Problem 24
Question
Determine whether the given vector field is conservative and/or incompressible. $$\left(\sin x, 2 y^{2}, \sqrt{2}\right)$$
Step-by-Step Solution
Verified Answer
The vector field given is neither conservative nor incompressible.
1Step 1: Expression of the Vector Field
First, express the vector field in a standard format. In 3 dimensions, the standard format often used is \(F = P(x,y,z) \hat{i} + Q(x,y,z) \hat{j} + R(x,y,z) \hat{k}\). For the task at hand, the given vector field is \(F = \sin x \hat{i} + 2y^2 \hat{j} + \sqrt{2} \hat{k}\)
2Step 2: Calculate the Curl of the Vector Field
Next, calculate the curl of the vector field to test for conservativeness. The curl of a vector field in three dimensions is given by \[\nabla \times F = \left[(R_y - Q_z) \hat{i} + (P_z - R_x) \hat{j} + (Q_x - P_y) \hat{k}\right]\] Applying to the given vector field: \[\nabla \times F = \left[(0 - 0) \hat{i} + (0 - \cos x) \hat{j} + (0 - 0) \hat{k}\] = -\cos x \hat{j}\] Since \(\nabla \times F\) is not zero, the vector field is not conservative.
3Step 3: Calculate the Divergence of the Vector Field
Lastly, calculate the divergence of the vector field to test for incompressibility. The divergence of a vector field in three dimensions is given by \(\nabla \cdot F = P_x + Q_y + R_z\). Applying to the given vector field: \[\nabla \cdot F = \cos x + 4y + 0 = \cos x + 4y\] Since \(\nabla \cdot F\) is not zero, the vector field is not incompressible.
Key Concepts
Conservative Vector FieldIncompressible Vector FieldCurl of a Vector FieldDivergence of a Vector Field
Conservative Vector Field
A vector field is called conservative if it can be expressed as the gradient of some scalar potential function. This is equivalent to saying that the vector field has a curl of zero. In mathematical terms, if we have a vector field \( \mathbf{F} = P(x,y,z) \hat{i} + Q(x,y,z) \hat{j} + R(x,y,z) \hat{k} \), it is conservative if \( abla \times \mathbf{F} = \mathbf{0} \).
Conservative vector fields have an important property: the line integral of the vector field over a closed path is zero. This means that the work done by the field along a closed loop is zero, making it path-independent. In simpler terms, it doesn't matter how you move through the field; the total effect is always the same.
However, in the original exercise, the vector field \( \left( \sin x, 2y^2, \sqrt{2} \right) \) was determined not to be conservative. This was shown by calculating the curl, \( abla \times \mathbf{F} = - \cos x \hat{j} \), which is not equal to zero. This non-zero curl means that the field is not the gradient of any scalar potential function.
Conservative vector fields have an important property: the line integral of the vector field over a closed path is zero. This means that the work done by the field along a closed loop is zero, making it path-independent. In simpler terms, it doesn't matter how you move through the field; the total effect is always the same.
However, in the original exercise, the vector field \( \left( \sin x, 2y^2, \sqrt{2} \right) \) was determined not to be conservative. This was shown by calculating the curl, \( abla \times \mathbf{F} = - \cos x \hat{j} \), which is not equal to zero. This non-zero curl means that the field is not the gradient of any scalar potential function.
Incompressible Vector Field
An incompressible vector field is one that has zero divergence. Divergence is a measure of how much the vector field "spreads out" from a given point. For a vector field \( \mathbf{F} \), the divergence is expressed as \( abla \cdot \mathbf{F} = P_x + Q_y + R_z \). If this expression equals zero throughout the entire field, the vector field is considered incompressible.
In physical terms, incompressibility often applies to fluid flows, where a fluid's density remains constant over time. An incompressible fluid doesn't change its volume as it flows, and this aspect is mathematically captured by having a divergence of zero.
