Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate each line integral. \(\int_{C} y d x+x d y ; C\) is the curve \(y=x^{2}, 0 \leq x \leq 1\).

Step-by-Step Solution

Answer

The value of the line integral is 1.

Step 1: Parameterize the Curve

The curve given is defined by the equation \[ y = x^2 \] for \( 0 \leq x \leq 1 \). We can parameterize this curve by letting \( x = t \) and \( y = t^2 \) for \( 0 \leq t \leq 1 \).

Step 2: Express dx and dy in terms of t

Differentiate the parameter functions with respect to \( t \):\[ dx = dt \]\[ dy = 2t \, dt \].

Step 3: Substitute into the Line Integral

Substitute the parameterized expressions into the line integral:\[ \int_{C} y \, dx + x \, dy = \int_{0}^{1} (t^2 \, dt + t \, (2t \, dt)) \].

Step 4: Simplify the Integral

Combine and simplify the terms inside the integral:\[ \int_{0}^{1} t^2 \, dt + 2t^2 \, dt = \int_{0}^{1} (t^2 + 2t^2) \, dt = \int_{0}^{1} 3t^2 \, dt \].

Step 5: Evaluate the Integral

Perform the integration:\[ \int_{0}^{1} 3t^2 \, dt = 3 \left[ \frac{t^3}{3} \right]_{0}^{1} = 3 \cdot \left( \frac{1^3}{3} - \frac{0^3}{3} \right) = 3 \cdot \frac{1}{3} = 1 \].

Key Concepts

Parameterization

To evaluate a line integral along a curve, we first need to describe the curve in a simple form. This is where parameterization comes in. Parameterizing a curve involves expressing the coordinates of the points on the curve using one or more parameters. For the curve given by the equation \( y = x^2 \), for \( 0 \leq x \leq 1 \), we can use \( t \) as a parameter. This lets us write \( x = t \) and consequently \( y = t^2 \). Here, the parameter \( t \) varies within \( 0 \leq t \leq 1 \). By parameterizing the curve, we simplify the equation, making it easier to perform subsequent steps like differentiation and integration.

Differentiation

After parameterizing the curve, the next step is to find the differentials \( dx \) and \( dy \), which are required for our line integral. Differentiate the parameterized expressions of \( x \) and \( y \) with respect to \( t \):

• Since \( x = t \), differentiating gives \( dx = dt \).
• Since \( y = t^2 \), differentiating gives \( dy = 2t \, dt \).

Using these differentials, we can substitute them into the line integral. Differentiation in this context helps us transition from parametric equations to expressions suitable for integration.

Integration

Once you have parameterized the curve and found the differentials, the next step is to perform integration. The line integral is now transformed into a standard integral with respect to \( t \).Substitute the expressions \( y = t^2 \), \( dx = dt \), and \( dy = 2t \ dt \) into the line integral:\[ \int_{C} y \, dx + x \, dy = \int_{0}^{1} (t^2 \, dt + t \, (2t \, dt)) \]Simplify it to:\[ \int_{0}^{1} (t^2 + 2t^2) \, dt \]Combine like terms:\[ \int_{0}^{1} 3t^2 \, dt \]This integral is now ready to be evaluated. Integration involves finding the antiderivative, which is simpler once the equation is fully parameterized and simplified.

Curve Equations

Understanding the equation of a curve is crucial when dealing with line integrals. The original problem involves a parabolic curve \( y = x^2 \). Interpreting this equation graphically, it depicts a shape that covers all points from \( (0,0) \) to \( (1,1) \) along the curve.Curve equations do more than describe static objects; they provide a path or a set of paths. This helps in calculating quantities along the curve, such as work done or flux. Describing the path in physical or geometrical terms assists one in visualizing and then conceptualizing the integral over a specific trajectory.Through parameterization, these curve equations transform into a simple linear parameter which describes the trajectory in relation to one variable. Recognizing this transformation is a key step in understanding how complex paths are represented practically in terms of integration.