Problem 4 - Advanced Engineering Mathematics

Question

Evaluate \(\int_{C} G(x, y) d x, \int_{C} G(x, y) d y\), and \(\int_{C} G(x, y) d s\) on the indicated curve \(C\). $$ G(x, y)=x^{2} / y^{3} ; 2 y=3 x^{2 / 3}, 1 \leq x \leq 8 $$

Step-by-Step Solution

Answer

\(\int_{C} G(x, y) dx = \frac{56}{27}\), \(\int_{C} G(x, y) dy = \frac{8}{9}\), \(\int_{C} G(x, y) ds\) requires numerical methods.

Step 1: Identify Components of the Problem

We are tasked with evaluating three integrals given a curve defined by the parametric relation \(2y = 3x^{2/3}\) for \(1 \leq x \leq 8\). The function to integrate is \(G(x, y) = \frac{x^2}{y^3}\). The curve \(C\) is traversed as \(x\) varies from 1 to 8.

Step 2: Express y in terms of x

First, we need to find \(y\) in terms of \(x\) from the given curve equation \(2y = 3x^{2/3}\). Solving for \(y\), we get: \[ y = \frac{3}{2} x^{2/3} \]

Step 3: Compute First Integral \(\int_{C} G(x, y) \, dx\)

Substitute \(y = \frac{3}{2} x^{2/3}\) into \(G(x, y)\), resulting in \(G(x, y) = \frac{x^2}{(\frac{3}{2} x^{2/3})^3} = \frac{x^{2}}{\frac{27}{8} x^{2}} = \frac{8}{27} x^0 = \frac{8}{27}\). Now integrate with respect to \(x\) over the range \(1\) to \(8\):\[ \int_{1}^{8} \frac{8}{27} \, dx = \frac{8}{27} \int_{1}^{8} \, dx = \frac{8}{27} [x]_{1}^{8} = \frac{8}{27} (8 - 1) = \frac{56}{27} \]

Step 4: Compute Second Integral \(\int_{C} G(x, y) \, dy\)

Differentiate \(y = \frac{3}{2} x^{2/3}\) to get \(\frac{dy}{dx} = x^{-1/3}\). Thus, \(dy = x^{-1/3} \, dx\). Substitute back into the integral and compute:\[ \int_{1}^{8} \frac{x^2}{\left(\frac{3}{2} x^{2/3}\right)^3} x^{-1/3} \, dx = \int_{1}^{8} \frac{8}{27} x^{-1/3} \, dx \]Apply the power rule for integration:\[ \int x^{-1/3} \, dx = \frac{3}{2} x^{2/3} + C \]Now compute:\[ \frac{8}{27} \times \frac{3}{2} [x^{2/3}]_{1}^{8} = \frac{4}{9} [x^{2/3}]_{1}^{8} = \frac{4}{9} (4 - 1) = \frac{8}{9} \]

Step 5: Compute Third Integral \(\int_{C} G(x, y) \, ds\)

The line element \(ds\) along the curve satisfies:\[ ds = \sqrt{(dx)^2 + (dy)^2} = \sqrt{1 + (\frac{dy}{dx})^2} \cdot dx \]Given \(\frac{dy}{dx} = x^{-1/3}\), \(ds = \sqrt{1 + x^{-2/3}} \, dx\). Thus, the integral becomes:\[ \int_{1}^{8} \frac{8}{27} \sqrt{1 + x^{-2/3}} \, dx \]This integral does not have a simple antiderivative, hence it needs to be evaluated using numerical methods.

Step 6: Summarize Results

The integrals along the curve yield:- \(\int_{C} G(x, y) \, dx = \frac{56}{27}\)- \(\int_{C} G(x, y) \, dy = \frac{8}{9}\)- \(\int_{C} G(x, y) \, ds\) requires numerical methods for evaluation.

Key Concepts

Parametric Equations

Parametric equations offer a way to describe a curve using parameters, often simplifying the calculations involved in line integrals. In the given problem, the curve is defined by the parametric equation \(2y = 3x^{2/3}\). This means both \(x\) and \(y\) are expressed in terms of another variable, but here it's in a more direct form where \(x\) controls the variation, and \(y\) is dependent on it.

• This creates an implicit relationship where \(y\) varies as \(x\) changes from 1 to 8.
• This approach simplifies the understanding of the curve's path since we can directly substitute \(x\) values to find \(y\).

By solving \(2y = 3x^{2/3}\) for \(y\), we isolate \(y\) to use in our integrals. This ability to manipulate parametric equations is pivotal, as it transforms a potentially complex curve into manageable parts which are much simpler to work with.

Integrals with respect to arc length

Line integrals with respect to arc length \(ds\) incorporate the curve's entire geometry in calculating something like the total "weight" along the path. This requires converting the typical \(dx\) or \(dy\) into \(ds\), the differential arc length. Arc length accounts not only for horizontal changes \(dx\) but also vertical \(dy\), presenting a more comprehensive path description.

• The formula \(ds = \sqrt{(dx)^2 + (dy)^2}\) is used, incorporating both \(dx\) and \(dy\).
• Here, solving for \(dy/dx = x^{-1/3}\) gives us the ability to express \(dy\) in terms of \(dx\), needed for the arc length.

The outcome \(ds = \sqrt{1 + x^{-2/3}} \cdot dx\) signifies the intricate curve's length, reflecting more than just a straight line between start and end points, requiring numerical methods for evaluation due to its complexity.

Differential Calculus

Differential calculus plays a crucial role in evaluating line integrals because we often need to differentiate functions to find rates of change, such as \(\frac{dy}{dx}\). In this exercise, differentiation determines how \(y\) changes with respect to \(x\), a key concept in moving from parametric to calculus-based geometry.

• The task involves finding \(\frac{dy}{dx} = x^{-1/3}\), revealing how quickly \(y\) rises or falls as \(x\) shifts.
• This derivative is essential to compute both \(dy\) and \(ds\).

Beyond just interpreting the rate of change, differential calculus helps in simplifying complex expressions within integrals, making them more straightforward to handle or setting them up for numerical methods.

Numerical Methods

Numerical methods provide practical solutions for integrals that cannot be solved analytically. While some line integrals simplify to basic forms, others, like those involving \(ds\) in complex curves, don't permit easy antiderivatives.

• The integral \(\int_{1}^{8} \frac{8}{27} \sqrt{1 + x^{-2/3}} \, dx\) lacks a straightforward antiderivative, pushing us to use numerical techniques.
• Methods such as the trapezoidal rule or Simpson's rule break down the curve into smaller, manageable segments, allowing approximation of the integral by summing these parts.

By using numerical methods, we gain the flexibility to approximate solutions for challenging integrals that arise in real-world problems, staying accurate within desired tolerances.