Choosing Bounds & Parametrizations - Calculus 3

1. Consider the curve $C$ that traces the intersection of the cylinder $x^2 + y^2 = 4$ and the plane $z = x + y$ exactly once in the direction of increasing $y$ when $x > 0$. Which parametrization correctly represents this curve?
   Answer: $\mathbf{r}(t) = (2\cos t, 2\sin t, 2\cos t + 2\sin t)$ for $-\pi/2 \leq t \leq \pi/2$ The curve is the intersection of $x^2 + y^2 = 4$ and $z = x + y$. Using $x = 2\cos t$, $y = 2\sin t$ gives $z = 2\cos t + 2\sin t$. For the curve to be traced exactly once with increasing $y$ when $x > 0$, we need $\frac{dy}{dt} = 2\cos t > 0$ when $x = 2\cos t > 0$. This occurs when $-\pi/2 < t < \pi/2$. Choice A traces the curve multiple times. Choice C has the wrong parameter range (gives decreasing $y$ when $x > 0$). Choice D uses a different parametrization that doesn't satisfy the directional requirement.
2. A surface $S$ is defined by $\mathbf{r}(u,v) = (u + v, u - v, uv)$ where $0 \leq u \leq 2$ and $0 \leq v \leq 1$. What is the relationship between the parameters and the coordinate bounds for this surface?
   Answer: The surface is contained within $0 \leq x \leq 3$, $-1 \leq y \leq 2$, $0 \leq z \leq 2$ but doesn't cover this entire region With $0 \leq u \leq 2$ and $0 \leq v \leq 1$: $x = u + v$ ranges from $0$ to $3$, $y = u - v$ ranges from $-1$ to $2$, and $z = uv$ ranges from $0$ to $2$. However, not every point $(x,y,z)$ in this box is on the surface. For example, the point $(3, 2, 2)$ would require $u + v = 3$, $u - v = 2$, and $uv = 2$, giving $u = 2.5$ and $v = 0.5$, but then $uv = 1.25 \neq 2$. The surface is contained within the box but doesn't fill it completely due to the constraint $z = xy/2$ (since $z = uv$ and $u = (x+y)/2$, $v = (x-y)/2$). Choice A and B incorrectly claim the entire region is covered. Choice D has wrong coordinate bounds.
3. To evaluate $\iint_R (x^2 + y^2) \, dA$ over the region $R$ bounded by $x^2 + y^2 = 4x$, a student considers several approaches. Which transformation leads to the simplest bounds?
   Answer: $x = 2 + r\cos\theta, y = r\sin\theta$ with $0 \leq r \leq 2, 0 \leq \theta \leq 2\pi$ When evaluating double integrals, choosing the right coordinate transformation can dramatically simplify your bounds of integration. The key is to analyze the region's geometry and select coordinates that naturally match its shape. First, let's understand the region $R$. The boundary $x^2 + y^2 = 4x$ can be rewritten by completing the square: $x^2 - 4x + y^2 = 0$ becomes $(x-2)^2 + y^2 = 4$. This is a circle with center $(2,0)$ and radius $2$. Option D gives us the simplest bounds because it uses polar coordinates centered at $(2,0)$ - exactly where the circle is centered. The transformation $x = 2 + r\cos\theta, y = r\sin\theta$ shifts the origin to the circle's center, making the region simply $0 \leq r \leq 2, 0 \leq \theta \leq 2\pi$. Option A uses standard polar coordinates, but since the circle isn't centered at the origin, you get the awkward bound $r \leq 4\cos\theta$, which only works for $-\pi/2 \leq \theta \leq \pi/2$ (the cosine becomes negative outside this range). Option B incorrectly assumes the region is a circle of radius 2 centered at the origin - this doesn't match our actual region at all. Option C uses a translation that centers the circle correctly, but then describes it in Cartesian coordinates with the constraint $u^2 + v^2 \leq 4$, which doesn't simplify the integration of $x^2 + y^2$. Study tip: Always complete the square to identify the center and radius of circular regions, then choose coordinates that place the origin at the natural center of your region.
4. A solid region $E$ is defined by $x^2 + y^2 + z^2 \leq 16$, $z \geq 0$, and $x^2 + y^2 \geq 4$. To evaluate $\iiint_E f(x,y,z) \, dV$ most efficiently, which coordinate system and bounds should be used?
