Choosing Parameterizations - Multivariable Calculus

1. A curve in 3D space is defined as the intersection of the surfaces $x^2 + y^2 = 4z$ and $x + y + z = 6$. To parameterize this curve efficiently for computing a line integral, which single-parameter approach is most appropriate?
   Answer: Use the constraint to write $z = 6 - x - y$, substitute into $x^2 + y^2 = 4z$ to get $(x+2)^2 + (y+2)^2 = 32$, then parameterize this circle Substituting $z = 6 - x - y$ into $x^2 + y^2 = 4z$ gives $x^2 + y^2 = 4(6-x-y) = 24 - 4x - 4y$. Rearranging: $x^2 + 4x + y^2 + 4y = 24$. Completing the square: $(x+2)^2 + (y+2)^2 = 24 + 4 + 4 = 32$. This is a circle with center $(-2, -2)$ and radius $4\sqrt{2}$, parameterized as $x = -2 + 4\sqrt{2}\cos t$, $y = -2 + 4\sqrt{2}\sin t$, $z = 6 - x - y = 10 - 4\sqrt{2}(\cos t + \sin t)$. Choice A incorrectly assumes the projection is a circle of radius 2. Choice B introduces an unnecessary second parameter. Choice D has an artificial and complicated relationship between parameters that doesn't simplify the problem.
2. The surface $S$ is defined by $z = \ln(x^2 + y^2)$ for $1 \leq x^2 + y^2 \leq e^2$. For evaluating $\iint_S f(x,y,z)\,dS$ where $f$ depends significantly on the distance from the z-axis, which parameterization strategy provides the most computational advantage?
   Answer: Cylindrical: $\mathbf{r}(r,\theta) = (r\cos\theta, r\sin\theta, 2\ln r)$ with $1 \leq r \leq e$, $0 \leq \theta \leq 2\pi$ Since $f$ depends significantly on the distance from the z-axis (which is $\sqrt{x^2 + y^2} = r$), cylindrical coordinates are most appropriate. The surface becomes $z = \ln(r^2) = 2\ln r$, so the parameterization is $\mathbf{r}(r,\theta) = (r\cos\theta, r\sin\theta, 2\ln r)$ with $1 \leq r \leq e$ (since $1 \leq x^2 + y^2 \leq e^2$ means $1 \leq r^2 \leq e^2$, so $1 \leq r \leq e$) and $0 \leq \theta \leq 2\pi$. This makes the dependence on distance from the z-axis explicit through the parameter $r$. Choice A uses Cartesian coordinates which obscures the radial structure. Choice B incorrectly writes $z = \ln r^2$ without simplifying to $2\ln r$. Choice D introduces an unnecessarily complex substitution that doesn't simplify the problem.
3. A space curve is defined implicitly as the intersection of $x^2 + z^2 = 9$ and $y^2 + z^2 = 9$. For parameterizing this curve to compute its arc length, which single-parameter approach avoids issues with undefined derivatives at specific points?
   Answer: Use $z = 3\cos t$ as the parameter, then $x = \pm 3\sin t$ and $y = \pm 3\sin t$, choosing signs to maintain continuity The intersection of $x^2 + z^2 = 9$ and $y^2 + z^2 = 9$ gives us $x^2 = y^2$, so $x = \pm y$. This means the curve consists of two components: one where $x = y$ and one where $x = -y$. Both lie on circles of radius 3. Using $z = 3\cos t$ as parameter gives $x^2 = 9 - z^2 = 9\sin^2 t$ and $y^2 = 9 - z^2 = 9\sin^2 t$, so $x = \pm 3\sin t$ and $y = \pm 3\sin t$. To avoid derivative issues, we choose signs consistently for each component. Choice B only captures one component ($x = y$). Choice C introduces absolute value functions that create derivative discontinuities at $t = \frac{\pi}{2}, \frac{3\pi}{2}$. Choice D also only captures one component and has the same limitation as Choice B.
4. Consider the surface defined implicitly by $x^2 + y^2 - z^2 = 1$ for $z \geq 0$. You need to parameterize the portion of this surface where $1 \leq x^2 + y^2 \leq 4$. Which approach provides the most efficient parameterization that avoids singularities?
   Answer: Use cylindrical coordinates: $\mathbf{r}(r,\theta) = (r\cos\theta, r\sin\theta, \sqrt{r^2-1})$ with $1 \leq r \leq 2$, $0 \leq \theta \leq 2\pi$ This is a hyperboloid of one sheet. The constraint $1 \leq x^2 + y^2 \leq 4$ naturally suggests cylindrical coordinates where $r = \sqrt{x^2+y^2}$, so $1 \leq r \leq 2$. From the surface equation $x^2 + y^2 - z^2 = 1$, we get $z = \sqrt{r^2-1}$ (taking positive root since $z \geq 0$). This parameterization is well-defined and smooth throughout the domain. Choice B incorrectly uses $\rho$ in place of $r$ and has wrong bounds. Choice C uses hyperbolic functions which are unnecessarily complex for this region and the bounds are incorrect. Choice D is problematic because it doesn't clearly specify the domain and can lead to computational difficulties when integrating.
