Question

Find the area of the indicated surface. Make a sketch in each case. The part of the cylinder \(x^{2}+z^{2}=9\) that is directly over the rectangle in the \(x y\) -plane with vertices \((0,0),(2,0),(2,3)\), and \((0,3)\).

Short answer

The area is approximately 9 units.

Step-by-step solution

Understand the Geometry

The equation of the cylinder is \( x^2 + z^2 = 9 \), which describes a circular cylinder centered on the y-axis with radius 3. The given rectangle in the xy-plane has vertices \((0,0), (2,0), (2,3), (0,3)\). The goal is to find the area of the cylinder that lies directly above this rectangular region.

Project the Region onto the Cylinder

The cylinder's equation \( x^2 + z^2 = 9 \) implies the circle's projection radius in the xz-plane will be maintained. Since the projection is made over the rectangle in the xy-plane, the values of x range from 0 to 2. This defines the range on the cylinder over which we are calculating the surface area.

Parameters for Surface Area

Parameterize the surface of the cylinder. Given the cylindrical coordinates, express the surface as a vector function: \( \mathbf{r}(y, \theta) = (3\cos(\theta), y, 3\sin(\theta)) \), where \( y \) ranges from 0 to 3 and \( x = 3\cos(\theta) \), and \( z = 3\sin(\theta) \). The values of \( \theta \) can be deduced from the range of x (between 0 and 2). Use \( \cos(\theta) = \frac{x}{3} \), hence \( \theta = \cos^{-1}\left(\frac{x}{3}\right) \).

Find the Bounds for \( \theta \)

Calculate bounds for \( \theta \) by setting \( x = 0 \) and \( x = 2 \):- \( x = 0 \rightarrow \cos(\theta) = 0 \rightarrow \theta = \frac{\pi}{2} \).- \( x = 2 \rightarrow \cos(\theta) = \frac{2}{3} \rightarrow \theta = \cos^{-1}\left(\frac{2}{3}\right) \).Thus, \( \theta \) ranges from \( \cos^{-1}\left(\frac{2}{3}\right) \) to \( \frac{\pi}{2} \).

Calculate the Surface Area

To compute the surface area of the cylindrical section, use the surface area element derived from the parametric description:\[ dS = |\mathbf{T}_y \times \mathbf{T}_\theta| \, dy \, d\theta \]\( \mathbf{T}_y = (0, 1, 0) \) and \( \mathbf{T}_\theta = (-3\sin(\theta), 0, 3\cos(\theta)) \).So, \(|\mathbf{T}_y \times \mathbf{T}_\theta| = 3 \). Thus, the area \( A \) is:\[A = \int_{0}^{3} \int_{\cos^{-1}(\frac{2}{3})}^{\frac{\pi}{2}} 3 \, d\theta \, dy = 3 \times 3 \times \left( \frac{\pi}{2} - \cos^{-1}\left(\frac{2}{3}\right) \right)\]Calculate the inner integral first.

Calculate the Numerical Value

Solve the inner integral:\[= 9 \left( \frac{\pi}{2} - \cos^{-1}\left(\frac{2}{3}\right) \right) \]Using a calculator, approximate \( \cos^{-1}\left(\frac{2}{3}\right) \) and compute the final value. This would require the approximation of the inverse cosine.

Key concepts

Cylindrical Coordinates

Cylindrical coordinates are a way to describe a point in three-dimensional space. They are particularly useful for problems involving circular symmetry, like our cylinder with the equation \(x^2 + z^2 = 9\). In cylindrical coordinates, any point is represented as \((r, \theta, y)\). Here, \(r\) is the radial distance from the origin in the \(x-z\) plane, \(\theta\) is the angle in the \(x-z\) plane, analogous to the angle in polar coordinates, and \(y\) is the height along the \(y\)-axis.
For our cylinder, the radial distance \(r\) is constant at 3 since the cylinder's radius is 3. In this system, we simplify the representation of surfaces and volumes that involve circles or parts of circles. Instead of directly engaging with \(x\) and \(z\), which can be daunting in circular problems, you deal with \(r\) and \(\theta\) more intuitively. By aligning the problem with cylindrical coordinates, the mathematics often simplifies, making integration and surface calculations more straightforward.

Parametric Surfaces

A parametric surface in mathematics is a surface defined using a parametric equation. This means expressing the coordinates of each point on the surface in terms of two parameters, rather than just using the coordinates \(x, y, z\).
For the cylindrical surface described by \(x^2 + z^2 = 9\), we use a parametric form where \(x\) and \(z\) are expressed with help of trigonometric functions, parameterized by \(\theta\) and \(y\):

• \(x = 3\cos(\theta)\)
• \(z = 3\sin(\theta)\)

These parameters map out the entire curved surface of the cylinder above the rectangle in the \(xy\)-plane.
Parametrization allows us to formally describe the shape and coverage of complex surfaces, facilitating calculations of characteristics like area and volume.

Integration

Integration is the process of calculating the area, volume, or other accumulations of quantities over a specific range. In calculus, integration can be understood as the reverse operation of differentiation.
For the calculation of surface area on our cylinder, we perform a double integral over a parametrized region. We integrate over \(y\) and \(\theta\), which represent the parametric bounds of our surface.

• The integral format is \( \int \int f(y, \theta) \ dy \ d\theta \)
• For our problem, the function inside the integral is a constant: 3, multiplied by the area element \(dy \, d\theta\).
• Specifically, the surface area \(A\) involves integrating \(3\) over given bounds.

This systematic approach allows us to calculate the exact size of the surface above the rectangle by considering tiny patches over the entire area and summing them up.

Calculus Problems

Calculus can be challenging, but it's pivotal for solving complex real-world problems involving change and motion. In this exercise, we tackled a classic calculus problem that combines geometry with integral calculus to find the surface area of a cylindrical section.
This problem illustrates several fundamental tasks:

• Understanding and visualizing geometric constraints, like the radius and orientation of the cylinder.
• Transitioning between coordinate systems—here, changing from Cartesian to cylindrical coordinates.
• Parametrizing a surface to express it in terms of independent variables \(y\) and \(\theta\).
• Setting up and evaluating integrals, first by identifying bounds and then computing the definite integral for a precise quantitative result.

By approaching calculus problems step by step, breaking them into manageable parts, and using appropriate mathematical tools, problems become more approachable, leading to accurate solutions.