   Answer: Cylindrical: $\int_0^{2\pi} \int_2^4 \int_0^{\sqrt{16-r^2}} f(r\cos\theta, r\sin\theta, z) r \, dz \, dr \, d\theta$ The region is a hemispherical shell: upper half of sphere of radius 4 with cylindrical hole of radius 2. Cylindrical coordinates work well since the constraint $x^2 + y^2 \geq 4$ becomes $r \geq 2$, and $x^2 + y^2 + z^2 \leq 16$ with $z \geq 0$ gives $0 \leq z \leq \sqrt{16-r^2}$. So $2 \leq r \leq 4$. Choice A uses incorrect $\rho$ bounds. Choice C attempts to handle the cylindrical hole in spherical coordinates with $\rho \geq 2/\sin\phi$, but this is overly complex. Choice D has incorrect $r$ bounds starting from 0 and incorrect lower $z$ bound.
5. A vector field $\mathbf{F}$ is to be integrated over the helical curve $\mathbf{r}(t) = (\cos t, \sin t, t)$ from the point $(1, 0, 0)$ to $(1, 0, 4\pi)$. What parameter bounds ensure the curve is traversed exactly twice around the cylinder in the correct direction?
   Answer: $0 \leq t \leq 4\pi$ The helix starts at $(1, 0, 0)$ when $t = 0$ and reaches $(1, 0, 4\pi)$ when $t = 4\pi$. Each complete revolution around the cylinder ($x^2 + y^2 = 1$) corresponds to $t$ increasing by $2\pi$. To go from $z = 0$ to $z = 4\pi$ requires $t$ to go from $0$ to $4\pi$, which gives exactly $\frac{4\pi}{2\pi} = 2$ complete revolutions. Choice B only gives one revolution, ending at $(1, 0, 2\pi)$. Choice C gives four revolutions, ending at $(1, 0, 8\pi)$. Choice D gives two revolutions but starts from the wrong point and ends at the wrong height.
6. Consider the surface $z = x^2 + y^2$ above the region in the $xy$-plane bounded by $x^2 + y^2 = 4$ and $x^2 + y^2 = 9$, where $x \geq 0$ and $y \geq 0$. Which parametrization is most appropriate for computing $\iint_S f(x,y,z) \, dS$?
   Answer: $\mathbf{r}(r,\theta) = (r\cos\theta, r\sin\theta, r^2)$ where $2 \leq r \leq 3$ and $0 \leq \theta \leq \pi/2$ The surface $z = x^2 + y^2$ is a paraboloid, and the domain is an annular sector in the first quadrant. Cylindrical coordinates are natural here. Since $x^2 + y^2 = r^2$ ranges from 4 to 9, we need $r$ from 2 to 3. The first quadrant corresponds to $0 \leq \theta \leq \pi/2$. Choice B uses Cartesian coordinates, making the bounds more complex. Choice C incorrectly states $4 \leq r^2 \leq 9$ as bounds for $r$ and $\theta$ rather than proper parameter ranges. Choice D uses spherical coordinates, which is inappropriate for a paraboloid surface.
7. Consider the surface $z = \sqrt{x^2 + y^2}$ above the region where $1 \leq x^2 + y^2 \leq 4$ and $y \geq |x|$. For computing $\iint_S f(x,y,z) \, dS$, which parametrization and bounds are most appropriate?
   Answer: $\mathbf{r}(r,\theta) = (r\cos\theta, r\sin\theta, r)$ with $1 \leq r \leq 2$ and $\pi/4 \leq \theta \leq 3\pi/4$ When setting up surface integrals using parametrization, you need to carefully analyze both the surface equation and the region constraints to determine appropriate parameter bounds. The surface $z = \sqrt{x^2 + y^2}$ is a cone, and the cylindrical parametrization $\mathbf{r}(r,\theta) = (r\cos\theta, r\sin\theta, r)$ is ideal since $z = r = \sqrt{x^2 + y^2}$. The condition $1 \leq x^2 + y^2 \leq 4$ translates directly to $1 \leq r \leq 2$. The key challenge is interpreting $y \geq |x|$. This inequality describes the region where $y$ is at least as large as the absolute value of $x$. In polar coordinates, this becomes $r\sin\theta \geq |r\cos\theta|$, which simplifies to $\sin\theta \geq |\cos\theta|$. This condition is satisfied when $\pi/4 \leq \theta \leq 3\pi/4$, making option D correct. Option A uses $\pi/3 \leq \theta \leq 2\pi/3$, which is too narrow and misses parts of the region near the lines $y = x$ and $y = -x$. Option B spans $\pi/6 \leq \theta \leq 5\pi/6$, which is too wide and includes regions where $y < |x|$. Option C covers $3\pi/4 \leq \theta \leq 5\pi/4$, which extends into the third quadrant where $y < 0$ and violates $y \geq |x|$. Study tip: When dealing with absolute value constraints like $y \geq |x|$, sketch the region first. The boundary lines $y = x$ and $y = -x$ correspond to $\theta = \pi/4$ and $\theta = 3\pi/4$ respectively, making the angular bounds clear.

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