Question

Find the surface area of the surface having the given parametric equations: a) \(x=(b+a \cos v) \cos u, y=(b+a \cos v) \sin u, z=a \sin v, 0 \leq u \leq 2 \pi, 0 \leq v \leq\) \(2 \pi\), where \(a\) and \(b\) are constants, \(0

Short answer

Question: Find the surface area of (a) the torus with parametric surface representation \(x = (b+a\cos v)\cos u\), \(y = (b+a\cos v)\sin u\), \(z = a\sin v\), \(0 \leq u, v \leq 2\pi\); and (b) the portion of the saddle surface represented by \(x=u\cos v\), \(y = u\sin v\), \(z=uv^2\), \(0 \leq u \leq 1\), \(0 \leq v \leq \frac{\pi}{2}\) Answer: (a) The surface area of the torus is \(4\pi^2 a^2 (2b^2+a^2)\). (b) The surface area of the portion of the saddle surface can be found by evaluating the following integral: \(A = \int_{0}^{1} \int_{0}^{\pi/2} \sqrt{u^2(1+4\sin^2 v\cos^2v)} dudv\).

Step-by-step solution

(Step 1: Find Partial Derivatives)

Firstly, find the partial derivatives of the parametric equations with respect to u and v: \(x_u = -(b+a\cos v)\sin u\) \(y_u = (b+a\cos v)\cos u\) \(z_u = 0\) \(r_u = ⟨-(b+a\cos v)\sin u , (b+a\cos v)\cos u , 0⟩\) \(x_v = -a\sin v \cos u\) \(y_v = -a\sin v \sin u\) \(z_v = a\cos v\) \(r_v = ⟨-a\sin v \cos u , -a\sin v \sin u , a\cos v⟩\)

(Step 2: Compute the Cross Product)

Next, compute the cross product of the partial derivatives \(r_u\) and \(r_v\): \(r_u \times r_v = ⟨(b+a\cos v)\cos u * a \cos v, -(b+a\cos v)\sin u * a \cos v, -a\sin v * -(b+a\cos v)\sin u * -a\sin v * (b+a\cos v) \cos u⟩\) \(r_u \times r_v = ⟨a(b+a\cos v)\cos v \cos u, a(b+a\cos v)\cos v\sin u, -a^2\sin^2v(b+a\cos v)⟩\)

(Step 3: Calculate Magnitude of Cross Product)

Now, compute the magnitude of the cross product: \(||r_u \times r_v|| = \sqrt{(a(b+a\cos v)\cos v \cos u)^2 + (a(b+a\cos v)\cos v\sin u)^2 + (-a^2 \sin ^2 v(b+a\cos v))^2}\) \(||r_u \times r_v|| = \sqrt{a^2(b+a\cos v)^2 \cos^2 v}\)

(Step 4: Integrate Magnitude of Cross Product)

Finally, integrate the magnitude of the cross product over the domain T: \(A = \int_{0}^{2\pi} \int_{0}^{2\pi} ||r_u \times r_v|| dudv\) \(A = a^2 \cos^2 v \int_{0}^{2\pi} \int_{0}^{2\pi} (b+a\cos v)^2 dudv\) \(A = 4\pi^2 a^2 (2b^2+a^2)\) The surface area of the torus is \(4\pi^2 a^2 (2b^2+a^2)\). ##Part b - Portion of the Saddle Surface##

(Step 1: Find Partial Derivatives)

First, find the partial derivatives of the parametric equations with respect to u and v: \(x_u = \cos v\) \(y_u = \sin v\) \(z_u = 2u \sin 2v\) \(r_u = ⟨\cos v , \sin v , 2u\sin 2v⟩\) \(x_v = -u\sin v\) \(y_v = u\cos v\) \(z_v = u^2 \cos 2v\) \(r_v = ⟨-u\sin v , u\cos v , u^2 \cos 2v⟩\)

(Step 2: Compute the Cross Product)

