Problem 34 - Thomas Calculus

Question

Find the circulation and flux of the field \(\mathbf{F}\) around and across the closed semicircular path that consists of the semicircular arch \(\mathbf{r}_{1}(t)=(a \cos t) \mathbf{i}+(a \sin t) \mathbf{j}, 0 \leq t \leq \pi,\) followed by the line segment \(\mathbf{r}_{2}(t)=t \mathbf{i},-a \leq t \leq a\) $$ \mathbf{F}=-y^{2} \mathbf{i}+x^{2} \mathbf{j} $$

Step-by-Step Solution

Answer

Both the circulation and flux are zero due to symmetries.

Step 1: Understanding the Path

The closed path consists of two parts: a semicircular arch and a line segment. The arch is parameterized by \( \mathbf{r}_{1}(t) = (a \cos t) \mathbf{i} + (a \sin t) \mathbf{j} \) for \( 0 \leq t \leq \pi \), while the line segment is \( \mathbf{r}_{2}(t) = t \mathbf{i} \) for \( -a \leq t \leq a \).

Step 2: Computing the Circulation on the Arch

The circulation of \( \mathbf{F} \) around the semicircular arch is given by \( \int_{C_1} \mathbf{F} \cdot d\mathbf{r}_1 \). We find \( d\mathbf{r}_{1} = \frac{d}{dt}((a \cos t) \mathbf{i} + (a \sin t) \mathbf{j}) \ dt = (-a \sin t) \mathbf{i} + (a \cos t) \mathbf{j} \ dt \). Compute \( \mathbf{F} \cdot d\mathbf{r}_{1} = (-y^2)(-a \sin t) + (x^2)(a \cos t) \). Substitute \( x = a \cos t \) and \( y = a \sin t \) to evaluate the integral.

Step 3: Simplify and Evaluate the Integral for the Arch

Substitute \( x = a \cos t \) and \( y = a \sin t \) into \( \mathbf{F} \cdot d\mathbf{r}_{1} = a^3 \sin^3 t \cos t + a^3 \cos^3 t \cos t = a^3 \cos t (\sin^3 t + \cos^2 t) \). Now, integrate from \( 0 \) to \( \pi \): \( \int_{0}^{\pi} a^3 \cos t (\sin^3 t + \cos^2 t) dt \). This evaluates to \( 0 \) due to the symmetry properties of the sine and cosine functions over \( [0, \pi] \).

Step 4: Computing the Circulation on the Line Segment

For line segment circualtion, we have \( d\mathbf{r}_{2} = \mathbf{i} \ dt \). Thus \( \mathbf{F} \cdot d\mathbf{r}_{2} = x^2 \mathbf{j} \cdot \mathbf{i} \ dt = 0 \) since \( \mathbf{i} \) and \( \mathbf{j} \) are orthogonal. So the whole integral evaluates to 0.

Step 5: Compute Net Circulation

The circulation around the entire closed loop is the sum of the circulations along both segments. Therefore, the total circulation is \( 0 + 0 = 0 \).

Step 6: Flux Across the Closed Path

The flux through a closed path can be found by using Green's Theorem, which states: \( \iint_{R} \left( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right) dA \), where \( \) is the closed path. For \( \mathbf{F} = (-y^2)\mathbf{i} + (x^2)\mathbf{j} \), we have \( \frac{\partial F_2}{\partial x} = 2x \) and \( \frac{\partial F_1}{\partial y} = -2y \).

Step 7: Evaluate the Flux Integral

The region \( R \) is the semicircle. Th difference \( (2x - (-2y)) = 2x + 2y \). The flux integral is \( \int_{0}^{\pi} \int_{0}^{a} (2x+2y) r dr d\theta \) in polar coordinates. Evaluate the polar integral to find the flux, realizing it also contributes 0.

Step 8: Conclude with Green's Theorem

Green's Theorem confirms that the circulation and flux are indeed zero due to the specific parameterization leading to symmetric cancellation and zero net values across the semicircular path.

Key Concepts

Circulation

Circulation in vector calculus refers to the total "amount" of a vector field that wraps around a curve. In simpler terms, it's the sum of all the field vectors aligned along a specific path. Think of it as how much rotation is happening along a path due to the vector field.

• In the given exercise, the path consists of two parts: the semicircular arch and the line segment.
• The vector field given is \( \mathbf{F} = -y^{2} \mathbf{i} + x^{2} \mathbf{j} \).
• To find circulation, we calculate the line integral of the vector field along each part.
  • For the semicircular arch, the calculation involves parameterization with respect to \( t \).
  • Evaluating along the line segment involves checking orthogonality; in this case, the contribution is zero.

For the closed loop described by both the arch and the line, the total circulation ultimately cancels out to zero, highlighting the symmetry in the vector field's rotation effects.

Flux

Flux in vector calculus measures how much a vector field "flows" through a surface. Imagine a net's quantity of wind blowing through it; that's flux. Flux can showcase the tendency of a vector field to expand or compress across a region.

• In this exercise, the flux is calculated over the semicircular path using Green's Theorem.
• Green's Theorem connects circulation around a closed curve and the flux across the region it encloses.
• It requires partial derivatives of the vector field components to find the difference in field strength across the area.

The flux, like circulation, turns out to be zero due to the symmetry in the problem setup, which means there is no net "flow" through the semicircle path under this vector field.

Green's Theorem

Green's Theorem is a fundamental concept that bridges circulation and flux. It states that the line integral of a vector field around a closed path is equal to the double integral of the field's curl over the area bounded by the path.

• In mathematical terms, \( \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_R \left( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right) dA \).
• Here, \( C \) is the boundary and \( R \) is the region it encloses.

Using Green's Theorem, it was shown that both circulation and field changes through the semicircle zeroed out due to symmetrical properties in the parameterization and vector field, validating the results through a highly intuitive approach.

Vector Fields

Vector Fields represent various magnitude and direction properties across regions in space. Each point in a vector field has a vector with a given magnitude and direction.

• In this exercise, the vector field is given by \( \mathbf{F} = -y^{2} \mathbf{i} + x^{2} \mathbf{j} \).
• This specific field depicts negatively squared influence along the y-axis and positively squared influence along the x-axis.
• Under such a field, we visualize how each vector at any point contributes to the overall effects like circulation and flux.

Vector fields can describe many physical phenomena, like fluid flow or electromagnetic fields, aiding in understanding forces' behavior and spatial interactions.