Problem 38
Question
Evaluate the flux integral \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d S\) \(\mathbf{F}=\langle y,-x, 1\rangle, S\) is the portion of \(z=x^{2}+y^{2}\) below \(z=4\) (n downward)
Step-by-Step Solution
Verified Answer
The flux through the surface S is \(-4π\).
1Step 1: Parameterization of the Surface S
The surface S, being a simple cone, can be parameterized in polar coordinates (r, θ). Here \(x=r cos(θ)\), \(y=r sin(θ)\), and \(z=r^{2}\) is the standard parameterization via polar coordinates of the surface defined by \( z=x^{2}+y^{2} \) below \( z=4 \). The polar coordinates r and θ will range over \(0≤r≤2\) and \(0≤θ≤2π\), respectively, since \(z=4\) corresponds to \( r=2 \) in the equation of the surface S.
2Step 2: Calculation of the Surface Integral
First, take the cross product of the partial derivatives of the parameterization to obtain the normal vector \( \mathbf{n}. \n \(\mathbf{n}= \frac{∂(x, y, z)}{∂r} × \frac{∂(x, y, z)}{∂θ}\) = \( \langle -r, -r ,1 \rangle \) (pointing downwards). Substitute \(x=r cos(θ)\) and \(y=r sin(θ)\) into \( \mathbf{F} \) to write the vector field \( \mathbf{F}=\langle y,-x, 1\rangle \) in terms of polar coordinates. The flux integral is then recognized to be the double integral: \( \iint_{S} \mathbf{F} \cdot \mathbf{n} d S = \int_{0}^{2\pi} \int_{0}^{2} (r sinθ,r cosθ,1) \cdot (-r,-r,1) rdrdθ\).
3Step 3: Evaluation of the Integral
Now, calculate the dot product and carry out the double integral: \( = \int_{0}^{2\pi} \int_{0}^{2} -2r^{2} +1 rdrdθ. \) Separating the integrals gives \( -2 \int_{0}^{2\pi} dθ \int_{0}^{2} r^{3} dr + \int_{0}^{2\pi} dθ \int_{0}^{2} rdr. \) Taking the integral results in \( =-2 \cdot 2π [ \frac{1}{4}r^{4} ]_{0}^{2} + π [ \frac{1}{2}r^{2} ]_{0}^{2} \) which simplifies to \( = -8π + 4π = -4π. \)
Key Concepts
Surface ParameterizationSurface Integral CalculationPolar Coordinates
Surface Parameterization
Understanding surface parameterization is essential when evaluating flux integrals, as it allows us to describe a surface in a way that we can perform calculus on it. In our example, the surface S is part of a paraboloid, and to parameterize it, we employ polar coordinates, where each point on the surface can be represented by two parameters, typically denoted as r (the radial distance from the origin) and θ (the angle in radians, which describes the position around the axis).
For the surface given by the equation \( z = x^2 + y^2 \) below \( z=4 \), the parameterization in polar coordinates is such that \( x=r\cos(\theta) \), \( y=r\sin(\theta) \), and \( z=r^2 \). This method simplifies the complexity of the surface to just two variables and allows us to work with a two-dimensional integral instead of a three-dimensional one.
Parameterization can be visualized as projecting the given three-dimensional surface onto a two-dimensional plane, where each point on the plane corresponds to a point on the surface. This technique is at the core of converting complex surface integrals into more manageable integrals that can be solved using familiar methods.
For the surface given by the equation \( z = x^2 + y^2 \) below \( z=4 \), the parameterization in polar coordinates is such that \( x=r\cos(\theta) \), \( y=r\sin(\theta) \), and \( z=r^2 \). This method simplifies the complexity of the surface to just two variables and allows us to work with a two-dimensional integral instead of a three-dimensional one.
Parameterization can be visualized as projecting the given three-dimensional surface onto a two-dimensional plane, where each point on the plane corresponds to a point on the surface. This technique is at the core of converting complex surface integrals into more manageable integrals that can be solved using familiar methods.
Surface Integral Calculation
Once a surface is parameterized, we can move on to the surface integral calculation. The aim of this step is to sum up the effects of a vector field across the entire surface. In the example with our vector field \( \mathbf{F} = \langle y, -x, 1 \rangle\) and the parameterized surface S, the calculation involves taking the dot product of \( \mathbf{F} \) and the normal vector \( \mathbf{n} \) at each point, integrating this dot product over the entire surface.
To find the normal vector, we take the cross product of the partial derivatives of the parameterized surface with respect to r and θ. The result, \( \mathbf{n} = \langle -r, -r, 1 \rangle \), is then used to compute the flux integral, which assesses how much of the vector field 'flows' through the surface.
In our exercise, the flux integral turns into the double integral \( \iint_{S} \mathbf{F} \cdot \mathbf{n} dS \) over the polar coordinates, combining the effects of parameterization and the vector field into a single value that captures the net flow across the given surface. This value can provide insights into various physical phenomena such as the flow of fluids or electromagnetic fields.
To find the normal vector, we take the cross product of the partial derivatives of the parameterized surface with respect to r and θ. The result, \( \mathbf{n} = \langle -r, -r, 1 \rangle \), is then used to compute the flux integral, which assesses how much of the vector field 'flows' through the surface.
In our exercise, the flux integral turns into the double integral \( \iint_{S} \mathbf{F} \cdot \mathbf{n} dS \) over the polar coordinates, combining the effects of parameterization and the vector field into a single value that captures the net flow across the given surface. This value can provide insights into various physical phenomena such as the flow of fluids or electromagnetic fields.
Polar Coordinates
The use of polar coordinates in our flux integral problem plays a crucial role in simplifying the mathematics required to find our solution. Polar coordinates are an alternative to the more common Cartesian coordinates, focusing on the distance from a central point and the angle from a reference direction, such as the positive x-axis.
In the context of surface integrals, polar coordinates are particularly useful when the geometry of the surface or the vector field has a natural circular symmetry, as is the case with the paraboloid defined by \( z = x^2 + y^2 \). When we express x and y in terms of polar coordinates \( x=r\cos(\theta) \) and \( y=r\sin(\theta) \), we can take advantage of this symmetry to perform the integration.
As polar coordinates represent points on a plane using radius and angle, they directly relate to the parameters used to describe the surface in our problem. It's crucial to remember that the radius and angle variables must cover the full area of the surface for the integral to represent the entire flux.
In the context of surface integrals, polar coordinates are particularly useful when the geometry of the surface or the vector field has a natural circular symmetry, as is the case with the paraboloid defined by \( z = x^2 + y^2 \). When we express x and y in terms of polar coordinates \( x=r\cos(\theta) \) and \( y=r\sin(\theta) \), we can take advantage of this symmetry to perform the integration.
As polar coordinates represent points on a plane using radius and angle, they directly relate to the parameters used to describe the surface in our problem. It's crucial to remember that the radius and angle variables must cover the full area of the surface for the integral to represent the entire flux.
Other exercises in this chapter
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