Problem 38
Question
Step-by-Step Solution
Verified Answer
The surface area is \(4\pi^2 ab\).
1Step 1: Identify the parameters
The circle is defined parametrically by the equations \(x = a \cos t\) and \(y = a \sin t\), where \(0 \leq t \leq 2\pi\). It is revolved around the line \(x = b\) to form a torus. Here, \(a\) is the radius of the circle, and \(b\) is the distance from the center of the circle to the line of revolution. Given \(0 < a < b\), the circle lies strictly to the left of the line \(x = b\).
2Step 2: Formula for the Surface Area of a Torus
The surface area \(A\) of a torus formed by revolving a circle of radius \(a\) around a line \(b\) units from the center of the circle, is given by: \[ A = 4\pi^2 ab \] This formula is derived from considering the circle tracing a circular path of length \(2\pi b\) through the revolution.
3Step 3: Substitute the Parameters
Substitute \(a\) and \(b\) into the formula:\[ A = 4\pi^2 ab \] This represents substituting into the known formula as all circles with specified parameters will generate a torus with this area.
Key Concepts
Parametric EquationsCircle RevolutionTorus GeometryCalculus Applications
Parametric Equations
Parametric equations are a powerful tool in mathematics, often used to define geometric figures using parameters. In the given exercise, a circle is described by the parametric equations:
This representation separates complex shapes into simple components, encoding geometry through equations rather than relying solely on Cartesian coordinates. Parametric equations give us flexibility, allowing us to represent curves that are difficult to express with standard equation forms.
- \(x = a \cos t\)
- \(y = a \sin t\)
This representation separates complex shapes into simple components, encoding geometry through equations rather than relying solely on Cartesian coordinates. Parametric equations give us flexibility, allowing us to represent curves that are difficult to express with standard equation forms.
Circle Revolution
Revolving a geometric figure around a line in 3D space is fascinating, and it helps in understanding real-world structures. In this exercise, the circle is revolved around the line \(x = b\). This process is known as circle revolution.
When revolved, every point on the circle sweeps a circular path around this line, resulting in a toroid shape. It's akin to holding a hoop and spinning it about a central line.
Key points to note:
When revolved, every point on the circle sweeps a circular path around this line, resulting in a toroid shape. It's akin to holding a hoop and spinning it about a central line.
Key points to note:
- The line of revolution (\(x = b\)) does not intersect the circle, as \(a < b\).
- The entire circle traces a circular path in space, forming what we call a torus.
Torus Geometry
A torus, often seen in doughnut shapes, is a surface of revolution. Its geometry can be challenging but intriguing. Here, the key is to maintain a constant equidistant path from the axis of revolution.
Characteristics of a torus like the one in this exercise:
Characteristics of a torus like the one in this exercise:
- Two radii: the radius of the original circle \(a\), and the distance \(b\) from the circle's center to the line of revolution.
- Its surface can be imagined as the continuous circular sweep of the initial circle in a plane, tracking around an axis at a fixed range.
Calculus Applications
Calculus plays a crucial role in calculating areas of shapes like the torus. Here, we're focusing on finding the surface area, which is derived using integral calculus concepts.
The provided formula for the torus's surface area, \(A = 4\pi^2 ab\), is obtained through a process related to wrapping the circular path along its own axis.
Why calculus?
The provided formula for the torus's surface area, \(A = 4\pi^2 ab\), is obtained through a process related to wrapping the circular path along its own axis.
Why calculus?
- Integrals help calculate total areas and volumes where simple arithmetic fails due to curvature or non-linear paths.
- The use of \(2\pi\) in the formula reflects the circular path's involvement—the circumference of the path that the circle follows during revolution.
Other exercises in this chapter
Problem 37
The expected value of a function \(g(X)\) of a continuous random variable \(X\) having PDF \(f(x)\) is defined to be \(E[g(X)]=\) \(\int_{A}^{B} g(x) f(x) d x\)
View solution Problem 38
Find the area of the region trapped between \(y=x e^{-x^{2}}\) and \(y=x / 4\). Hint: There are two separate regions.
View solution Problem 38
A continuous random variable \(X\) has \(\mathrm{PDF} f(x)=\) \(\frac{3}{256} x(8-x), 0 \leq x \leq 8\). Find \(E\left(X^{2}\right)\) and \(E\left(X^{3}\right)\
View solution Problem 39
Use the Parabolic Rule with \(n=8\) to approximate the area of the region trapped between \(y=1-e^{-x^{2}}\) and \(y=e^{-x^{2}}\).
View solution