91Ó°ÊÓ

When working with line integrals, we often deal with a path or curve along which the integration is performed. To evaluate a line integral, the first crucial step is to parameterize the curve. This means expressing the curve in terms of one or more parameters. For a line segment in a 3-dimensional space, such as the one given from point \((0, 1, 0)\) to \((1, 0, 0)\), the parameterization simplifies the problem by transforming it into a function of a single variable, typically denoted by \(t\).

In the given exercise, we parameterize the curve by finding functions for \(x\), \(y\), and \(z\) in terms of \(t\). Here, \((x(t), y(t), z(t)) = (t, 1-t, 0)\), where \(t\) varies from 0 to 1.

• The parameter \(t\) goes from the start point \((0,1,0)\) at \(t=0\) to the endpoint \((1,0,0)\) at \(t=1\).
• For each value of \(t\), the corresponding \((x, y, z)\) position on the line is calculated, ensuring that the path is traced accurately.

Once the curve is parameterized, we need to find the differential element along the curve, often referred to as the "line element." In calculus, the line element \(ds\) represents a small segment of the curve. It is calculated by determining the rate of change of each coordinate with respect to the parameter \(t\).

For a parametric curve \(\mathbf{r}(t) = (x(t), y(t), z(t))\), the line element is given by the expression:\[d s = \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2} \, dt\]In the step-by-step example, with \(x = t\), \(y = 1-t\), and \(z = 0\), the derivatives are \(\frac{dx}{dt} = 1\), \(\frac{dy}{dt} = -1\), and \(\frac{dz}{dt} = 0\).

• Plugging these into the line element formula, we get:\[d s = \sqrt{1^2 + (-1)^2 + 0^2} \, dt = \sqrt{2} \, dt\]
• This line element reflects the uniform distance along the curve between the bounds of integration.

Evaluating a line integral involves substituting the parameterized expressions and the line element into the integral. In the problem, the integral we need to evaluate is \(\int_{C} (x+y) \ ds\).

First, we substitute the parameterized equations \(x(t)\) and \(y(t)\) into the integral. Simplifying, we find that \((x + y) = (t + (1-t)) = 1\). We then substitute this into the integral as follows:\[\int_{0}^{1} 1 \cdot \sqrt{2} \, dt\]With \(\sqrt{2}\) being constant, it can be factored out of the integral:\[\sqrt{2} \int_{0}^{1} 1 \, dt\]

• Integrating \(1\) from 0 to 1 gives \([t]_{0}^{1} = 1 - 0 = 1\).
• Multiplying by the constant factor, the solution becomes \(\sqrt{2} \times 1 = \sqrt{2}\).
• The definite integral reveals the total contribution along the path when the function is integrated with respect to the line element.