Problem 36 - Calculus Early Transcendentals

Question

Evaluate \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) along the curve \(C\) $$ \begin{array}{l}{\mathbf{F}(x, y, z)=z \mathbf{i}+x \mathbf{j}+y \mathbf{k}} \\\ {C: \mathbf{r}(t)=\sin t \mathbf{i}+3 \sin t \mathbf{j}+\sin ^{2} t \mathbf{k} \quad(0 \leq t \leq \pi / 2)}\end{array} $$

Step-by-Step Solution

Answer

The value of the integral is \( \frac{17}{6} \).

Step 1: Write down the line integral expression

The expression for the line integral of a vector field \( \mathbf{F} \) along a curve \( C \) is given by \( \int_{C} \mathbf{F} \cdot d\mathbf{r} \). Here, \( \mathbf{F}(x,y,z) = z \mathbf{i} + x \mathbf{j} + y \mathbf{k} \) and the curve \( C \) is parameterized by \( \mathbf{r}(t) = \sin t \mathbf{i} + 3\sin t \mathbf{j} + \sin^2 t \mathbf{k} \), with \( t \) going from 0 to \( \frac{\pi}{2} \).

Step 2: Compute \( \mathbf{F} \) along the curve

Substitute \( x = \sin t \), \( y = 3\sin t \), and \( z = \sin^2 t \) into the expression for \( \mathbf{F} \). This gives \( \mathbf{F}(t) = \sin^2 t \mathbf{i} + \sin t \mathbf{j} + 3\sin t \mathbf{k} \).

Step 3: Compute the derivative \( \frac{d\mathbf{r}}{dt} \)

Take the derivative of \( \mathbf{r}(t) \) with respect to \( t \). \( \frac{d\mathbf{r}}{dt} = \cos t \mathbf{i} + 3 \cos t \mathbf{j} + 2 \sin t \cos t \mathbf{k} \).

Step 4: Set up the dot product \( \mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} \)

Calculate the dot product: \((\sin^2 t \mathbf{i} + \sin t \mathbf{j} + 3\sin t \mathbf{k}) \cdot (\cos t \mathbf{i} + 3 \cos t \mathbf{j} + 2\sin t\cos t \mathbf{k})\), which simplifies to \( \sin^2 t \cos t + 3\sin t \cos t + 6 \sin^2 t \cos t \).

Step 5: Simplify the expression for integration

Combine terms: \( \sin^2 t \cos t + 6 \sin^2 t \cos t = 7\sin^2 t \cos t \) and \( 3\sin t \cos t \). Total expression for integration becomes \( 7\sin^2 t \cos t + 3\sin t \cos t \).

Step 6: Integrate with respect to \( t \) from 0 to \( \frac{\pi}{2} \)

The integral is \( \int_0^{\frac{\pi}{2}} (7 \sin^2 t \cos t + 3 \sin t \cos t) \, dt \). Break into two integrals: \( 7\int_0^{\frac{\pi}{2}} \sin^2 t \cos t \, dt + 3\int_0^{\frac{\pi}{2}} \sin t \cos t \, dt \).

Step 7: Solve the integrals

The first integral \( 7\int_0^{\frac{\pi}{2}} \sin^2 t \cos t \, dt \) can be done by substitution \( u = \sin t \), giving result \( \frac{7}{3} \). The second integral \( 3\int_0^{\frac{\pi}{2}} \sin t \cos t \, dt \) can be done by substitution \( w = \sin t \), giving result \( \frac{3}{2} \).

Step 8: Calculate the final result

Adding the two results, the total integral value is \( \frac{7}{3} + \frac{3}{2} = \frac{17}{6} \).

Key Concepts

Line Integral

A line integral extends the concept of an integral to integrating over a curve in a vector field. It is often used to find the work done by a force field when moving an object along a path. In the context of vector calculus, the line integral of a vector field \( \mathbf{F} \) along a curve \( C \) is given by the integral \( \int_{C} \mathbf{F} \cdot d \mathbf{r} \). This expression involves the dot product of the vector field \( \mathbf{F} \) and the differential element of the curve \( d \mathbf{r} \).

Why is line integral important?

• It helps determine the work done by a force along a path.
• Used in various fields including physics for electromagnetism and gravity fields.

In our example, the curve \( C \) is parameterized with a specific vector function, which makes calculating the line integral feasible through substitution and integral calculus.

Vector Field

A vector field assigns a vector to every point in space. Think of it as a force field where each point is associated with a vector indicating force's direction and magnitude. In this exercise, \( \mathbf{F}(x, y, z) = z \mathbf{i} + x \mathbf{j} + y \mathbf{k} \) represents the vector field.

Key aspects of vector fields:

• Provide a comprehensive snapshot of how forces or velocities vary across space.
• Common examples include gravitational fields, magnetic fields, and velocity fields in fluid dynamics.

In our example, this vector field is evaluated along the curve \( C \), which means substituting the parameterized expressions into the vector field's formula. This simplifies the problem to integrating a vector field along a curve.

Parameterization

Parameterization is the technique of representing a curve or surface in terms of a single parameter (or parameters), usually denoted by \( t \). This is essential in evaluating line integrals as it allows us to express the vector field and the curve in a manageable form using the parameter.

Significance of parameterization:

• Facilitates the conversion of complex geometries into simpler mathematical expressions.
• Enables the use of calculus for curves and surfaces that aren't easily expressed otherwise.

In our exercise, the curve \( C \) is parameterized as \( \mathbf{r}(t) = \sin t \mathbf{i} + 3 \sin t \mathbf{j} + \sin^2 t \mathbf{k} \). It describes the path over which the integral is calculated and provides the needed expressions for substitution into the vector field.

Dot Product

The dot product is a way to multiply two vectors to get a scalar (a single number) result. It is fundamental in determining the line integral of a vector field. For two vectors \( \mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k} \) and \( \mathbf{b} = b_1 \mathbf{i} + b_2 \mathbf{j} + b_3 \mathbf{k} \), the dot product \( \mathbf{a} \cdot \mathbf{b} \) is computed as \( a_1b_1 + a_2b_2 + a_3b_3 \).

Why is it used?

• To measure how much of one vector goes in the direction of another.
• Key in calculating work done as it measures the component of force in the direction of motion.

In our exercise, the line integral \( \int_{C} \mathbf{F} \cdot d \mathbf{r} \) involves calculating the dot product \( \mathbf{F}(t) \cdot \frac{d \mathbf{r}}{dt} \) at each point along the curve. This product is integrated over the interval of \( t \) to find the line integral's value.