Common Setup Pitfalls - Calculus 3

1. A student parametrizes the curve of intersection of $x^2 + y^2 = 1$ and $z = xy$ as $\mathbf{r}(t) = \langle \cos t, \sin t, \cos t \sin t \rangle$ for $t \in [0, 2\pi]$. When computing $\int_C z \, ds$, they use $ds = |\mathbf{r}'(t)| dt$ and find $|\mathbf{r}'(t)| = \sqrt{\sin^2 t + \cos^2 t + (\cos^2 t - \sin^2 t)^2}$. What should they check before proceeding?
   Answer: Whether the parametrization traces the curve exactly once, since $\cos t \sin t$ has period $\pi$, not $2\pi$ The function $z = xy = \cos t \sin t = \frac{1}{2}\sin(2t)$ has period $\pi$, meaning the $z$-values repeat after $t = \pi$. However, the $(x,y)$ coordinates $(\cos t, \sin t)$ have period $2\pi$. This means the curve $\mathbf{r}(t)$ traces the same spatial curve twice as $t$ goes from $0$ to $2\pi$. For a line integral $\int_C z \, ds$, this would double-count the integral. The student should use $t \in [0, \pi]$ or $t \in [0, 2\pi]$ depending on whether they want to traverse the curve once or twice. Choice B misunderstands the derivative calculation. Choice C incorrectly focuses on the curve being closed. Choice D has the right period observation but wrong reasoning about bounds.
2. A student attempts to parametrize the surface $z = \sqrt{x^2 + y^2}$ for $1 \leq x^2 + y^2 \leq 4$ using $\mathbf{r}(u,v) = \langle u \cos v, u \sin v, u \rangle$ where $1 \leq u \leq 2$ and $0 \leq v \leq 2\pi$. What is wrong with this parametrization?
   Answer: Nothing is wrong; this correctly parametrizes the cone $z = \sqrt{x^2 + y^2}$ for the given domain The parametrization is actually correct. We have $x^2 + y^2 = u^2\cos^2 v + u^2\sin^2 v = u^2$, so $\sqrt{x^2 + y^2} = u$ (since $u > 0$). The constraint $1 \leq x^2 + y^2 \leq 4$ becomes $1 \leq u^2 \leq 4$, giving $1 \leq u \leq 2$. The full range $0 \leq v \leq 2\pi$ is needed to cover the complete annular region. Choice B incorrectly suggests $z = u^2$. Choices A and D identify non-existent problems.
3. When computing $\iint_S \mathbf{F} \cdot d\mathbf{S}$ where $\mathbf{F} = \langle x, y, z \rangle$ and $S$ is the surface $z = 1 - x^2 - y^2$ for $z \geq 0$, a student parametrizes using $\mathbf{r}(x,y) = \langle x, y, 1-x^2-y^2 \rangle$ and computes $\mathbf{r}_x \times \mathbf{r}_y = \langle 2x, 2y, 1 \rangle$. What error will this likely lead to?
   Answer: The normal vector points inward rather than outward, which will give the flux integral the wrong sign For $\mathbf{r}(x,y) = \langle x, y, 1-x^2-y^2 \rangle$, we have $\mathbf{r}_x = \langle 1, 0, -2x \rangle$ and $\mathbf{r}_y = \langle 0, 1, -2y \rangle$. Thus $\mathbf{r}_x \times \mathbf{r}_y = \langle 2x, 2y, 1 \rangle$. However, this normal points inward (toward the concave side of the paraboloid). For a surface flux integral, we typically want the outward normal $\langle -2x, -2y, -1 \rangle$. Choice A gives the correct cross product but wrong reasoning. Choice B correctly identifies the domain but misses the orientation issue. Choice D suggests an unnecessary change of coordinates that doesn't address the fundamental orientation problem.
4. A student uses Stokes' theorem to evaluate $\iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{n} \, dS$ where $S$ is the part of $z = x^2 + y^2$ with $x^2 + y^2 \leq 1$, and $\mathbf{F} = \langle yz, xz, xy \rangle$. They parametrize the boundary circle as $\mathbf{r}(t) = \langle \cos t, \sin t, 1 \rangle$ for $t \in [0, 2\pi]$. What's problematic about this approach?
   Answer: Stokes' theorem requires the surface to be oriented consistently with the boundary curve orientation, which needs verification Stokes' theorem requires that the orientation of the surface $S$ be consistent with the orientation of its boundary curve $C$. The student parametrized the boundary but didn't verify that the surface orientation (determined by the normal vector) follows the right-hand rule relative to the boundary orientation. If $\mathbf{r}(t)$ traverses the boundary counterclockwise when viewed from above, then the surface normal should point upward. The student needs to check that their surface parametrization gives the correct normal direction. Choice A correctly verifies the boundary parametrization. Choice C mentions computational complexity but that's not a fundamental error. Choice D incorrectly states that Stokes' theorem doesn't apply - it does apply to this simply connected surface with a well-defined boundary.
