Surface Parameterization & Area - Calculus 3

1. A surface $S$ is parameterized by $\mathbf{r}(u,v) = \langle u + v, u - v, 2uv \rangle$ for $0 \leq u \leq 1$ and $0 \leq v \leq 1$. At which point $(u,v)$ in the parameter domain is the local surface area element $||\mathbf{r}_u \times \mathbf{r}_v||$ minimized?
   Answer: $(0, 0)$ When you're asked to find where a surface area element is minimized, you need to compute the magnitude of the cross product of the partial derivatives of the parameterization. This measures how much the surface is "stretched" at each point. First, find the partial derivatives: $\mathbf{r}_u = \langle 1, 1, 2v \rangle$ and $\mathbf{r}_v = \langle 1, -1, 2u \rangle$. Next, compute their cross product: $\mathbf{r}_u \times \mathbf{r}_v = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 2v \\ 1 & -1 & 2u \end{vmatrix} = \langle 2u + 2v, -2u + 2v, -2 \rangle$ The magnitude is: $||\mathbf{r}_u \times \mathbf{r}_v|| = \sqrt{(2u + 2v)^2 + (-2u + 2v)^2 + 4}$ Expanding: $(2u + 2v)^2 + (-2u + 2v)^2 = 4u^2 + 8uv + 4v^2 + 4u^2 - 8uv + 4v^2 = 8u^2 + 8v^2$ So $||\mathbf{r}_u \times \mathbf{r}_v|| = \sqrt{8u^2 + 8v^2 + 4} = 2\sqrt{2u^2 + 2v^2 + 1}$ This is minimized when $u^2 + v^2$ is minimized, which occurs at $(0,0)$, making C correct. Option A $(0.5, 0.5)$ gives a larger value since both coordinates are positive. Option B $(1,1)$ gives the maximum value in the domain. Option D is wrong because the expression clearly depends on $u$ and $v$. Study tip: For surface area problems, always compute $||\mathbf{r}_u \times \mathbf{r}_v||$ and look for where this expression reaches its extrema by treating it as an optimization problem in the parameter variables.
2. Consider the surface $S$ parameterized by $\mathbf{r}(u,v) = \langle u\cos v, u\sin v, u^2 \rangle$ where $0 \leq u \leq 2$ and $0 \leq v \leq \pi$. If the surface area element is given by $dS = ||\mathbf{r}_u \times \mathbf{r}_v|| \, du \, dv$, what is $||\mathbf{r}_u \times \mathbf{r}_v||$?
   Answer: $u\sqrt{1 + 4u^2}$ First, compute the partial derivatives: $\mathbf{r}_u = \langle \cos v, \sin v, 2u \rangle$ and $\mathbf{r}_v = \langle -u\sin v, u\cos v, 0 \rangle$. The cross product is $\mathbf{r}_u \times \mathbf{r}_v = \langle -2u^2\cos v, -2u^2\sin v, u \rangle$. Taking the magnitude: $||\mathbf{r}_u \times \mathbf{r}_v|| = \sqrt{4u^4\cos^2 v + 4u^4\sin^2 v + u^2} = \sqrt{4u^4 + u^2} = u\sqrt{4u^2 + 1}$. Choice B incorrectly computes the coefficient inside the square root. Choice C factors incorrectly, missing the factor of $u$ outside. Choice D omits the $u$ factor entirely.
3. Consider the surface $z = xy$ over the triangular region in the $xy$-plane with vertices at $(0,0)$, $(2,0)$, and $(0,3)$. Using the parameterization $\mathbf{r}(x,y) = \langle x, y, xy \rangle$, what is the surface area element $dS$ in terms of $dx$ and $dy$?
   Answer: $\sqrt{1 + x^2 + y^2} \, dx \, dy$ When you encounter a surface area problem with a parametric surface, you need to find the surface area element $dS$ using the cross product of partial derivatives. This tests your understanding of how surfaces "stretch" in 3D space. For the parameterization $\mathbf{r}(x,y) = \langle x, y, xy \rangle$, you first find the partial derivatives: $\mathbf{r}_x = \langle 1, 0, y \rangle$ and $\mathbf{r}_y = \langle 0, 1, x \rangle$. The surface area element is $dS = |\mathbf{r}_x \times \mathbf{r}_y| \, dx \, dy$. Computing the cross product: $\mathbf{r}_x \times \mathbf{r}_y = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & y \\ 0 & 1 & x \end{vmatrix} = \langle -y, -x, 1 \rangle$ The magnitude is $|\mathbf{r}_x \times \mathbf{r}_y| = \sqrt{(-y)^2 + (-x)^2 + 1^2} = \sqrt{x^2 + y^2 + 1}$. Therefore, $dS = \sqrt{1 + x^2 + y^2} \, dx \, dy$, which is answer C. Answer A, $\sqrt{1 + x + y}$, incorrectly uses linear terms instead of squared terms—this would come from forgetting to square the components when finding the magnitude. Answer B, $\sqrt{1 + x^2 + y^2 + 2xy}$, includes an extra $2xy$ term that would result from incorrectly expanding $(x+y)^2$ instead of computing $x^2 + y^2$. Answer D, $\sqrt{1 + xy}$, mistakenly uses the surface equation $z = xy$ directly rather than the cross product magnitude. Remember: surface area elements always involve the magnitude of $\mathbf{r}_x \times \mathbf{r}_y$—never use the surface equation directly in your calculation.
