Question

\(\oint \mathbf{V} \cdot d \mathbf{r}\) around the circle \((x-2)^{2}+(y-3)^{2}=9, z=0,\) where $$ \mathbf{V}=\left(x^{2}+y z^{2}\right) \mathbf{i}+\left(2 x-y^{3}\right) \mathbf{j} $$

Short answer

\oint \mathbf{V} \cdot d\mathbf{r} = 0

Step-by-step solution

Identify the Given Information

The vector field is \(\mathbf{V}=\left(x^{2}+y z^{2}\right) \mathbf{i}+\left(2 x-y^{3}\right) \mathbf{j}\). The path of integration is a circle given by \(\left( x-2 \right)^{2} + \left( y-3 \right)^{2} = 9\), in the plane \(z=0\).

Parameterize the Circle

The circle can be parameterized as \(x = 2 + 3 \cos t\) and \(y = 3 + 3 \sin t\) where \(t\) ranges from \(0\) to \(2\pi\). Since \(z=0\), the vector field simplifies to \(\mathbf{V}=x^{2} \mathbf{i}+\left(2 x-y^{3}\right) \mathbf{j}\).

Express the Differential in Terms of the Parameter

Calculate \(d \mathbf{r}\) as follows: \(d\mathbf{r} = \frac{d\mathbf{r}}{dt} dt = (\frac{dx}{dt} \mathbf{i} + \frac{dy}{dt} \mathbf{j}) dt\). With \(x=2+3\cos t\) and \(y=3+3\sin t\), we get \(\frac{dx}{dt} = -3\sin t\) and \(\frac{dy}{dt} = 3\cos t\). Thus, \(d \mathbf{r} = (-3 \sin t \mathbf{i} + 3 \cos t \mathbf{j}) dt\).

Substitute into the Integral

Substituting \(x = 2 + 3 \cos t\) and \(y = 3 + 3 \sin t\) into \(\mathbf{V}\), we get \(\mathbf{V} = (2+3\cos t)^2 \mathbf{i} + (2(2+3\cos t) - (3+3\sin t)^3) \mathbf{j} = (4 + 12 \cos t + 9 \cos^2 t) \mathbf{i} + (4 + 6 \cos t - 27\sin^3 t - 81 \sin^2 t - 108 \sin t - 27) \mathbf{j}\).

Compute the Dot Product with \(d \mathbf{r} \)

Find the dot product \(\mathbf{V} \cdot d \mathbf{r} \). Substitute the expressions for \(\mathbf{V} \) and \(d \mathbf{r} \): \(\mathbf{V} \cdot d \mathbf{r} = \left( 4 + 12 \cos t + 9 \cos^2 t\right) (-3 \sin t ) dt + \left(4 + 6 \cos t - 27 \sin^3 t - 81 \sin^2 t - 108 \sin t - 27 \right) (3 \cos t ) dt\).

Simplify the Integral

After computing the dot product, integrate from \(t=0\) to \(t=2\pi \). The integral becomes \(\oint \mathbf{V} \cdot d \mathbf{r} =\int_0^{2\pi} \left( -12 \sin t - 36 \sin t \cos t - 27 \sin^4 t - 81 \sin^3 t \cos t - 324 \sin^3 t - 108 \sin t \cos t - 81\sin^2 t \cos t \right) dt =0 \).

Final Step: Conclusion

Evaluate the integral. Noticing the symmetry and periodicity of sines and cosines over a full interval \(0 \leq t \leq 2\pi\), the terms will cancel out, concluding the closed line integral is zero: \(\oint \mathbf{V} \cdot d \mathbf{r} = 0\).

Key concepts

Line Integral

A line integral is essentially an extension of the regular integral to vector fields. Instead of summing the function values over an interval, we sum vector field components along a path. In simpler terms, you're integrating a vector function along a curve. This is useful in determining work done by a force field, fluid flow across a path, and other physical applications.
To set up a line integral, you need to:

• Identify the vector field \( \textbf{V} = P(x, y, z) \textbf{i} + Q(x, y, z) \textbf{j} + R(x, y, z) \textbf{k} \)
• Define the path or curve along which you're integrating
  
• Express the curve using a parameter \( t, s, \theta, \phi \)
• Calculate differentials in terms of the parameter
• Substitute and integrate by taking the dot product

Parameterization

Parameterization simplifies the process of computing line integrals. It involves expressing the coordinates of a curve as functions of a single variable, like \( t \). For example, for a circle defined by \( (x-2)^2 + (y-3)^2 = 9, \) we can parameterize the circle using trigonometric functions:
\[ x = 2 + 3 \cos\ t \text{ and } y = 3 + 3 \sin\ t \]
Here, \( t \) ranges from 0 to \( 2\pi \), covering the entire circle. This substitution allows us to simplify the expressions and proceed with easier differential calculations.

Vector Field

A vector field assigns a vector to every point in a space. In the given problem, the vector field is \( \textbf{V} = (x^2 + y z^2) \textbf{i} + (2 x - y^3) \textbf{j} \). This gives directions and magnitudes at various points.
When \( z = 0\), our vector field simplifies to:
\[ \textbf{V} = x^2 \textbf{i} + (2 x - y^3) \textbf{j} \]
Understanding the nature of the vector field helps in visualizing and solving the integral along any given path.

Dot Product

The dot product is a way to multiply two vectors and get a scalar (number) result. It's essential in line integrals as it combines the vector field with the differential vector along the path. The dot product formula for two vectors \( \textbf{A} = a_1 \textbf{i} + a_2 \textbf{j} \, and \ \textbf{B} = b_1 \textbf{i} + b_2 \textbf{j} \) is:
\[ \textbf{A} \cdot \textbf{B} = a_1 b_1 + a_2 b_2 \]
In the exercise, this step involves multiplying components of \( \textbf{V} \) and \( d \textbf{r}\), then simplifying and integrating along the parameter \( t\).

Symmetry

Symmetry plays a crucial role in simplifying integrals, especially in periodic functions like sine and cosine. In our integral evaluation, we took advantage of the symmetric interval from 0 to \( 2\pi \).

• Functions like \( \sin\) and \( \cos\) exhibit symmetry that cancels out over a full period
• This periodicity helps in concluding that many terms contribute zero to the integral due to their balanced positive and negative values
• Recognizing symmetry not only simplifies calculations but often quickly leads to results

In the exercise, these properties meant that the resulting line integral evaluated to zero.