Evaluate The Line Integral $\int_C F \cdot D\mathbf{r}$, Where $F(x, Y, Z) = -\sin(x) \mathbf{i} - 5\cos(y) \mathbf{j} - 3xz \mathbf{k}$ And $C$ Is Given By The Vector Function $\mathbf{r}(t) = T^5 \mathbf{i} - T^4

Introduction

In this article, we will evaluate the line integral of a vector field $F(x, y, z) = -\sin(x) \mathbf{i} - 5\cos(y) \mathbf{j} - 3xz \mathbf{k}$ along a curve $C$ given by the vector function $\mathbf{r}(t) = t^5 \mathbf{i} - t^4 \mathbf{j} + 2t^3 \mathbf{k}$. The line integral of a vector field is a fundamental concept in vector calculus, and it has numerous applications in physics, engineering, and other fields.

The Line Integral

The line integral of a vector field $F(x, y, z)$ along a curve $C$ is defined as:

$$\int_C F \cdot d\mathbf{r} = \int_a^b F(\mathbf{r}(t)) \cdot \mathbf{r}'(t) dt$$

where $a$ and $b$ are the limits of integration, and $\mathbf{r}(t)$ is the vector function that parameterizes the curve $C$.

The Vector Field

The vector field $F(x, y, z)$ is given by:

$$F(x, y, z) = -\sin(x) \mathbf{i} - 5\cos(y) \mathbf{j} - 3xz \mathbf{k}$$

This vector field has three components, each of which depends on the coordinates $x$, $y$, and $z$.

The Curve

The curve $C$ is given by the vector function:

$$\mathbf{r}(t) = t^5 \mathbf{i} - t^4 \mathbf{j} + 2t^3 \mathbf{k}$$

This vector function parameterizes the curve $C$ in three-dimensional space.

Evaluating the Line Integral

To evaluate the line integral, we need to find the derivative of the vector function $\mathbf{r}(t)$:

$$\mathbf{r}'(t) = 5t^4 \mathbf{i} - 4t^3 \mathbf{j} + 6t^2 \mathbf{k}$$

Now, we can substitute the vector field $F(x, y, z)$ and the derivative of the vector function $\mathbf{r}'(t)$ into the line integral formula:

$$\int_C F \cdot d\mathbf{r} = \int_a^b F(\mathbf{r}(t)) \cdot \mathbf{r}'(t) dt$$

$$= \int_a^b (-\sin(t^5) \mathbf{i} - 5\cos(t^4) \mathbf{j} - 3t^5 \cdot 2t^3 \mathbf{k}) \cdot (5t^4 \mathbf{i} - 4t^3 \mathbf{j} + 6t^2 \mathbf{k}) dt$$

$$= \int_a^b (-5t^4 \sin(t^5) - 20t^3 \cos(t^4) - 18t^8) dt$$

Solving the Integral

To solve the integral, we can use the following substitutions:

$$u = t^5$$

$$dv = \sin(u) du$$

$$du = 5t^4 dt$$

$$v = -\cos(u)$$

Using these substitutions, we can rewrite the integral as:

$$\int_a^b (-5t^4 \sin(t^5) - 20t^3 \cos(t^4) - 18t^8) dt$$

$$= \int_{t^5=a}^{t^5=b} (-\cos(u) - 4t^3 \cos(t^4) - 18t^8) du$$

$$= \left[ \sin(u) + 4 \int t^3 \cos(t^4) dt - 18 \int t^8 dt \right]_{t^5=a}^{t^5=b}$$

Evaluating the Integrals

To evaluate the integrals, we can use the following formulas:

$$\int \cos(u) du = \sin(u) + C$$

$$\int t^3 \cos(t^4) dt = \frac{1}{4} t^3 \sin(t^4) + \frac{1}{16} t^4 \cos(t^4) + C$$

$$\int t^8 dt = \frac{1}{9} t^9 + C$$

Using these formulas, we can rewrite the integral as:

$$= \left[ \sin(u) + 4 \left( \frac{1}{4} t^3 \sin(t^4) + \frac{1}{16} t^4 \cos(t^4) \right) - 18 \left( \frac{1}{9} t^9 \right) \right]_{t^5=a}^{t^5=b}$$