4. A surface $S$ is parameterized by $\mathbf{r}(s,t) = \langle s + t, s - t, st \rangle$ for $0 \leq s \leq 1$ and $0 \leq t \leq 1$. Which expression correctly represents the surface area element for integration?
   Answer: $\sqrt{2s^2 + 2t^2 + 8} \, ds \, dt$ Computing the partial derivatives: $\mathbf{r}_s = \langle 1, 1, t \rangle$ and $\mathbf{r}_t = \langle 1, -1, s \rangle$. The cross product is $\mathbf{r}_s \times \mathbf{r}_t = \langle s + t, s - t, -2 \rangle$. The magnitude is $||\mathbf{r}_s \times \mathbf{r}_t|| = \sqrt{(s+t)^2 + (s-t)^2 + 4} = \sqrt{s^2 + 2st + t^2 + s^2 - 2st + t^2 + 4} = \sqrt{2s^2 + 2t^2 + 8}$. Choice B misses a factor of 2. Choice C has the wrong constant term. Choice D doubles all coefficients incorrectly.
5. A surface is parameterized by $\mathbf{r}(u,v) = \langle u\cos v, u\sin v, u + v \rangle$ for $u > 0$. At what angle do the parameter curves (curves of constant $u$ and curves of constant $v$) intersect on the surface?
   Answer: $\cos^{-1}\left(\frac{1}{\sqrt{u^2 + 2}}\right)$ The angle between parameter curves is determined by the angle between the tangent vectors $\mathbf{r}_u$ and $\mathbf{r}_v$. We have $\mathbf{r}_u = \langle \cos v, \sin v, 1 \rangle$ and $\mathbf{r}_v = \langle -u\sin v, u\cos v, 1 \rangle$. The angle $\theta$ satisfies $\cos \theta = \frac{\mathbf{r}_u \cdot \mathbf{r}_v}{||\mathbf{r}_u|| \cdot ||\mathbf{r}_v||}$. Computing: $\mathbf{r}_u \cdot \mathbf{r}_v = -u\cos v \sin v + u\sin v \cos v + 1 = 1$, $||\mathbf{r}_u|| = \sqrt{\cos^2 v + \sin^2 v + 1} = \sqrt{2}$, and $||\mathbf{r}_v|| = \sqrt{u^2\sin^2 v + u^2\cos^2 v + 1} = \sqrt{u^2 + 1}$. Therefore, $\cos \theta = \frac{1}{\sqrt{2}\sqrt{u^2 + 1}} = \frac{1}{\sqrt{2(u^2 + 1)}} = \frac{1}{\sqrt{2u^2 + 2}}$. Choice A assumes orthogonality. Choice C omits the $u$-dependence. Choice D incorrectly suggests $v$-dependence.
6. Consider the surface $\mathbf{r}(u,v) = \langle u, v^2, uv \rangle$ where $0 \leq u \leq 2$ and $0 \leq v \leq 1$. If we reparameterize using $s = u + v$ and $t = u - v$, what constraint must be placed on the $(s,t)$ domain to cover the same surface region?
   Answer: $0 \leq s \leq 3$ and $-1 \leq t \leq 2$ with $s \geq |t|$ and $s + t \leq 4$ and $s - t \geq 0$ From $s = u + v$ and $t = u - v$, we get $u = (s+t)/2$ and $v = (s-t)/2$. The original constraints $0 \leq u \leq 2$ and $0 \leq v \leq 1$ become: $0 \leq (s+t)/2 \leq 2$ (so $0 \leq s+t \leq 4$) and $0 \leq (s-t)/2 \leq 1$ (so $0 \leq s-t \leq 2$). Additionally, we need $v \geq 0$, which gives $s \geq t$, and $u \geq 0$, which gives $s \geq -t$. Combined, this is $s \geq |t|$. The range of $s$ is $0 \leq s \leq 3$ (when $u=2, v=1$) and $t$ ranges from $-1$ (when $u=0, v=1$) to $2$ (when $u=2, v=0$). Choice A omits the additional constraints. Choice B uses incorrect bounds. Choice C is missing the constraint $s \geq |t|$.
7. A surface is given by the vector function $\mathbf{r}(u,v) = \langle u\cos v, u\sin v, v \rangle$ where $u \geq 0$ and $0 \leq v \leq 2\pi$. If we want to compute the surface area over the region where $1 \leq u \leq 3$ and $0 \leq v \leq \pi$, which integral should be evaluated?
   Answer: $\int_0^\pi \int_1^3 \sqrt{u^2 + 1} \, du \, dv$ Computing $\mathbf{r}_u = \langle \cos v, \sin v, 0 \rangle$ and $\mathbf{r}_v = \langle -u\sin v, u\cos v, 1 \rangle$. The cross product is $\mathbf{r}_u \times \mathbf{r}_v = \langle \sin v, -\cos v, u \rangle$. The magnitude is $||\mathbf{r}_u \times \mathbf{r}_v|| = \sqrt{\sin^2 v + \cos^2 v + u^2} = \sqrt{1 + u^2}$. The surface area integral is $\int_0^\pi \int_1^3 \sqrt{u^2 + 1} \, du \, dv$. Choice A incorrectly includes an extra factor of $u$. Choice C omits the square root entirely. Choice D has the wrong integrand and uses $\sqrt{2}$ instead of the correct expression.

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