Problem 25 - 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 \(\mathbf{F}=x y \mathbf{i}-z \mathbf{k}\) outward (normal away from the \(z\) -axis) through the cone \(z=\sqrt{x^{2}+y^{2}}, 0 \leq z \leq 1\)

Step-by-Step Solution

Answer

The flux through the cone is zero.

Step 1: Parameterize the Surface

The cone is given by the equation \(z = \sqrt{x^2 + y^2}\). To parameterize it, we switch to cylindrical coordinates where \((x, y, z) = (r\cos\theta, r\sin\theta, r)\), with \(r\) ranging from 0 to 1 and \(\theta\) from 0 to \(2\pi\). The parametrization is \(\mathbf{r}(r, \theta) = (r \cos \theta, r \sin \theta, r)\).

Step 2: Calculate the Normal Vector

The normal vector \(\mathbf{n}\) can be found via the cross product \(\mathbf{n} = \frac{\partial \mathbf{r}}{\partial r} \times \frac{\partial \mathbf{r}}{\partial \theta}\). Calculate these partial derivatives: \(\frac{\partial \mathbf{r}}{\partial r} = (\cos\theta, \sin\theta, 1)\) and \(\frac{\partial \mathbf{r}}{\partial \theta} = (-r\sin\theta, r\cos\theta, 0)\). Then, \(\mathbf{n} = \det \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \cos\theta & \sin\theta & 1 \ -r\sin\theta & r\cos\theta & 0 \end{vmatrix} = (-r\cos\theta, -r\sin\theta, r)\).

Step 3: Substitute into the Flux Integral

The flux is given by \( \iint_{S} \mathbf{F} \cdot \mathbf{n} \, d\sigma \) with \(\mathbf{F}(x, y, z) = xy\mathbf{i} - z\mathbf{k}\). Plug the parameterized vectors into \(\mathbf{F}\): \(xy = (r\cos\theta) (r\sin\theta) = r^2\cos\theta\sin\theta\), and \(z = r\). Then, \(\mathbf{F}(r, \theta) = r^2\cos\theta\sin\theta \mathbf{i} - r \mathbf{k}\).

Step 4: Compute the Dot Product

Find the dot product \(\mathbf{F} \cdot \mathbf{n} = (r^2\cos\theta\sin\theta \mathbf{i} - r \mathbf{k}) \cdot (-r\cos\theta \mathbf{i} - r\sin\theta \mathbf{j} + r \mathbf{k})\). Simplifying this gives \(-r^3\cos^2\theta\sin\theta + r^2\cos\theta\sin^2\theta + r^2\cos\theta\sin^2\theta = 0\).

Step 5: Evaluate the Integral

The result of the dot product is zero: \(\iint_{S} \mathbf{F} \cdot \mathbf{n} \, d\sigma = \int_{0}^{2\pi} \int_{0}^{1} 0 \, r \, dr \, d\theta = 0\). This means the flux through the cone is zero.

Key Concepts

Parametrization

Parametrization is the process of expressing a surface using parameters, typically to simplify calculations. In the case of the cone, converting cartesian coordinates (x, y, z) into cylindrical coordinates is an efficient choice.
Cylindrical coordinates make it easier to work with structures that have circular symmetry, like cones or cylinders. For the cone defined by the equation \(z = \sqrt{x^2 + y^2}\), we use cylindrical coordinates to parameterize it:

• \(x = r\cos\theta\)
• \(y = r\sin\theta\)
• \(z = r\)

The parameter \(r\) ranges from 0 to 1 (defining the radius), and \(\theta\) ranges from 0 to \(2\pi\) (defining the angle around the z-axis).
This parametrization \(\mathbf{r}(r, \theta) = (r \cos \theta, r \sin \theta, r)\) represents every point on the cone in terms of \(r\) and \(\theta\), simplifying the evaluation of surface integrals.

Normal Vector

Calculating the normal vector is essential for evaluating surface integrals, such as the flux integral in our exercise. A normal vector is perpendicular to the surface at a given point.
In this case, we calculate the normal vector \(\mathbf{n}\) through the cross product of two partial derivatives:

• \(\frac{\partial \mathbf{r}}{\partial r} = (\cos\theta, \sin\theta, 1)\)
• \(\frac{\partial \mathbf{r}}{\partial \theta} = (-r\sin\theta, r\cos\theta, 0)\)

The cross product \(\mathbf{n} = \frac{\partial \mathbf{r}}{\partial r} \times \frac{\partial \mathbf{r}}{\partial \theta}\) results in a vector normal to the surface of the cone.
By calculating this, we find \(\mathbf{n} = (-r\cos\theta, -r\sin\theta, r)\). This normal vector allows us to determine the direction of the flux across the cone's surface, planned outward away from the z-axis.

Cylindrical Coordinates

Cylindrical coordinates are a three-dimensional extension of polar coordinates and are especially useful for problems with cylindrical symmetry.
They are composed of the radius \(r\), angle \(\theta\), and height \(z\). For transforming from Cartesian to cylindrical coordinates, the transformations are:

• \(x = r\cos\theta\)
• \(y = r\sin\theta\)
• \(z = z\)

These transformations are particularly convenient for our cone \(z=\sqrt{x^2+y^2}\), allowing the height \(z\) to be directly mapped as \(z = r\).
Understanding these coordinates simplifies the computation of vector fields over surfaces like cones and cylinders. It shrinks the complexity of a flux integral by reducing it to integrations over \(r\) and \(\theta\), parameters that naturally fit the conical surface structure.

Dot Product

The dot product is a central operation in vector calculus, often used to project one vector onto another. It's crucial in calculating flux, as it measures how much of the vector field goes through a surface.
For the vector field \(\mathbf{F}(x, y, z) = xy \mathbf{i} - z \mathbf{k}\), expressed in parametrized form as \(\mathbf{F}(r, \theta) = r^2\cos\theta\sin\theta \mathbf{i} - r \mathbf{k}\), we calculate the dot product with the normal vector \((-r\cos\theta, -r\sin\theta, r)\).

Dot Product (Cont'd)

The process involves:

• Multiply corresponding components of the vectors.
• Sum up these products for all components.

The dot product \(\mathbf{F} \cdot \mathbf{n}\) is calculated as follows: \[-r^3\cos^2\theta\sin\theta + r^2\cos\theta\sin^2\theta + r^2\cos\theta\sin^2\theta = 0\].
The result of this dot product is zero, indicating that the vector field \(\mathbf{F}\) does not pass through the surface in the direction of \(\mathbf{n}\). This simplifies the flux integral over the surface to zero, showing that there is no net flow through the cone in the specified direction.