Problem 17 - Advanced Engineering Mathematics

Question

Evaluate \(\int_{C} 2 x^{3} y d x+(3 x+y) d y\), where \(C\) is given by \(x=y^{2}\) from \((1,-1)\) to \((1,1)\).

Step-by-Step Solution

Answer

The value of the integral is 3.

Step 1: Understand the Curve C

Since the curve \(C\) is given by \(x = y^2\), the parameter \(y\) will vary from \(-1\) to \(1\) as we move from the point \((1, -1)\) to \((1, 1)\). We'll express \(x\) in terms of \(y\) for the parameterization.

Step 2: Parameterize the Curve

The curve can be parameterized as \((x, y) = (y^2, y)\). Here, \(y\) varies from \(-1\) to \(1\). Thus, \(C\) is parameterized as \(x = y^2\) and \(y = y\).

Step 3: Find dx and dy in Terms of y

From the parameterization, calculate differentials: \(dx = 2y \, dy\) and \(dy = dy\). These will be substituted into the integral.

Step 4: Substitute and Reformulate the Integral

Substitute \(x = y^2\), \(dx = 2y \, dy\), and \(dy = dy\) into the integral \[\int_{-1}^{1} \left(2(y^2)^3 y (2y) + (3(y^2) + y)\right) \, dy\] which simplifies to \[\int_{-1}^{1} \left(4y^7 + 3y^2 + y\right) \, dy.\]

Step 5: Integrate with Respect to y

Evaluate the integral: \[\int_{-1}^{1} (4y^7 + 3y^2 + y) \, dy = \left[\frac{4y^8}{8} + \frac{3y^3}{3} + \frac{y^2}{2}\right]_{-1}^{1}\]Simplifying yields:\[\left[ \frac{y^8}{2} + y^3 + \frac{y^2}{2} \right]_{-1}^{1}.\]

Step 6: Evaluate the Definite Integral

Calculate the definite integral by substituting the bounds:\[\left( \frac{1^8}{2} + 1^3 + \frac{1^2}{2} \right) - \left( \frac{(-1)^8}{2} + (-1)^3 + \frac{(-1)^2}{2} \right)\]This results in\[\left( \frac{1}{2} + 1 + \frac{1}{2} \right) - \left( \frac{1}{2} - 1 + \frac{1}{2} \right)\]Which simplifies finally to "3".

Key Concepts

Parameterization of curves

Parameterization of curves is an essential concept in calculus, particularly when dealing with line integrals. It involves expressing a curve as a set of equations that represent the curve in terms of a single variable, often called the parameter. In the case of this problem, the curve \( C \) is described by the equation \( x = y^2 \). This implies that as \( y \) changes, \( x \) behaves in a predictable manner based on its relationship with \( y \).

To parameterize the given curve, we use \( y \) as the parameter. Therefore, as \( y \) ranges from -1 to 1, which are the limits of the integral, the corresponding \( x \) values follow the equation \( x = y^2 \).

• This approach allows us to express the curve as a vector function \((x, y) = (y^2, y)\).
• This vector function clearly shows the relationship between \( x \) and \( y \), paving the way to calculate the core components needed for the line integral.

Once expressed in parametric form, the problem is simplified into calculating the differentials \( dx \) and \( dy \) for integration.

Definite integrals

Definite integrals are used to calculate the accumulation of quantities over a particular interval, which is fundamental in evaluating the line integrals. In our task, we solve a definite integral from \( y = -1 \) to \( y = 1 \).

The substitution of the parameterized equations into the integral gives:

• \( \int_{-1}^{1} \left(4y^7 + 3y^2 + y\right) \, dy \).

This equation represents a new integral that depends solely on the parameter \( y \).

Now, for computing the definite integral, follow these steps:

• Integrate the expression \( 4y^7 + 3y^2 + y \).
• Apply the fundamental theorem of calculus by substituting the boundary values.

After performing these steps, the integral evaluates to 3, symbolizing the net accumulated change along the curve between the specified endpoints.

Differential forms

Differential forms play a crucial role in formulating and solving calculus problems, especially when dealing with line integrals. In this context, we focus on how they simplify integration along parameterized curves.

Initially, we express the line integral in terms of differentials \( dx \) and \( dy \). Because we have parameterized the curve \((x = y^2, y)\), we determine the differentials:

• \( dx = 2y \, dy \)
• \( dy = dy \)

These differentials are crucial as they adjust the integrals into a form compatible with our curve parameterization.

Replacing the original differential forms in the line integral, we translate it into an integral over \( y \):

• \( \int_{-1}^{1} (2(x^3)y(2y) + (3x + y)) \, dy \).

Substituting the parameter values, it eventually simplifies into an integral that is easier to evaluate.