Problem 26 - Thomas Calculus

Question

In Exercises \(19-28,\) use a parametrization to find the flux \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d \sigma\) across the surface in the specified direction. CONE \(\mathbf{F}=y^{2} \mathbf{i}+x z \mathbf{j}-\mathbf{k}\) outward (normal away from the \(z\) -axis) through the cone \(z=2 \sqrt{x^{2}+y^{2}}, 0 \leq z \leq 2\)

Step-by-Step Solution

Answer

The flux through the cone is evaluated by parametrizing the cone's surface, calculating the normal vector, and computing the surface integral of \( \mathbf{F} \cdot \mathbf{n} \).

Step 1: Parametrize the Cone's Surface

To parametrize the surface of the cone given by the equation \(z = 2 \sqrt{x^2 + y^2}\), we use the cylindrical coordinates. Let \(x = u \cos v\), \(y = u \sin v\), and \(z = 2u\), where \(0 \leq u \leq 1\) and \(0 \leq v < 2\pi\). This parametrization describes the surface of the cone within the specified limits.

Step 2: Calculate the Normal Vector

The normal vector \(\mathbf{n}\) to the surface can be found using the cross product of the partial derivatives of the parametrization with respect to \(u\) and \(v\). Compute the vectors: \(\frac{\partial \mathbf{r}}{\partial u}\) and \(\frac{\partial \mathbf{r}}{\partial v}\). Then, \(\mathbf{n} = \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}\). Make sure the normal vector points outward (away from the \(z\)-axis).

Step 3: Evaluate \( \mathbf{F} \cdot \mathbf{n} \)

Substitute the parametrized values \(x = u \cos v\), \(y = u \sin v\), \(z = 2u\) into \(\mathbf{F} = y^2 \mathbf{i} + xz \mathbf{j} - \mathbf{k}\) to express \(\mathbf{F}\) in terms of \(u\) and \(v\). Then compute \(\mathbf{F} \cdot \mathbf{n}\).

Step 4: Set Up the Surface Integral

The surface integral is \(\iint_S \mathbf{F} \cdot \mathbf{n} \, d\sigma = \int_0^{2\pi} \int_0^1 (\mathbf{F} \cdot \mathbf{n}) \, \|\mathbf{n}\| \, dudv\), where \(\|\mathbf{n}\|\) is the magnitude of the normal vector. Substitute the expression of \(\mathbf{F} \cdot \mathbf{n}\) from Step 3.

Step 5: Integrate to Find Flux

Evaluate the integral from Step 4. This involves integrating with respect to \(u\) from 0 to 1 and with respect to \(v\) from 0 to \(2\pi\). The result will give the total flux through the cone's surface.

Key Concepts

Surface Parametrization

Surface parametrization is an essential step when calculating flux across complex surfaces like a cone. The objective is to represent the surface using two parameters, in this case, through the cylindrical coordinate system which simplifies the approach for surfaces with circular symmetry.

For the given cone defined by the equation, \(z = 2\sqrt{x^2 + y^2}\), we can break down its features by introducing parameters \(u\) and \(v\). Here:

• \(x = u \cos v\),
• \(y = u \sin v\),
• \(z = 2u\),

where \(0 \leq u \leq 1\) and \(0 \leq v < 2\pi\). This transformation essentially "unwraps" our 3D surface onto a 2D parameter space, making further calculations easier to manage.

This kind of parametrization simplifies handling intricate surfaces as it aligns with the cylindrical symmetry, often seen in geometrical figures like cones and cylinders.

Cylindrical Coordinates

Cylindrical coordinates are a clever fusion of polar coordinates and the Cartesian system, enabling straightforward calculations for geometries like cones or cylinders. They are particularly useful when dealing with surfaces or volumes that have circular symmetry around an axis, such as the given cone.

In the cylindrical coordinate system, each point in space is represented by three values: <- \(r\), \(\theta\), and \(z\).

• \(r\) represents the radial distance from the origin to the projection of the point on the base plane (like a radius in polar coordinates).
• \(\theta\) is the angle in the base plane from a fixed direction (typically the positive x-axis, analogous to polar angle).
• \(z\) measures the height above the base plane, equivalent to the Cartesian z-coordinate.

In our problem, \(r\) can be replaced by the parameter \(u\), while angle \(\theta\) is represented by the parameter \(v\). This ensures the accurate representation of the cone’s surface using the equations we derived in the parameterization step.

Surface Integral

A surface integral allows us to calculate quantities that "flow across" or "act over" a curved surface. Specifically, it can measure the net flux, which is crucial in fields like electromagnetism and fluid dynamics.

When computing a surface integral for a vector field \(\mathbf{F}\) over a surface \(S\), the expression is given by \(\iint_{S} \mathbf{F} \cdot \mathbf{n} \, d\sigma\). Here, \(\mathbf{F} \cdot \mathbf{n}\) measures how much of the field is passing through the surface directed on \(\mathbf{n}\), the unit normal vector.

In this exercise, the surface integral encompasses finding the dot product \(\mathbf{F} \cdot \mathbf{n}\) alongside integrating this product over the specified domain of the parameters. The integration process is set as:\[ \int_{0}^{2\pi} \int_{0}^{1} (\mathbf{F} \cdot \mathbf{n}) \, \|\mathbf{n}\| \, dudv \]This encompasses understanding \(\mathbf{n}\) and ensuring it accurately reflects the outward direction as specified, emphasizing the vector nature of flux across surfaces.

Normal Vector Calculation

Calculating the normal vector is vital since it determines the direction in which we measure the flux or flow across a surface. The normal vector \(\mathbf{n}\) is a perpendicular vector on any given patch of the surface.

To find this vector, we compute the cross product of two tangent vectors derived from the surface parametrization. Specifically, these two vectors result from partial derivatives of the parametrization with respect to parameters \(u\) and \(v\):

• First, find \(\frac{\partial \mathbf{r}}{\partial u}\) and \(\frac{\partial \mathbf{r}}{\partial v}\).
• Next, compute \(\mathbf{n} = \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}\).

This cross product directly provides a vector normal to the surface. However, it’s crucial to ensure it points outward, as dictated by the problem. If not, a simple tweak by reversing the vector's direction will solve this.