Problem 38 - Thomas Calculus

Question

Find the flux of the field \(\mathbf{F}(x, y, z)=4 x \mathbf{i}+4 y \mathbf{j}+2 \mathbf{k}\) outward (away from the \(z\) -axis) through the surface cut from the bottom of the paraboloid \(z=x^{2}+y^{2}\) by the plane \(z=1\)

Step-by-Step Solution

Answer

The flux through the surface is \( 6\pi \).

Step 1: Understand the Surface and Region

The surface bounded by the paraboloid \( z = x^2 + y^2 \) and the plane \( z = 1 \) is the cap of the paraboloid. The region is cut at \( z = 1 \), creating a circle with radius \( r = 1 \), since the intersection sets \( x^2 + y^2 = 1 \).

Step 2: Set Up the Surface Integral

We are required to find the flux of \( \mathbf{F} = 4x\mathbf{i} + 4y\mathbf{j} + 2 \mathbf{k} \) through this surface. The flux \( \Phi \) across a surface \( S \) for a vector field \( \mathbf{F} \) is given by the integral \( \iint_{S} \mathbf{F} \cdot \mathbf{n} \, dS \), where \( \mathbf{n} \) is the outward normal vector to the surface.

Step 3: Parameterize the Surface

The surface can be parameterized using cylindrical coordinates given by \( x = r\cos\theta \), \( y = r\sin\theta \), and \( z = r^2 \). Here, since \( z = 1 \), we have \( r = 1 \). So, the parametrization simplifies to \( (x, y, z) = (\cos\theta, \sin\theta, 1) \).

Step 4: Compute the Normal Vector

Since the surface is part of a parabolical cap, the normal vector can be found from the parameterization: \( \mathbf{r}(r, \theta) = (r\cos\theta, r\sin\theta, r^2) \). The tangent vectors are \( \mathbf{t}_{r} = (\cos\theta, \sin\theta, 2r) \) and \( \mathbf{t}_{\theta} = (-r\sin\theta, r\cos\theta, 0) \). The cross product \( \mathbf{t}_r \times \mathbf{t}_{\theta} \) gives the normal vector, and evaluating this at \( r=1 \) yields \( \mathbf{n} = (2\cos\theta, 2\sin\theta, -1) \).

Step 5: Integrate over the Parameter Domain

The flux integral becomes: \[ \iint_{S} \mathbf{F} \cdot \mathbf{n} \, dS = \int_{0}^{2\pi} \int_{0}^{1} (4x, 4y, 2) \cdot (2\cos\theta, 2\sin\theta, -1) \, r \, dr \, d\theta \]. Since \( x = \cos\theta \) and \( y = \sin\theta \), compute the dot product: \( 8x\cos\theta + 8y\sin\theta + -2 \). Substituting values gives: \( 8\cos^2\theta + 8\sin^2\theta - 2 \). Simplifying using \( \cos^2\theta + \sin^2\theta = 1 \) results in \( 6 \).

Key Concepts

Vector Field

In vector calculus, a vector field is a function that assigns a vector to every point in space. Picture a vector as an arrow with both direction and magnitude. For example, the vector field in our problem is defined as \( \mathbf{F}(x, y, z) = 4x \mathbf{i} + 4y \mathbf{j} + 2 \mathbf{k} \). This means:

• Each arrow in the x-direction is scaled by \( 4x \).
• In the y-direction, arrows are scaled by \( 4y \).
• In the z-direction, all arrows have a constant value of 2.

The concept of a vector field is fundamental in physics and engineering because it can represent many kinds of physical quantities, such as velocity and force fields.

Surface Integral

A surface integral extends the idea of an integral to functions over a surface in three-dimensional space. When you integrate a vector field over a surface, you're essentially measuring the total "flow" or "flux" through that surface. Here's how it's set up in our exercise:
- We are calculating the flux of \( \mathbf{F} \) through the surface cut from the paraboloid by the plane. - The flux through a surface \( S \) is given by: \[\iint_{S} \mathbf{F} \cdot \mathbf{n} \, dS\] Here, \( \mathbf{n} \) is the outward normal vector to the surface which points away from the z-axis, and \( dS \) represents an infinitesimal piece of the surface area.
Calculating surface integrals is crucial for understanding behaviors like fluid flow across a surface, making this a key technique in fields like hydrodynamics.

Paraboloid

A paraboloid is a 3-dimensional shape resembling a parabola that has been spun around an axis. In this exercise, our paraboloid is defined by the equation \( z = x^2 + y^2 \).
Some key characteristics of this paraboloid include:

• It is symmetric around the z-axis.
• The cross-sections parallel to the xy-plane form circles.
• When intersected by the plane \( z = 1 \), the result is a circular cap with radius 1.

Understanding the shape of a paraboloid is vital for visualizing the problem setup, which helps in setting boundaries and performing integration correctly.

Cylindrical Coordinates

Cylindrical coordinates are an extension of 2D polar coordinates into three dimensions. They are particularly useful for surfaces that have rotational symmetry around one axis, like the paraboloid in this problem. In cylindrical coordinates:

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

For this particular exercise, since our paraboloid cap is defined by \( z = 1 \), and the intersection sets \( r = 1 \), we have \( (x, y, z) = (\cos \theta, \sin \theta, 1) \).
These coordinates make it far easier to manage integrations over circular regions, streamlining the calculations needed for the flux integral. Understanding when and how to switch to cylindrical coordinates is a powerful tool for tackling complex geometric shapes.