Next, compute the cross product of the partial derivatives \(r_u\) and \(r_v\): \(r_u \times r_v = ⟨\sin v * u^2 \cos 2v - 2u \sin 2v * u\cos v, -\cos v * u^2 \cos 2v + (-u\sin v) * 2u \sin 2v, (\cos v * u\cos v) - (\sin v * -u\sin v)⟩\) \(r_u \times r_v = ⟨u^2 \sin v \cos 2v - 2u^2 \cos v \sin 2v, -u^2 \cos v \cos 2v + 2u^2\sin v \sin 2v , u⟩\)

(Step 3: Calculate Magnitude of Cross Product)

Now, compute the magnitude of the cross product: \(||r_u \times r_v|| = \sqrt{(u^2 \sin v \cos 2v - 2u^2 \cos v \sin 2v)^2 + (-u^2 \cos v \cos 2v + 2u^2\sin v\sin 2v)^2 + u^2}\)

(Step 4: Integrate Magnitude of Cross Product)

Finally, integrate the magnitude of the cross product over the domain T: \(A = \int_{0}^{1} \int_{0}^{\pi/2} ||r_u \times r_v|| dudv\) Our integral will require some elementary integration techniques like handling trigonometric terms and around these ranges of integration. Here's the final integral expression: \(A = \int_{0}^{1} \int_{0}^{\pi/2} \sqrt{u^2(1+4\sin^2 v\cos^2v)} dudv\) And the surface area of the portion of the saddle surface can be found by evaluating this integral.

Key concepts

Parametric Equations

When you encounter a complex shape like a torus or a saddle surface, expressing the geometry using conventional Cartesian coordinates can be challenging. This is where parametric equations shine. They provide a way to define a surface in terms of two parameters, generally denoted as u and v. For both the torus and the saddle surface in the exercise, you're given equations linking various values of u and v to points in three-dimensional space.
Parametric equations are potent because they can neatly describe curves and surfaces that would otherwise be difficult to represent. Think of it as assigning each point on the surface a set of coordinates that depend on u and v, like a grid over the shape, where you can locate each point using just two numbers instead of three. This grid-like system simplifies the process of analyzing complex surfaces, such as when calculating their areas.

Surface Area Calculation

Calculating the surface area of a parametrically defined shape is a bit trickier than finding the area of a flat two-dimensional object. You must take into account the curvature and complexity of the surface. Essentially, you're approximating the surface by an infinite number of infinitesimal patches, and then adding up the areas of all those patches.
To do this, one generally uses a double integral over the domain of the parameters u and v. The integrand is the magnitude of the cross product of the partial derivatives of the parametric equations, which gives us the area of each infinitesimal patch. The exercise demonstrates this by breaking down the complex problem into manageable steps – finding partial derivatives, computing their cross product, and then integrating.

Cross Product of Vectors

The cross product of vectors is a powerful tool in multivariable calculus, especially when dealing with parametric surfaces. Given two vectors in three dimensions, their cross product yields another vector that is orthogonal to both of the original vectors, and its length corresponds to the area of the parallelogram they span.

Why is this useful?

It turns out that if you find the cross product of the partial derivatives of the parametric equations – effectively, the vectors tangent to the surface at each point – you get a vector whose magnitude directly relates to the infinitesimal area element of the surface at that point. The steps in the exercise use this property to find a formula for each tiny patch's area on the surfaces mentioned.

Partial Derivatives

The concept of partial derivatives is central to working with multivariable functions, such as the parametric equations for surfaces. A partial derivative represents the rate at which the function changes as one variable varies while the others remain fixed.
In the context of parametric surfaces, you compute partial derivatives with respect to each parameter independently (u and v). These derivatives, presented in the exercise steps, essentially give you the tangent vectors along the surface. When you have two such tangent vectors, their cross product, as mentioned earlier, helps determine the surface area. Therefore, understanding partial derivatives is not just about finding rates of change; it's also about understanding the shape and geometry of the surfaces you're studying.