5. You need to parameterize the part of the paraboloid $z = x^2 + y^2$ that lies between the cylinders $x^2 + y^2 = 1$ and $x^2 + y^2 = 4$, but only in the region where $x \geq |y|$. Which parameterization correctly captures this region and is suitable for surface integration?
   Answer: $\mathbf{r}(r,\theta) = (r\cos\theta, r\sin\theta, r^2)$ with $1 \leq r \leq 2$ and $-\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4}$ or $\frac{3\pi}{4} \leq \theta \leq \frac{5\pi}{4}$ The condition $x \geq |y|$ means $x \geq y$ when $y \geq 0$ and $x \geq -y$ when $y < 0$. In polar coordinates, this corresponds to two angular regions: the first quadrant and parts of the second and fourth quadrants where $|\cos\theta| \geq |\sin\theta|$. For $\theta \in [0, \frac{\pi}{4}]$, we have $\cos\theta \geq \sin\theta \geq 0$. For $\theta \in [-\frac{\pi}{4}, 0]$, we have $\cos\theta \geq -\sin\theta \geq 0$. By symmetry, we also need $\theta \in [\frac{3\pi}{4}, \frac{5\pi}{4}]$ to capture the region where $x \leq 0$ and $x \geq |y|$. Choice A misses the third quadrant portion. Choice B uses incorrect union notation that suggests simultaneous constraints rather than separate intervals. Choice C uses Cartesian parameterization which is less natural for the annular region and makes integration more complex.
6. Consider parameterizing the surface $z = xy$ over the region where $x^2 + y^2 \leq 4$ and $x \geq 0$. For computing the flux of the vector field $\mathbf{F} = z\mathbf{k}$ through this surface, which parameterization choice will result in the normal vector pointing generally upward (positive z-component)?
   Answer: $\mathbf{r}(x,y) = (x, y, xy)$ with $x^2 + y^2 \leq 4$, $x \geq 0$, computed as $\mathbf{r}_x \times \mathbf{r}_y$ For a surface parameterized as $\mathbf{r}(u,v)$, the normal vector is $\mathbf{r}_u \times \mathbf{r}_v$. For $\mathbf{r}(x,y) = (x, y, xy)$, we have $\mathbf{r}_x = (1, 0, y)$ and $\mathbf{r}_y = (0, 1, x)$. Thus $\mathbf{r}_x \times \mathbf{r}_y = (-y, -x, 1)$, which has a positive z-component. Choice A uses the correct region but the cross product $\mathbf{r}_r \times \mathbf{r}_{\theta}$ may not consistently point upward. Choice C computes $\mathbf{r}_v \times \mathbf{r}_u$ instead of $\mathbf{r}_u \times \mathbf{r}_v$, which reverses the orientation and gives a downward normal. Choice D parameterizes the wrong half of the disk ($x \leq 0$ instead of $x \geq 0$).
7. Consider the curve formed by the intersection of the ellipsoid $\frac{x^2}{4} + \frac{y^2}{9} + \frac{z^2}{16} = 1$ and the plane $x + 2y - z = 0$. To parameterize this curve for computing work done by a conservative vector field, which approach most efficiently handles the constraint while maintaining parameter bounds that are easy to determine?
   Answer: Parameterize the plane as $\mathbf{p}(u,v) = (u, v, u + 2v)$ and substitute into the ellipsoid equation to find the constraint relating $u$ and $v$, then use this constraint to eliminate one parameter Method C is most systematic. Parameterizing the plane as $(u, v, u + 2v)$ and substituting into the ellipsoid gives $\frac{u^2}{4} + \frac{v^2}{9} + \frac{(u + 2v)^2}{16} = 1$. Expanding: $\frac{u^2}{4} + \frac{v^2}{9} + \frac{u^2 + 4uv + 4v^2}{16} = 1$, which simplifies to $\frac{5u^2}{16} + \frac{uv}{4} + \frac{25v^2}{144} = 1$. This can be written in the form of an ellipse and parameterized trigonometrically. Choice A leads to a complicated quartic relationship that's difficult to solve explicitly. Choice B introduces two parameters unnecessarily and requires finding a complex relationship between them. Choice D assumes a specific radial form that may not match the actual intersection curve and could lead to $r$ being multi-valued as a function of $\theta$.

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