Problem 18 - Thomas Calculus

Question

In Exercises \(17-26,\) use a parametrization to express the area of the surface as a double integral. Then evaluate the integral. (There are many correct ways to set up the integrals, so your integrals may not be the same as those in the back of the book. They should have the same values, however.) $$ \begin{array}{l}{\text { Plane inside cylinder The portion of the plane } z=-x \text { inside }} \\ {\text { the cylinder } x^{2}+y^{2}=4}\end{array} $$

Step-by-Step Solution

Answer

The area is \( 4\pi\sqrt{2} \).

Step 1: Parametrize the Plane and Cylinder

To find the area of the surface, we need a parametrization. The given plane is described by the equation \( z = -x \), and the cylinder is \( x^2 + y^2 = 4 \). In polar coordinates, parametrize the region \( x = r \cos(\theta) \), \( y = r \sin(\theta) \), where \( 0 \le r \le 2 \) and \( 0 \le \theta \le 2\pi \). Thus, the parametrization is \((x,y,z) = (r \cos(\theta), r \sin(\theta), -r \cos(\theta)).\)

Step 2: Compute the Surface Normal

Calculate the surface normal using the cross product of the partial derivatives. Set \( \mathbf{r}(r, \theta) = (r \cos(\theta), r \sin(\theta), -r \cos(\theta)) \), then \( \frac{\partial \mathbf{r}}{\partial r} = (\cos(\theta), \sin(\theta), -\cos(\theta)) \) and \( \frac{\partial \mathbf{r}}{\partial \theta} = (-r \sin(\theta), r \cos(\theta), r \sin(\theta)) \). The cross product is obtained as \( \mathbf{n} = (r \cos^2(\theta) + r \sin^2(\theta), -r\cos(\theta), r \cos(\theta)) \). Simplify to \( \mathbf{n} = (r, -r\cos(\theta), r\cos(\theta)) \).

Step 3: Determine \(|\mathbf{n}|\), the magnitude of the normal

Compute the magnitude of the normal vector: \[ |\mathbf{n}| = \sqrt{r^2 + r^2\cos^2(\theta) + r^2\cos^2(\theta)} = r\sqrt{2} \] since \( |\cos(\theta)| \) sums to 1 inside the unit circle.

Step 4: Set Up the Double Integral

To find the area, integrate over the surface: \( \int \int_R |\mathbf{n}| \, dr \, d\theta \). Substitute in the magnitude \(|\mathbf{n}| = r\sqrt{2}\): \[ \int_0^{2\pi} \int_0^{2} r\sqrt{2} \, dr \, d\theta \].

Step 5: Evaluate the Double Integral

Evaluate the inner integral with respect to \( r \): \( \int_0^2 r \sqrt{2} \, dr = \sqrt{2} \int_0^2 r \, dr = \sqrt{2} \left[ \frac{r^2}{2} \right]_0^2 = \sqrt{2} \times 2 = 2\sqrt{2} \). Next, evaluate with respect to \( \theta \): \( \int_0^{2\pi} 2\sqrt{2} \, d\theta = 2\sqrt{2} \times 2\pi = 4\pi\sqrt{2} \).

Step 6: Interpret the Result

The area of the portion of the plane inside the cylinder is \( 4\pi\sqrt{2} \).

Key Concepts

Double Integral

A double integral allows us to compute the volume under a surface or, as in our exercise, the area of a particular surface. It involves integrating a function of two variables over a specified region. When we evaluate a double integral, we perform integration twice: first with respect to one variable and then the other.

In this exercise, the double integral is used to find the area of the portion of the plane within the cylinder. Initially, we integrate with respect to the radial coordinate, followed by the angular coordinate in polar coordinates. This approach simplifies the integration by aligning with the cylindrical symmetry of the region.

• An inner integral handles the variable transformation, such as from Cartesian to polar coordinates.
• An outer integral further refines the area calculation according to the geometric region of interest.

Applying double integrals in this manner can help solve complex geometry problems with ease.

Polar Coordinates

Utilizing polar coordinates is especially useful when dealing with circular or cylindrical regions. Instead of using the usual Cartesian approach with x and y, polar coordinates rely on the radius and angle.

Why Polar Coordinates?

In our exercise, we have a cylindrical boundary defined by \( x^2 + y^2 = 4 \). By switching to polar coordinates with \( x = r \cos(\theta), y = r \sin(\theta) \), we can better align with the circular symmetry.

• The variable \( r \) represents the radius, ranging from 0 to 2, which corresponds to the cylinder's radius.
• The angle \( \theta \) ranges from 0 to \( 2\pi \), representing a full circle around the cylinder.

This switch simplifies the process of integration because our expressions naturally adapt to the shape of the region.

Surface Integral

A surface integral extends the concept of a double integral to account for the curvature or orientation of a surface in three-dimensional space. When dealing with a surface integral, we consider both the geometry of the surface and the function defined over it.

Area Calculation

In our example, the surface integral evaluates the area of the plane segment within the cylindrical region. Surface integrals can handle non-flat surfaces, which are more complex than flat planes. The process involves:

• Defining a parametrization of the surface.
• Calculating the surface's normal vector and its magnitude \(|\mathbf{n}|\).
• Setting up and evaluating the integral based on \(|\mathbf{n}|\).

Surface integrals are practical when you need to compute areas or fluxes through surfaces in three-dimensional spaces.

Parametrization

Parametrization involves expressing a surface using parameter variables, which simplifies calculations and describes the surface in terms of these parameters. This method uses a chosen coordinate system to define the surface comprehensively.

Application in Exercise

For the given task, the planar surface is parametrized using polar coordinates. We express the coordinates as \( (x, y, z) = (r \cos(\theta), r \sin(\theta), -r \cos(\theta)) \), where \( r \) and \( \theta \) are the parameters.

This representation simplifies the integration and perfectly aligns with the constraints given by the cylindrical boundary. It transforms the surface into a parameter space, making it easier to analyze and integrate.Consider parametrization as a bridge between complex geometries and their more manageable abstract representations.

Normal Vector

The normal vector is a perpendicular vector to a surface that plays a key role in understanding the surface orientation. It helps determine the magnitude used in surface integrals.

Importance of Normal Vectors

In computations involving surface integrals, calculating the normal vector is essential. It is obtained through the cross product of partial derivatives of the parametrization:

• First, compute the partial derivatives \( \frac{\partial \mathbf{r}}{\partial r} \) and \( \frac{\partial \mathbf{r}}{\partial \theta} \).
• Perform a cross product to find the normal vector \( \mathbf{n} \).
• Calculate \(|\mathbf{n}|\), the magnitude, needed for the surface integral calculation.

The normal vector simplifies deriving surface properties, such as area or flux, by offering a clear directional component to use in mathematical models.