Problem 18 - Advanced Engineering Mathematics

Question

Evaluate \(\int_{C} 4 x d x+2 y d y\), where \(C\) is given by \(x=y^{3}+1\) from \((0,-1)\) to \((9,2)\).

Step-by-Step Solution

Answer

The value of the integral is 165.

Step 1: Parameterize the Curve

The curve is given by the equation \( x = y^3 + 1 \). We'll use \( y \) as the parameter.When \( y = -1 \), \( x = (-1)^3 + 1 = 0 \), which corresponds to the starting point \((0, -1)\).When \( y = 2 \), \( x = 2^3 + 1 = 9 \), which corresponds to the endpoint \((9, 2)\).Thus, the parameterization of the curve \( C \) is \( x(y) = y^3 + 1 \) for \( y \) ranging from \(-1\) to \(2\).

Step 2: Express the Integral with Parameterization

Substitute the parameterization into the line integral.The differential \( dx = \frac{dx}{dy} \, dy = 3y^2 \, dy \).Therefore, the given line integral becomes:\[ \int_{C} (4x \, dx + 2y \, dy) = \int_{-1}^{2} \left(4(y^3 + 1) \, 3y^2 + 2y\right) \, dy. \]

Step 3: Simplify the Expression for Integration

Simplify the expression inside the integral:\[ 4(y^3 + 1) \, 3y^2 + 2y = 12y^5 + 12y^2 + 2y. \]So, the integral becomes:\[ \int_{-1}^{2} (12y^5 + 12y^2 + 2y) \, dy. \]

Step 4: Integrate each term

Integrate each term separately:\[ \int 12y^5 \, dy = 2y^6, \]\[ \int 12y^2 \, dy = 4y^3, \]\[ \int 2y \, dy = y^2. \]Combine these results to obtain:\[ 2y^6 + 4y^3 + y^2. \]

Step 5: Evaluate the Definite Integral

Evaluate the antiderivative from \(-1\) to \(2\):\[ \left[ 2y^6 + 4y^3 + y^2 \right]_{-1}^{2} = \left( 2(2)^6 + 4(2)^3 + (2)^2 \right) - \left( 2(-1)^6 + 4(-1)^3 + (-1)^2 \right). \]Simplify each term:For \( y = 2 \):\[ = 2(64) + 4(8) + 4 = 128 + 32 + 4 = 164. \]For \( y = -1 \):\[ = 2(1) - 4 + 1 = -1. \]

Step 6: Compute the Final Result

Subtract the evaluation at \(-1\) from that at \(2\):\[ 164 - (-1) = 165. \]

Key Concepts

Parameterization of Curves

When dealing with line integrals, the parameterization of curves is a crucial step to simplify the problem. Imagine that instead of working with a complex path, you can describe it with a single parameter. In this exercise, the curve is defined by the equation \( x = y^3 + 1 \). This means that we use \( y \) as our parameter.

• For \( y = -1 \), the corresponding \( x \) is \((0, -1)\), providing our starting point.
• When \( y = 2 \), the endpoint is \((9, 2)\). This range from \(-1\) to \(2\) for \( y \) describes the full path \( C \).

The parameterization simplifies our calculations by converting the path into a more manageable form, allowing us to rewrite other variables such as \( dx \) in terms of \( dy \). This conversion is necessary in this step for moving forward with the integration process. Once parameterized, it becomes easier to setup the integral with limits that match precisely with the original curve's description.

Definite Integration

Definite integration is the process of finding the integral of a function within specific boundaries. In our exercise, this corresponds to computing the integral from \( y = -1 \) to \( y = 2 \). This step is where mathematics confirms the exact amount of whatever it is you're measuring over that specific range.
To carry out a definite integral, you first need the antiderivative of the function involved. The process involves calculating the integral function over a range of values, then evaluating the antiderivative at the upper and lower bounds. Here, the evaluated expression is:\[\left[ 2y^6 + 4y^3 + y^2 \right]_{-1}^{2},\]which is evaluated from \( y = -1 \) to \( y = 2 \).
The definite integration leads us to determine the exact value of the integral over this range. The subtraction of the values resulting from substituting the upper and lower bounds into the antiderivative gives the final, accurate answer.

Polynomial Integration

Polynomial integration involves finding the integral of polynomial terms, which is often a straightforward step in calculus. Each term in a polynomial, such as \( y^5 \), \( y^2 \), or \( y \), can be integrated individually by increasing the power by one and then dividing by this new exponent.
For example, in our solution:

• \( \int 12y^5 \, dy = 2y^6 \) involves increasing the exponent by one to 6 and dividing by 6.
• Similarly, \( \int 12y^2 \, dy = 4y^3 \) results from adding one to the power and dividing by the new power 3.
• Finally, \( \int 2y \, dy = y^2 \) is computed using the same steps.

By integrating each term separately, then combining them, you derive the antiderivative. Polynomial integration is often one of the simpler tasks in calculus because the rules are intuitive and repetitive. Once each term is integrated, you can easily piece them back together to form the complete solution.