Problem 28 - Thomas Calculus

Question

Use a parametrization to find the flux \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d \sigma\) across the surface in the specified direction. \(\begin{array}{l}{\text { Paraboloid } \mathbf{F}=4 x \mathbf{i}+4 y \mathbf{j}+2 \mathbf{k} \text { outward (normal away from }} \\ {\text { the } z \text { -axis through the surface cut from the bottom of the paraboloid }} \\\ {\text z=x^{2}+y^{2} \text { by the plane } z=1}\end{array}\)

Step-by-Step Solution

Answer

The flux across the surface is 4π.

Step 1: Parametrize the Surface

The surface of the paraboloid is given by the equation \(z = x^2 + y^2\). Since it is cut off by the plane \(z = 1\), we can set up the parametrization. Let \(x = u\) and \(y = v\). Then the equation for the surface becomes \(z = u^2 + v^2\). The parametrization is \[ \mathbf{r}(u,v) = \langle u, v, u^2 + v^2 \rangle \] with the domain \(u^2 + v^2 \leq 1\).

Step 2: Find the Normal Vector

To find the flux, we need the normal vector to the surface. Compute the partial derivatives \( \frac{\partial \mathbf{r}}{\partial u} \) and \( \frac{\partial \mathbf{r}}{\partial v} \). \[ \frac{\partial \mathbf{r}}{\partial u} = \langle 1, 0, 2u \rangle \]\[ \frac{\partial \mathbf{r}}{\partial v} = \langle 0, 1, 2v \rangle \]The normal vector is the cross product \( \mathbf{n} = \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} \):\[ \mathbf{n} = \langle -2u, -2v, 1 \rangle \]

Step 3: Calculate the Dot Product \(\mathbf{F} \cdot \mathbf{n}\)

Given \( \mathbf{F} = 4x \mathbf{i} + 4y \mathbf{j} + 2 \mathbf{k} \), substitute the parametrization values:\[ \mathbf{F}(u,v) = 4u \mathbf{i} + 4v \mathbf{j} + 2 \mathbf{k} \]Compute the dot product with the normal:\[ \mathbf{F} \cdot \mathbf{n} = \langle 4u, 4v, 2 \rangle \cdot \langle -2u, -2v, 1 \rangle = -8u^2 - 8v^2 + 2 \]

Step 4: Set Up the Double Integral

The flux through the surface is given by the integral \[ \iint_{S} \mathbf{F} \cdot \mathbf{n} \, d\sigma = \iint_{D} \mathbf{F} \cdot \mathbf{n} \, \|\mathbf{n}\| \, du \, dv \]Use the surface element \(d\sigma = \|\mathbf{n}\| \ du \ dv\) where \(\|\mathbf{n}\| = \sqrt{(-2u)^2 + (-2v)^2 + 1^2} = \sqrt{4u^2 + 4v^2 + 1}\). The region \(D: u^2 + v^2 \leq 1\) is a circle.

Step 5: Convert to Polar Coordinates

To simplify, convert the double integral to polar coordinates. Let \(u = r \cos\theta\), \(v = r \sin\theta\). The Jacobian gives \(dudv = r \, dr \, d\theta\). The region becomes \(0 \leq r \leq 1\) and \(0 \leq \theta < 2\pi\). Substitute in the integrand:\[ -8(u^2 + v^2) + 2 = -8r^2 + 2 \]\[ \iint \sqrt{4r^2 + 1} \ (-8r^2 + 2) \, r \, dr \, d\theta \]

Step 6: Evaluate the Integral

Integrate in polar coordinates:\[ \int_0^{2\pi} \int_0^1 (-8r^2 + 2) \sqrt{4r^2 + 1} \ r \, dr \, d\theta \]First, compute the \(r\) integral:\[ \int_0^1 (-8r^3 + 2r) \sqrt{4r^2 + 1} \, dr \]Using substitution, set \(u = 4r^2 + 1\), \(du = 8r \, dr\), and adjust the limits accordingly.Solve the integral and multiply by the \(\theta\) limit to get the total flux.

Key Concepts

Parametrization

Parametrizing a surface involves expressing the coordinates of points on the surface in terms of two parameters. For the paraboloid described by the equation \(z = x^2 + y^2\), cut off by the plane \(z = 1\), we adopt a straightforward approach. By setting \(x = u\) and \(y = v\), the expression for \(z\) becomes \(u^2 + v^2\). Thus, we parametrize the surface using the vector function:

• \(\mathbf{r}(u,v) = \langle u, v, u^2 + v^2 \rangle\)

The domain for this parametrization is determined by the boundary \(u^2 + v^2 \leq 1\), which defines a circular region in the \(uv\) plane. This straightforward parametrization translates our surface in 3D into a manageable form for further calculations, playing a crucial role in establishing subsequent vector operations needed for flux computation.

Normal Vector

The normal vector to a surface is essential for determining the orientation of the surface. It allows us to calculate quantities like flux across the surface. For our parametrized surface \(\mathbf{r}(u,v) = \langle u, v, u^2 + v^2 \rangle\), the normal vector is found using the cross product of the surface's partial derivatives.
First, calculate the partial derivative with respect to \(u\):

• \(\frac{\partial \mathbf{r}}{\partial u} = \langle 1, 0, 2u \rangle\)

Next, with respect to \(v\):

• \(\frac{\partial \mathbf{r}}{\partial v} = \langle 0, 1, 2v \rangle\)

The normal vector, \(\mathbf{n}\), is the cross product of these two vectors:

• \(\mathbf{n} = \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} = \langle -2u, -2v, 1 \rangle\)

This normal vector points outward away from the \(z\)-axis, which aligns correctly with the specified direction in the problem.

Polar Coordinates

Switching from Cartesian to polar coordinates simplifies the evaluation of integrals over circular regions, which are frequent in flux calculations. The conversion to polar coordinates involves:

• \(u = r \cos\theta\)
• \(v = r \sin\theta\)

With the Jacobian of this transformation being \(r\), the differential area element becomes \(dudv = r \, dr \, d\theta\). This conversion changes the region defined by \(u^2 + v^2 \leq 1\) in Cartesian coordinates to \(0 \leq r \leq 1\) and \(0 \leq \theta < 2\pi\) in polar coordinates.
By altering the coordinate system, the integral of a complex function over a circular area transforms into a more approachable double integral with a simpler form. This method is powerful in symmetric problems and is foundational in vector calculus to find solutions that would be cumbersome in regular Cartesian coordinates.

Double Integral

A double integral is used to calculate the flux of a vector field through a surface. In our context, the double integral integrates the dot product of the vector field \(\mathbf{F}\) and the normal vector \(\mathbf{n}\) over the given surface. We convert the integral into polar coordinates for easier computation:

• Initial expression: \(\iint_{S} \mathbf{F} \cdot \mathbf{n} \, d\sigma\),
• Transform the surface integral into \(\iint_{D} \mathbf{F} \cdot \mathbf{n} \, \|\mathbf{n}\| \, dudv\)

By identifying \(d\sigma\) as \(\|\mathbf{n}\| \, dudv\) with \(\|\mathbf{n}\| = \sqrt{4u^2 + 4v^2 + 1}\), and converting to polar coordinates, we simplify:

• \(\int_0^{2\pi} \int_0^1 (-8r^2 + 2) \sqrt{4r^2 + 1} \ r \, dr \, d\theta\)

This step combines previous transformations and ensures accurate evaluation over the entire surface, playing a vital role in determining the final flux result.