Problem 15 - Advanced Engineering Mathematics

Question

Evaluate \(\int_{C}\left(6 x^{2}+2 y^{2}\right) d x+4 x y d y\), where \(C\) is given by \(x=\sqrt{t}, y=t, 4 \leq t \leq 9\).

Step-by-Step Solution

Answer

The evaluated integral is 460.

Step 1: Parametrize the Curve

The curve is parametrized by the given equations: \[ x = \sqrt{t} \text{ and } y = t. \]The parameter \(t\) ranges from 4 to 9. Therefore, the parametrization of the curve is:\[(x(t), y(t)) = (\sqrt{t}, t)\text{ for } 4 \leq t \leq 9. \]

Step 2: Compute Derivatives

Compute the derivatives of \(x\) and \(y\) with respect to \(t\):\[ \frac{dx}{dt} = \frac{1}{2\sqrt{t}}, \quad \frac{dy}{dt} = 1. \]

Step 3: Rewrite the Differential

Substitute the derivatives into the differential:\[ dx = \frac{1}{2\sqrt{t}} dt, \quad dy = dt. \]

Step 4: Substitute into the Integral

Substitute \(x = \sqrt{t}\), \(y = t\) and the differentials into the line integral:\[ \int_{4}^{9} \left(6(\sqrt{t})^{2} + 2t^{2}\right) \cdot \frac{1}{2\sqrt{t}} dt + 4(\sqrt{t})(t) dt. \]

Step 5: Simplify the Expression

Simplify the integrand: - For the first part: \[ \left(\frac{6t + 2t^{2}}{2\sqrt{t}}\right) = 3\sqrt{t} + t^{2}/\sqrt{t} = 3\sqrt{t} + t^{3/2}. \]- The second part is:\[ 4t\sqrt{t} = 4t^{3/2}. \]

Step 6: Combine and Simplify Further

Combine the two parts of the integral:\[ \int_{4}^{9} (3\sqrt{t} + t^{3/2} + 4t^{3/2}) dt = \int_{4}^{9} (3\sqrt{t} + 5t^{3/2}) dt. \]

Step 7: Integrate Each Term

Integrate each term separately:1. \(3\sqrt{t} = 3t^{1/2}\) integrates to \[ \frac{3}{1/2 + 1}t^{1/2 + 1} = 2t^{3/2}. \]2. \(5t^{3/2}\) integrates to \[ \frac{5}{3/2 + 1}t^{3/2 + 1} = \frac{10}{5}t^{5/2} = 2t^{5/2}. \]

Step 8: Evaluate the Integrals

Combine the integrated terms:\[ \left[ 2t^{3/2} + 2t^{5/2} \right]_{4}^{9}. \]This results in evaluating:\[ \left(2(9^{3/2}) + 2(9^{5/2})\right) - \left(2(4^{3/2}) + 2(4^{5/2})\right). \]

Step 9: Calculate the Final Result

Perform the calculations:- \(9^{3/2} = 27, \, 9^{5/2} = 243\)- \(4^{3/2} = 8, \, 4^{5/2} = 32\)Thus, the integral evaluates to:\[ (54 + 486) - (16 + 64) = 540 - 80 = 460. \]

Key Concepts

Curve Parametrization

When dealing with line integrals, curve parametrization is a key step, as it allows for the representation of a curve using a parameter, usually denoted by \( t \). In this exercise, the curve \( C \) is given by equations that relate \( x \) and \( y \) to \( t \), specifically \( x = \sqrt{t} \) and \( y = t \), where \( t \) ranges from 4 to 9.
This means that instead of dealing with \( x \) and \( y \) as independent coordinates, they are expressed in terms of \( t \). This transformation simplifies the calculation of line integrals, as it reduces the problem to a single-variable calculus problem where we can integrate with respect to \( t \) over the interval from 4 to 9.
Parametrization is especially useful in scenarios where the curve is not easily describable using standard Cartesian coordinates, allowing for an elegant and simplified method to manage curves in the context of line integrals.

Integral Calculus

Integral calculus is the core mathematical framework used to evaluate line integrals. It revolves around the idea of integrating functions to find quantities like area, volume, or, in this case, movement along a curve.
The line integral presented in the exercise involves a vector field over a curve \( C \). It represents the sum of a function evaluated along this curve. In integral calculus, these types of integrals generalize the concept of an integral over a line or curve rather than a simple path or straight line.

• The starting point of evaluating a line integral is parametrizing the curve, as discussed previously.
• Next, each part of the line integral, usually involving \( dx \) and \( dy \), is transformed using the derivatives of the parametrized function. This turns the multivariable integral into a more manageable single-variable integral.
• The direct evaluation of these integrals often involves basic techniques and rules from single-variable calculus, such as finding antiderivatives and calculating definite integrals on the given interval.

By using these principles and techniques in integral calculus, the integration along a given curve is translated into operations that are more straightforward to solve.

Differential Equations

Differential equations appear in line integral problems when we express differentials of \( x \) and \( y \) with respect to another parameter, such as \( t \) in this context.
In our exercise, this involves calculating the derivatives of the parametrizations: \( \frac{dx}{dt} = \frac{1}{2\sqrt{t}} \) and \( \frac{dy}{dt} = 1 \). After finding these derivatives, the differentials used in the original line integral expression \( dx \) and \( dy \) are expressed as:

• \( dx = \frac{1}{2\sqrt{t}} dt \)
• \( dy = dt \)

This adjustment from differentials in terms of \( x \) and \( y \) to a parameterized form using \( t \) is crucial to solve line integrals. It shifts the problem from a vector calculus situation to one more aligned with techniques used in differential equations.
Differential equations help bridge the gap between parametric curves and their practical applications in evaluating the integral along these curves.

Definite Integral

The concept of the definite integral is fundamental in solving the line integral, representing the exact numerical value of an integral over a given interval.
In this exercise, after transforming the original line integral into a single-variable calculus problem, we end up with a definite integral with bounds from 4 to 9. At this point:

• Each term resulting from substituting the parameterized form and differentials is integrated over the interval \([4, 9]\).
  
• Individual components of the integrand such as \( 3t^{1/2} \) and \( 5t^{3/2} \) are integrated separately to find their antiderivatives.
• The definite limits of integration allow us to calculate the exact value by evaluating the antiderivatives at these bounds and finding the difference.

Through this process, the theoretical construct of a line integral is grounded in tangible numerical results. The definite integral gives the solution context, providing the concrete value, which in this problem was computed to be 460. This step significantly emphasizes both the power and importance of the definite integral in integral calculus, guiding us to find real-world quantities and measurements.