In the given exercise, the divergence of the vector field \( \left( \sin x, 2y^2, \sqrt{2} \right) \) was calculated as \( \cos x + 4y \), which is not zero. Hence, this vector field is compressible, meaning it can have sources or sinks of the field lines.
In physical terms, incompressibility often applies to fluid flows, where a fluid's density remains constant over time. An incompressible fluid doesn't change its volume as it flows, and this aspect is mathematically captured by having a divergence of zero.
In the given exercise, the divergence of the vector field \( \left( \sin x, 2y^2, \sqrt{2} \right) \) was calculated as \( \cos x + 4y \), which is not zero. Hence, this vector field is compressible, meaning it can have sources or sinks of the field lines.
Curl of a Vector Field
The curl of a vector field measures the rotation, or "twisting," of the field in a given region. For a vector field \( \mathbf{F} = P \hat{i} + Q \hat{j} + R \hat{k} \), the curl is defined as \( abla \times \mathbf{F} = \left( R_y - Q_z \right) \hat{i} + \left( P_z - R_x \right) \hat{j} + \left( Q_x - P_y \right) \hat{k} \).
Understanding the curl can be vital in physics for studying phenomena like electromagnetic fields and fluid vortices. High curl implies a strong rotational effect at that point in the vector field.
In this example, the vector field \( \left( \sin x, 2y^2, \sqrt{2} \right) \) had a computed curl of \( -\cos x \hat{j} \), which shows a rotational component in the negative y-direction. Since the curl is not zero, it confirmed the vector field is not conservative.
Understanding the curl can be vital in physics for studying phenomena like electromagnetic fields and fluid vortices. High curl implies a strong rotational effect at that point in the vector field.
In this example, the vector field \( \left( \sin x, 2y^2, \sqrt{2} \right) \) had a computed curl of \( -\cos x \hat{j} \), which shows a rotational component in the negative y-direction. Since the curl is not zero, it confirmed the vector field is not conservative.
Divergence of a Vector Field
The divergence of a vector field is a scalar measure that assesses how much a vector field is "flowing out" from a given point. For a vector field \( \mathbf{F} = P \hat{i} + Q \hat{j} + R \hat{k} \), it is calculated by \( abla \cdot \mathbf{F} = P_x + Q_y + R_z \).
Divergence is a critical concept in various fields of physics and engineering. For instance, in fluid dynamics, positive divergence at a point suggests a source, while negative divergence indicates a sink.
For the vector field \( \left( \sin x, 2y^2, \sqrt{2} \right) \) in the exercise, calculating the divergence \( \cos x + 4y \) yielded a non-zero result, which means the field is not incompressible. This allows for changes in field line density across the vector field, indicating areas where field lines may originate or converge.
Divergence is a critical concept in various fields of physics and engineering. For instance, in fluid dynamics, positive divergence at a point suggests a source, while negative divergence indicates a sink.
For the vector field \( \left( \sin x, 2y^2, \sqrt{2} \right) \) in the exercise, calculating the divergence \( \cos x + 4y \) yielded a non-zero result, which means the field is not incompressible. This allows for changes in field line density across the vector field, indicating areas where field lines may originate or converge.
Other exercises in this chapter
Problem 23
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Use Stokes' Theorem to evaluate \(\int c \mathbf{F} \cdot d \mathbf{r}\). \(C\) is the intersection of \(z=x^{2}+y^{2}-4\) and \(z=y-1\) oriented clockwise as v
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Find the flux of \(\mathbf{F}\) over \(\partial Q\). \(Q\) is bounded by \(y=4-x^{2}-z^{2}\) and the \(x z\) -plane, \(\mathbf{F}=\left\langle z^{2} x, x^{2} y,
View solution Problem 24
If \(h\) has continuous partial derivatives and \(S\) is a closed surface enclosing a solid \(Q,\) show that \(\iint_{S}(h \nabla h) \cdot \mathbf{n} d S=\iiint
View solution