Simplifying the Expression

To simplify the expression, we can substitute back $u = t^5$:

$$= \left[ \sin(u) + t^3 \sin(t^4) + \frac{1}{4} t^4 \cos(t^4) - 2t^9 \right]_{t^5=a}^{t^5=b}$$

Evaluating the Limits

To evaluate the limits, we can substitute $a$ and $b$ into the expression:

$$= \left[ \sin(b) + b^3 \sin(b^4) + \frac{1}{4} b^4 \cos(b^4) - 2b^9 \right] - \left[ \sin(a) + a^3 \sin(a^4) + \frac{1}{4} a^4 \cos(a^4) - 2a^9 \right]$$

The Final Answer

The final answer is:

$$\left[ \sin(b) + b^3 \sin(b^4) + \frac{1}{4} b^4 \cos(b^4) - 2b^9 \right] - \left[ \sin(a) + a^3 \sin(a^4) + \frac{1}{4} a^4 \cos(a^4) - 2a^9 \right]$$

This is the final answer to the line integral problem.

Introduction

In our previous article, we evaluated the line integral of a vector field $F(x, y, z) = -\sin(x) \mathbf{i} - 5\cos(y) \mathbf{j} - 3xz \mathbf{k}$ along a curve $C$ given by the vector function $\mathbf{r}(t) = t^5 \mathbf{i} - t^4 \mathbf{j} + 2t^3 \mathbf{k}$. In this article, we will answer some common questions related to the line integral problem.

Q: What is the line integral of a vector field?

A: The line integral of a vector field $F(x, y, z)$ along a curve $C$ is a measure of the work done by the vector field on an object moving along the curve. It is defined as:

$$\int_C F \cdot d\mathbf{r} = \int_a^b F(\mathbf{r}(t)) \cdot \mathbf{r}'(t) dt$$

Q: How do I evaluate the line integral?

A: To evaluate the line integral, you need to follow these steps:

1. Find the derivative of the vector function $\mathbf{r}(t)$.
2. Substitute the vector field $F(x, y, z)$ and the derivative of the vector function $\mathbf{r}'(t)$ into the line integral formula.
3. Evaluate the integral using the appropriate techniques, such as substitution or integration by parts.

Q: What are some common techniques for evaluating line integrals?

A: Some common techniques for evaluating line integrals include:

• Substitution: This involves substituting a new variable into the integral to simplify it.
• Integration by parts: This involves breaking down the integral into smaller parts and integrating each part separately.
• Parametric differentiation: This involves differentiating the vector function $\mathbf{r}(t)$ to find the derivative of the curve.

Q: How do I choose the limits of integration?

A: The limits of integration are typically chosen based on the problem statement. For example, if the curve $C$ is parameterized by the vector function $\mathbf{r}(t)$, then the limits of integration are typically $a$ and $b$, where $a$ and $b$ are the values of $t$ that correspond to the endpoints of the curve.

Q: What are some common applications of line integrals?

A: Line integrals have numerous applications in physics, engineering, and other fields. Some common applications include:

• Calculating the work done by a force on an object moving along a curve.
• Finding the potential energy of an object at a given point.
• Determining the electric field or magnetic field at a given point.

Q: How do I check my work when evaluating a line integral?

A: To check your work, you can follow these steps:

1. Verify that the vector field $F(x, y, z)$ and the derivative of the vector function $\mathbf{r}'(t)$ are correctly substituted into the line integral formula.
2. Check that the integral is evaluated correctly using the appropriate techniques.
3. Verify that the final answer is reasonable and makes sense in the context of the problem.

Conclusion

In this article, we have answered some common questions related to the line integral problem. We have discussed the definition of the line integral, how to evaluate it, and some common techniques for doing so. We have also discussed some common applications of line integrals and how to check your work when evaluating a line integral.