5. When setting up a double integral to find the area of the region where $x^2 + y^2 \leq 4$ and $y \geq x^2 - 1$, a student writes $\int_{-2}^2 \int_{x^2-1}^{\sqrt{4-x^2}} dy \, dx$. What assumption makes this setup potentially incorrect?
   Answer: The student assumed $x^2 - 1 \leq \sqrt{4-x^2}$ for all $x \in [-2,2]$, which fails when $|x| > \sqrt{3}$ For the integral setup to be valid, we need $x^2 - 1 \leq \sqrt{4-x^2}$ throughout the integration region. Squaring both sides (valid since both sides are positive when $x^2 - 1 > 0$), we get $x^4 - 2x^2 + 1 \leq 4 - x^2$, which simplifies to $x^4 - x^2 - 3 \leq 0$. This fails when $|x| > \sqrt{3} \approx 1.73$. For $|x| > \sqrt{3}$, the parabola lies above the circle, making the integral limits invalid. Choice B is wrong because symmetry isn't required. Choice C misses the real issue about limit ordering. Choice D incorrectly focuses on the lower semicircle.
6. When finding the mass of a lamina with density $\rho(x,y) = x^2 + y^2$ over the region $R = \{(x,y) : x^2 + y^2 \leq 4, y \geq |x|\}$, a student converts to polar coordinates and writes $\int_0^{2\pi} \int_0^2 r^2 \cdot r \, dr \, d\theta$. What constraint did they ignore?
   Answer: The region $R$ doesn't cover the full disk $x^2 + y^2 \leq 4$ due to the constraint $y \geq |x|$ The constraint $y \geq |x|$ significantly restricts the region $R$ to only part of the disk $x^2 + y^2 \leq 4$. In polar coordinates, $y \geq |x|$ means $r\sin\theta \geq |r\cos\theta|$, which simplifies to $\sin\theta \geq |\cos\theta|$ (for $r > 0$). This gives $\theta \in [\pi/4, 3\pi/4]$, not $[0, 2\pi]$. The student's integral covers the entire disk instead of just the sector where $y \geq |x|$. Choice A correctly identifies the density conversion. Choice C incorrectly suggests wrong coordinate system. Choice D incorrectly interprets how the constraint affects the radial variable.
7. When applying the divergence theorem to $\iiint_E \nabla \cdot \mathbf{F} \, dV = \iint_{\partial E} \mathbf{F} \cdot \mathbf{n} \, dS$ where $E$ is the region inside $x^2 + y^2 + z^2 = 1$ and outside $x^2 + y^2 = \frac{1}{4}$ (cylindrical hole), a student only computes the flux through the outer sphere. What's the issue?
   Answer: The boundary $\partial E$ includes both the sphere and the cylindrical surface, so flux through both surfaces must be computed When applying the divergence theorem, you must carefully identify the complete boundary of your region. The divergence theorem relates the triple integral of divergence over a region to the flux through the entire boundary of that region. Here, your region $E$ sits inside the unit sphere but has a cylindrical hole carved out (where $x^2 + y^2 = \frac{1}{4}$ within the sphere). This creates a region with two distinct boundary surfaces: the outer spherical surface $x^2 + y^2 + z^2 = 1$ and the inner cylindrical surface $x^2 + y^2 = \frac{1}{4}$ (bounded by $-1 \leq z \leq 1$). The divergence theorem requires computing flux through both surfaces. Answer D is correct because $\partial E$ consists of both the sphere and the cylindrical surface. The student's error was computing only the spherical flux while ignoring the cylindrical contribution. Answer A is wrong because the cylindrical surface is bounded by the sphere's $z$-limits ($-1 \leq z \leq 1$), so it doesn't extend infinitely. Answer B is incorrect—while the region isn't simply connected due to the hole, the divergence theorem still applies to regions with holes; you just need to account for all boundary surfaces. Answer C misses the point—the normal direction issue is handled by using outward-pointing normals consistently, but the real problem is the missing surface entirely. Study tip: Always sketch the region when applying the divergence theorem. Identify every surface that bounds your region—holes create additional boundary surfaces that contribute to the total flux.
8. A student wants to use Green's theorem to evaluate $\oint_C \mathbf{F} \cdot d\mathbf{r}$ where $\mathbf{F} = \langle \frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2} \rangle$ and $C$ is the boundary of the region $1 \leq x^2 + y^2 \leq 4$, oriented counterclockwise. They compute $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0$ and conclude the integral equals zero. What's wrong?
   Answer: The region is not simply connected due to the hole at $x^2 + y^2 < 1$, so Green's theorem requires modification The vector field $\mathbf{F}$ has a singularity at the origin, but since $1 \leq x^2 + y^2 \leq 4$, the origin lies in the hole of the annulus, not in the region itself. However, the region is not simply connected (it has a hole), so Green's theorem in its basic form doesn't apply directly. The student needs to use the generalized version for multiply connected regions, which involves adding the circulation around the inner boundary. The calculation $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0$ is actually correct where $\mathbf{F}$ is defined. Choice A is wrong about the calculation. Choices B and C both misstate where the origin lies relative to the region.

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