Problem 24 - Thomas Calculus

Question

Evaluate $$\int_{C} x^{2} d x+y z d y+\left(y^{2} / 2\right) d z$$ along the line segment \(C\) joining \((0,0,0)\) to \((0,3,4)\).

Step-by-Step Solution

Answer

The integral evaluates to 18.

Step 1: Parameterize the Curve

The line segment from \((0,0,0)\) to \((0,3,4)\) can be parameterized as follows: \(x = 0\), \(y = 3t\), \(z = 4t\). Let \(t\) range from 0 to 1 to trace the entire line segment.

Step 2: Compute the Differentials

Using the parameterization from Step 1, compute the differentials: \(dx = 0\), \(dy = 3 \, dt\), and \(dz = 4 \, dt\).

Step 3: Substitute into the Integral

Substitute the parameterization and differentials into the integral:\[\int_{0}^{1} \left(0\right) d x + y z d y + \frac{y^{2}}{2} d z = \int_{0}^{1} \left((3t)(4t)(3) \, dt + \frac{(3t)^{2}}{2}(4) \, dt \right)\] Simplifies to: \[\int_{0}^{1} \left(36t^2 + 18t^2\right) dt\] .

Step 4: Simplify the Integral

Simplify the expression inside the integral:\[\int_{0}^{1} (36t^2 + 18t^2) \, dt = \int_{0}^{1} 54t^2 \, dt\].

Step 5: Evaluate the Integral

Evaluate the integral: Calculate \[\int_{0}^{1} 54t^2 \, dt = 54 \int_{0}^{1} t^2 \, dt\].The antiderivative of \(t^2\) is \(\frac{t^3}{3}\), thus:\[54 \cdot \frac{t^3}{3}\Bigg|_0^1 = 54 \times \frac{1^3}{3} - 54 \times \frac{0^3}{3} = 54 \times \frac{1}{3} = 18.\]

Key Concepts

Vector Calculus

Vector calculus is a powerful branch of mathematics that extends single-variable calculus to higher dimensions.
It deals with vector fields and integrates functions over curves, surfaces, and volumes.
In the context of line integrals, vector calculus allows the calculation of integrals over a curve in a vector field.
Line integrals are useful for computing quantities like work done by a force field or the flow of a fluid along a path.
For any given vector field, a line integral can accumulate the values of field along a certain path. This integration process involves breaking a vector field down into smaller parts and calculating how these infinitesimally small parts contribute to the entire path.
Understanding vector calculus is essential for working with line integrals, as it requires a grasp of both vector operations and integration techniques.

Parametrization

Parametrization is the process of expressing the coordinates of the points that make up a geometric shape, or in this case, a curve, in terms of a single parameter.
Parametrization is particularly helpful when dealing with line integrals, as it converts what's effectively a multi-dimensional path into a single-dimensional curve.

• In the exercise, the line segment from r(0,0,0) to (0,3,4) is parametrized by setting x = 0, y = 3t and z = 4t.
• This parametrization captures the entire path as the parameter t varies from 0 to 1.

With parametrization, complex geometrical paths can be expressed in terms of simpler variables, making integration more straightforward.

Antiderivative

An antiderivative is a function whose derivative is the given function.
In calculus, finding an antiderivative is equivalent to performing the indefinite integral of a function.
Antiderivatives are crucial in evaluating line integrals, as they provide a way to calculate the integral over a path.

• When solving the problem, the antiderivative of the function \(t^2\) is \(\frac{t^3}{3}\).
• This result is then used to evaluate the definite integral from 0 to 1 for the function 54\(t^2\).

Understanding how to find an antiderivative allows you to move from differential forms to actual value computations in calculus problems.

Differential Forms

Differential forms are mathematical objects in calculus used to generalize the concepts of functions and differentials.
They offer a way to integrate over paths, surfaces, and more in higher-dimensional spaces.
For line integrals, differential forms are written in a manner that breaks down an integral across infinitesimal components of the path.

• In the exercise, the form \(x^2 \, dx + yz \, dy + \frac{y^2}{2} \, dz\) represents the differentials along the path.
• By substituting the parametrized expressions for \(dx\), \(dy\), and \(dz\), these differentials adapt to the specific path being considered from (0,0,0) to (0,3,4).

Differential forms are valuable as they allow the decomposition of integrals into sums of simpler integrals, enhancing the understanding and solving of calculus problems in multiple dimensions.