Question

The area of the parametrized surface \[ S = \left\{ \left( (2 + \cos u) \cos v, (2 + \cos u) \sin v, \sin u \right) \in \mathbb{R}^3 \mid 0 \leq u \leq \frac{\pi}{2}, 0 \leq v \leq \frac{\pi}{2} \right\} \] is ..................  (correct up to two decimal places). 
 

Hint:

For surface area calculations, use the formula involving the cross product of the partial derivatives of the parametric equations.

Answer 1

Correct Answer: 6.3 - 6.7

To find the area of the parametrized surface \( S \), we use the formula for the area of a parametrized surface: 
\[ A = \int \int_D \left\| \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} \right\| dudv \] where \( \mathbf{r}(u,v) = ((2+\cos u)\cos v,(2+\cos u)\sin v,\sin u) \) and \( D = \{ (u,v) \mid 0 \leq u \leq \frac{\pi}{2}, 0 \leq v \leq \frac{\pi}{2} \} \).

First, compute the partial derivatives: 
\(\frac{\partial \mathbf{r}}{\partial u} = (-\sin u \cos v, -\sin u \sin v, \cos u)\) 
\(\frac{\partial \mathbf{r}}{\partial v} = (-(2+\cos u)\sin v, (2+\cos u)\cos v, 0)\).

Now, find the cross product: 
\(\frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -\sin u \cos v & -\sin u \sin v & \cos u \\ -(2+\cos u)\sin v & (2+\cos u)\cos v & 0 \end{vmatrix}\). 
This evaluates to: 
\(((2+\cos u)\cos u \cos v, (2+\cos u)\cos u \sin v, \sin u(2+\cos u))\).

Calculate the magnitude: 
\(\left\| \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} \right\| = \sqrt{((2+\cos u)\cos u \cos v)^2 + ((2+\cos u)\cos u \sin v)^2 + (\sin u(2+\cos u))^2}\). 
This simplifies to: 
\(\sqrt{(2+\cos u)^2 \cos^2 u + \sin^2 u (2+\cos u)^2}\) 
\(\Rightarrow \sqrt{(2+\cos u)^2}\) since \(\cos^2 u + \sin^2 u = 1\).

The expression is \( (2+\cos u) \) since it is non-negative for \( 0 \leq u \leq \frac{\pi}{2} \). Thus, the area integral becomes: 
\[ A = \int_0^{\pi/2} \int_0^{\pi/2} (2+\cos u) \, dv \, du. \]

Evaluate the integral: 
\[ = \int_0^{\pi/2} \left[(2+\cos u)v\right]_0^{\pi/2} \, du \] 
\[ = \int_0^{\pi/2} \frac{\pi}{2}(2+\cos u) \, du \] 
\[ = \frac{\pi}{2} \int_0^{\pi/2} (2+\cos u) \, du \] 
\[ = \frac{\pi}{2} \left[ 2u + \sin u \right]_0^{\pi/2} \] 
\[ = \frac{\pi}{2} \left[ 2\cdot\frac{\pi}{2} + 1 - (0 + 0) \right] \] 
\[ = \frac{\pi}{2} \left[ \pi + 1 \right] \].

Thus, the area is \(\frac{\pi^2}{2} + \frac{\pi}{2}\). Using \(\pi \approx 3.14159\), this evaluates to approximately:
\((3.14159)^2/2 + 3.14159/2 \approx 6.505\).

Answer: 6.505