Question

Area,find the area of the surface over the given region. Use a computer algebra system to verify your results. The torus \(\mathbf{r}(u, v)=(a+b \cos v) \cos u \mathbf{i}+(a+b \cos v) \sin u \mathbf{j}+\) \(b \sin v \mathbf{k},\) where \(a>b, 0 \leq u \leq 2 \pi,\) and \(0 \leq v \leq 2 \pi\)

Short answer

The area of the torus is \(4 \pi^2 ab\).

Step-by-step solution

Calculate the partial derivatives

Firstly, calculate the partial derivatives of the parametric equations with respect to u and v. The derivative with respect to u (\(r_u\)) is \((a+b \cos v) (-\sin u) \mathbf{i} + (a + b \cos v)(\cos u) \mathbf{j}\). The derivative with respect to v (\(r_v\)) is \(-b \sin v \cos u \mathbf{i} - b \sin v \sin u \mathbf{j} + b \cos v \mathbf{k}\)

Compute the cross product of \(r_u\) and \(r_v\)

The next step is to compute the cross product of the two derivative vectors (\(r_u\) and \(r_v\)). The cross product |\(r_u \times r_v\)| is \((a+b \cos v) \sqrt{1 + \sin^2 v}\)

Integrate over the region

Now, you need to compute the double integral over the region S (which is from 0 to \(2\pi\) for both u and v) of |\(r_u \times r_v\)|.Doing this will give you the total area of the torus: \(A = \int\int_S (a+b \cos v) \sqrt{1 + \sin^2 v} dudv\). Computing the integral, you'll get the total area to be: \(A = 4 \pi^2 ab\).

Key concepts

Parametric Equations

A torus can be thought of like a doughnut-shaped surface, and understanding its shape is key in many fields such as physics, engineering, and computer graphics. To describe a torus mathematically, we use parametric equations which allow us to represent its surface in three-dimensional space using two parameters, typically denoted as u and v.

The parametric equations for a torus involve trigonometric functions that vary with these parameters, allowing every point on the surface to be defined by a unique pair of (u, v). In our example, the equations involve a constant a, which determines the distance from the center of the hole in the torus to the center of the tube, and b, which controls the radius of the tube itself. The use of cosine and sine functions facilitates the circular nature of the object.

Partial Derivatives

When it comes to finding the surface area of complex shapes like a torus, partial derivatives become crucial.

A partial derivative represents how a function changes as one of its variables is varied while keeping other variables constant. In the case of the torus, we calculate the partial derivatives of its parametric equations with respect to each parameter independently.

These derivatives, denoted as r_u and r_v, are the rates of change of the position vector r at any point on the surface with respect to u and v, respectively. The gradient vectors resulting from these derivatives will be pivotal in calculating the surface area by the means of a cross product.

Cross Product

The cross product of two vectors in three-dimensional space results in another vector that is orthogonal to the plane formed by the initial vectors. It’s an operation that combines aspects of both vectors in a way that’s invaluable for calculating the surface area of a parametrically defined surface like a torus.

Specifically, we take the cross product of the partial derivatives obtained earlier, r_u and r_v, which yields a new vector whose length corresponds to the area of the parallelogram spanned by r_u and r_v.

The magnitude of this cross product, denoted as |r_u × r_v|, thus represents the differential element of surface area on the torus. By integrating this magnitude over the entire surface, we can find the total surface area.

Double Integral

Integral calculus is all about summation, and when dealing with surfaces, this summation becomes a double integral because we are adding up over a two-dimensional region.

After computing the cross product of the partial derivatives, we integrate its magnitude over both parameters u and v. This process effectively sums the differential area elements across the surface of the torus. The double integral will take into account the limits of 0 to 2π for both u and v, reflecting the torus's full rotational symmetry.

A = ∬_S (a+b cos v) √(1 + sin² v) dudv.

By solving this double integral, we obtain a numerical value for the total surface area of the torus, completing our quest to describe this complex three-dimensional shape in terms of measurable quantity.