Problem 83 - Calculus Volume 3

Question

For the following exercises, use a CAS to evaluate the given line integrals. \([\mathrm{T}] \quad\) Evaluate \(\quad \int_{C} \mathbf{F} \cdot d \mathbf{r},\) where \(\mathbf{F}(x, y, z)=x^{2} y \mathbf{i}+(x-z) \mathbf{j}+x y z \mathbf{k} \quad\) and \(\mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}+2 \mathbf{k}, 0 \leq t \leq 1\)

Step-by-Step Solution

Answer

The value of the integral is \( \frac{-11}{15} \).

Step 1: Understand the Components

First, we need to identify the components of the vector field \( \mathbf{F}(x, y, z) = x^2 y \mathbf{i} + (x-z) \mathbf{j} + xy z \mathbf{k} \) and the parameterization of the curve \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + 2 \mathbf{k} \). The parameter \( t \) moves from 0 to 1.

Step 2: Differentiate the Parameterized Curve

Find the derivative of \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + 2 \mathbf{k} \) with respect to \( t \). This gives us \( \mathbf{r}'(t) = \frac{d}{dt}(t) \mathbf{i} + \frac{d}{dt}(t^2) \mathbf{j} + \frac{d}{dt}(2) \mathbf{k} = \mathbf{i} + 2t \mathbf{j} + 0 \mathbf{k} = \mathbf{i} + 2t \mathbf{j} \).

Step 3: Substitute Curve into Vector Field

Substitute \( x = t \), \( y = t^2 \), and \( z = 2 \) into \( \mathbf{F} \) to get the vector field along the curve: \( \mathbf{F}(t) = (t^2 \cdot t^2) \mathbf{i} + (t - 2) \mathbf{j} + (t \cdot t^2 \cdot 2) \mathbf{k} = t^4 \mathbf{i} + (t - 2) \mathbf{j} + 2t^3 \mathbf{k} \).

Step 4: Compute the Dot Product

Compute the dot product of \( \mathbf{F}(t) \) and \( \mathbf{r}'(t) \): \[ \mathbf{F}(t) \cdot \mathbf{r}'(t) = (t^4 \mathbf{i} + (t - 2) \mathbf{j} + 2t^3 \mathbf{k}) \cdot (\mathbf{i} + 2t \mathbf{j}) \]This simplifies to \( t^4 \cdot 1 + (t-2) \cdot 2t + 2t^3 \cdot 0 = t^4 + 2t^2(t - 2) = t^4 + 2t^3 - 4t^2 \).

Step 5: Integrate the Expression

Evaluate the definite integral from 0 to 1:\(\int_{0}^{1} \left(t^4 + 2t^3 - 4t^2\right) dt\)This integral can be evaluated term by term. The antiderivatives are: - \( \int t^4 \, dt = \frac{1}{5}t^5 \) - \( \int 2t^3 \, dt = \frac{1}{2}t^4 \) - \( \int -4t^2 \, dt = -\frac{4}{3}t^3 \)evaluating from 0 to 1.

Step 6: Compute Definite Integral

Substitute the limits of the integral (0 and 1) into the antiderivative result:- \( \frac{1}{5}(1)^5 + \frac{1}{2}(1)^4 - \frac{4}{3}(1)^3 \)- Substitute \(0\) into each: \( \frac{1}{5}(0)^5 + \frac{1}{2}(0)^4 - \frac{4}{3}(0)^3 = 0\). Final result is:\[ \left(\frac{1}{5} + \frac{1}{2} - \frac{4}{3}\right) = \frac{3}{30} + \frac{15}{30} - \frac{40}{30} = \frac{-22}{30} = \frac{-11}{15}. \]

Key Concepts

Vector Field

Understanding vector fields is crucial when working with line integrals. A vector field assigns a vector to every point in a space, and in three dimensions, this is usually a function of three variables, say, \(x\), \(y\), and \(z\). For example, \(\mathbf{F}(x, y, z) = x^2 y \mathbf{i} + (x-z) \mathbf{j} + xy z \mathbf{k}\) is a vector field.

Each of these components has a specific role where \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\) are the unit vectors along the \(x\), \(y\), and \(z\) axes, respectively. The expressions like \(x^2 y\) define the vector's contribution in the direction of each axis at any point \((x, y, z)\).

• \(x^2 y\) - influences the \(x\)-axis direction.
• \(x-z\) - affects the \(y\)-axis.
• \(xy z\) - determines the \(z\)-axis direction.

The resulting vector is different at each point in space, characterizing the vector field.

Curve Parameterization

Parameterizing a curve is a process of expressing the curve via a parameter, usually denoted as \(t\). This helps in converting multi-dimensional problems into a form that can be graphed or integrated more easily.

In the problem, the curve is parameterized as \(\mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + 2 \mathbf{k}\), which represents a curve in three-dimensional space. As \(t\) ranges from 0 to 1, this parameterization describes how the curve is traversed.

• \(\mathbf{i}\) follows a linear path as it's just \(t\).
• \(\mathbf{j}\) follows a quadratic path due to \(t^2\).
• \(\mathbf{k}\) remains constant at 2.

This parameterization allows us to substitute into the vector field, evaluating it along the curve.

Dot Product

The dot product is a mathematical operation that takes two vectors and returns a scalar. In line integrals, we take the dot product of the vector field \(\mathbf{F}(t)\) and the derivative of the curve \(\mathbf{r}'(t)\).

With the exercise we simplified: \[ \mathbf{F}(t) \cdot \mathbf{r}'(t) = t^4 + 2t^3 - 4t^2.\]

The dot product helps combine the vector field values in a direction that's aligned with the movement along the parameterized curve. This makes the results from different directions into one single value that can be integrated.

• Unchecked factors like \(\mathbf{k}\) in \(\mathbf{r}'(t)\) cancels out any \(\mathbf{k}\) contribution.
• The resulting function \(t^4 + 2t^3 - 4t^2\) is now ready for integration.

Using dot products in this way is vital for evaluating line integrals.

Definite Integral

Definite integrals provide the accumulated effect along an interval. In the context of line integrals, they help compute the total value of the chosen path from \(t = 0\) to \(t = 1\).

The goal is to find the integral:\[ \int_{0}^{1} (t^4 + 2t^3 - 4t^2) \, dt. \]

This integral is evaluated by finding the antiderivatives separately. Each term is integrated over this interval to find the contribution:

• \( \int t^4 \, dt = \frac{1}{5}t^5 \)
• \( \int 2t^3 \, dt = \frac{1}{2}t^4 \)
• \( \int -4t^2 \, dt = -\frac{4}{3}t^3 \)

After calculation and substitution of the limits, the integral provides the result. This method helps us determine the cumulative effect along our specified path.