Question

Evaluate each of the integrals as either a volume integral or a surface integral, whichever is easier. \(\iint \mathbf{r} \cdot \mathbf{n} d \sigma\) over the whole surface of the cylinder bounded by \(x^{2}+y^{2}=1, z=0\), and \(z=3 ; \mathbf{r}\) means \(\mathbf{i} x+\mathbf{j} y+\mathbf{k} z\).

Short answer

9\pi

Step-by-step solution

Understanding the Problem

The integral \(\iint \mathbf{r} \cdot \mathbf{n} \, d \sigma\) needs to be evaluated over the whole surface of a cylinder defined by \(x^2 + y^2 = 1\) from \(z = 0\) to \(z = 3\). Here, \(\mathbf{r} = \mathbf{i} x + \mathbf{j} y + \mathbf{k} z\) and \(\mathbf{n}\) is the unit normal vector.

Identify the Surfaces

The cylinder is bounded by three surfaces: the curved surface \(x^2 + y^2 = 1\), and the top and bottom caps at \(z = 3\) and \(z = 0\) respectively.

Set Up the Curved Surface Integral

For the curved surface, the outward normal vector \(\mathbf{n}\) can be expressed as \(\mathbf{n} = x \mathbf{i} + y \mathbf{j}\). Therefore, \(\mathbf{r} \cdot \mathbf{n} = x^2 + y^2\).

Evaluate the Curved Surface Integral

On the curved surface \(x^2 + y^2 = 1\), this reduces to \(\iint (x^2 + y^2) \, d \sigma = \iint 1 \, d \sigma\). The integral of 1 over the surface is equal to the surface area of the curved part of the cylinder. The surface area of the cylinder is \(2 \pi r h\), so in this case, it equals \(2 \pi \times 1 \times 3 = 6 \pi\).

Set Up Top and Bottom Cap Integrals

For the top and bottom caps, the normal vectors are \(\mathbf{n} = \mathbf{k}\) and \(-\mathbf{k}\) respectively. Thus, \(\mathbf{r} \cdot \mathbf{n}\) on the top cap at \(z = 3\) is \(z = 3\) and on the bottom cap at \(z = 0\) is \(z = 0\).

Evaluate Top and Bottom Cap Integrals

The integral at the top cap is: \(\iint_{top} 3 \, dA = 3 \times \pi r^2 = 3 \pi\). The integral at the bottom cap is: \(\iint_{bottom} 0 \, dA = 0\).

Add Up All Results

Add the contributions from the curved surface and the top surface: \(6 \pi + 3 \pi + 0 = 9 \pi\).

Key concepts

Understanding Cylindrical Coordinates

Cylindrical coordinates are a three-dimensional coordinate system that extends polar coordinates by adding a height (z) dimension. They are particularly useful for problems involving symmetry around an axis, such as cylinders or circular regions. In this system, any point in space is represented by three values: \(r, \theta, z\), where:

• r: the radial distance from the origin to the projection of the point on the xy-plane


• \(\theta\): the angle measured counterclockwise from the positive x-axis to the line segment from the origin to the projection of the point on the xy-plane


• z: the height above the xy-plane

These coordinates simplify integration over cylindrical surfaces and volumes, making it easier to set up and evaluate integrals.

Introduction to Vector Calculus

Vector calculus is a branch of mathematics that deals with vector fields and differentiates and integrates functions involving vectors. It has various applications in physics and engineering, especially in fluid dynamics and electromagnetism.

Key operations in vector calculus include:

• Gradient: Measures the rate and direction of change in a scalar field.


• Divergence: A measure of the 'outflowing-ness' of a vector field from a point, often denoted as \( abla \cdot \mathbf{F} \).


• Curl: Identifies the rotation or the 'circulation' in a vector field, denoted as \( abla \times \mathbf{F} \).

Integrals in vector calculus include:

• Line integrals: Integrate along a curve.


• Surface integrals: Integrate over a surface.


• Volume integrals: Integrate within a volume.

These integrals often involve dot products or cross products, which help in evaluating the components along different directions.

Integral Evaluation in Vector Calculus

Evaluating integrals in vector calculus can involve surface and volume integrals. In surface integrals, we integrate functions over a surface in three-dimensional space, which often involves a dot product with a normal vector. For volume integrals, we integrate over a three-dimensional region.

When solving surface integrals like the one in the exercise, here’s a simplified approach:

• Identify the surfaces involved and their normal vectors.


• Set up the integral with appropriate limits.


• Evaluate each surface integral separately.

For the specific exercise:

• The curved surface integral becomes manageable by noting symmetry and translating it into simpler boundary definitions.


• The top and bottom caps are calculated separately, applying the specific normal vectors for each cap.

Understanding these principles helps tackle more complex problems in vector calculus with confidence.