partial derivatives, we concluded that our potential function works since its gradient gives us back the original vector field. This makes solving line integrals easier using the potential function instead of directly.

Gradient Theorem

The Gradient Theorem, also known as the Fundamental Theorem for Line Integrals, tells us that if a vector field is conservative (can be expressed as the gradient of a potential function), then the line integral over this field depends only on the values at the endpoints of the curve.
This simplifies computations because, if \(F\backslashcdot dr = abla f\backslashcdot dr\), then \(\backslashint_C F\backslashcdot dr = f(B) - f(A)\).
In our example, it allowed us to evaluate the potential function at the points \(A = (1,0,0)\) and \(B = (1,2,3)\), leading to \(f(B) - f(A) = \backslashln(14)\).

Independence of Path

Path independence means that the value of the line integral is the same no matter which path is taken between two points, as long as the vector field is conservative.
In the exercise, this concept was used to show that the integral would have the same result regardless of the path taken from point A to B.
By verifying that the vector field is conservative (meaning a potential function \(f(x,y,z)\) exists), we proved that \(\backslashint_{C1} = \backslashint_{C2}\) for any two paths C1 and C2 between points A and B.