Problem 18 - Calculus: Early Transcendental Functions

Question

Evaluate the line integral. \(\int_{C}(x+y) d y,\) where \(C\) is the portion of \(x=y^{2}\) from (1,1) to (1,-1)

Step-by-Step Solution

Answer

The value of the line integral is 0

Step 1: Parametrize Curve

Parametrize the curve \(x=y^2\) as \(x=t^2\), \(y=t\), where \(t\) changes from 1 to -1. Therefore, \(dx = 2t dt\) and \(dy = dt\).

Step 2: Rewrite Integral

Substitute \(x\), \(y\) and \(dy\) in the given line integral with the parametrized curve values: \(\int_{C}(t+t) dt\) from 1 to -1

Step 3: Evaluate Integral

Evaluate the new integral from -1 to 1: \(\int_{-1}^{1}2tdt\). This creates \([t^{2}]_{-1}^{1}\), which is 0.

Key Concepts

Curve Parametrization

When dealing with line integrals, one essential step is to parametrize the curve. This means expressing the curve in terms of a single variable, called the parameter.
For instance, if a curve is described by the equation \(x = y^2\), we can parametrize it using \(t\) by setting \(x = t^2\) and \(y = t\). This way, the parameter \(t\) allows us to trace the curve as \(t\) varies.

• In this specific example, the range of \(t\) is from 1 to -1, representing the starting and ending points of the curve.
• Through parametrization, differential elements like \(dx\) and \(dy\) can be expressed in terms of the derivative of the parameter, making them \(dx = 2t\, dt\) and \(dy = dt\).

This step is crucial because it simplifies the integration process and sets up for evaluating the line integral effectively.

Integral Evaluation

Once the curve is parametrized, the next task is to evaluate the line integral by substituting parametrized expressions into the original integral. In doing so, we transform the problem into an integration issue over the parameter.
Consider the integral \(\int_{C}(x+y)dy\) given, where using the parametrized values, we substitute \(x = t^2\), \(y = t\), and \(dy = dt\). Consequently, the integral transforms to \(\int_{1}^{-1} (t^2+t) dt\).

• Notice how life's now simplified because everything's in terms of \(t\), allowing for a straightforward evaluation.

This step prepares the integral for evaluation, which reflects understanding of how the curve interacts with the function within the integral.

Vector Calculus

Vector calculus comes into play when dealing with line integrals as it bridges the gap between scalar fields and the paths we integrate over. Line integrals often evaluate how a vector field influences an object moving along a path.
In this scenario, while the original integral \(\int_{C} (x+y) dy\) might seem straightforward, it reflects vector calculus concepts because it assesses a scalar function along a curve.

• Vector calculus uses these integrals to encapsulate the act of summing infinitesimal contributions from a field along a path.
• It is integral (pun intended!) in courses dealing with physics and engineering because it accurately models real-world phenomena like work done by forces or flux across a surface.

This part of vector calculus finds extensive applications, bridging theoretical math with practical scenarios.