Selecting Methods & Approaches - Calculus 3

1. When evaluating $\int_C \vec{F} \cdot d\vec{r}$ where $\vec{F} = \langle -y/(x^2+y^2), x/(x^2+y^2) \rangle$ and $C$ is any simple closed curve in the plane not containing the origin, which theorem provides the most direct approach?
   Answer: Green's Theorem directly, since $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0$ everywhere the field is defined Computing the partial derivatives: $P = -y/(x^2+y^2)$ gives $\frac{\partial P}{\partial y} = -(x^2+y^2-y(2y))/(x^2+y^2)^2 = -(x^2-y^2)/(x^2+y^2)^2$. And $Q = x/(x^2+y^2)$ gives $\frac{\partial Q}{\partial x} = (x^2+y^2-x(2x))/(x^2+y^2)^2 = (y^2-x^2)/(x^2+y^2)^2$. Therefore $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = (y^2-x^2)/(x^2+y^2)^2 - (-(x^2-y^2)/(x^2+y^2)^2) = 0$. Since the curve doesn't contain the origin, Green's Theorem applies and gives the integral equals zero. Choice B is incorrect because the domain is not simply connected (origin is removed). Choice C incorrectly describes the potential function. Choice D unnecessarily avoids Green's Theorem when it can be applied.
2. To find the center of mass of a solid hemisphere of radius $a$ with density $\rho(x,y,z) = z$, which combination of coordinate system and symmetry arguments would minimize computational effort?
   Answer: Use spherical coordinates and recognize that $\bar{x} = \bar{y} = 0$ by symmetry, then compute only $\bar{z}$ The hemisphere has symmetry about the z-axis, and the density function $\rho(x,y,z) = z$ preserves this symmetry (it depends only on the z-coordinate, not on x or y). Therefore, $\bar{x} = \bar{y} = 0$ by symmetry arguments. Only $\bar{z}$ needs to be computed. Spherical coordinates are ideal because the hemisphere boundary is $\rho = a$, $0 \leq \phi \leq \pi/2$, and the density becomes $\rho\cos\phi$ which integrates nicely. Choice B incorrectly claims density breaks symmetry. Choice C incorrectly suggests the center of mass is at the geometric center (it's not, due to the variable density). Choice D incorrectly claims all coordinates must be computed despite the clear symmetry.
3. When computing the circulation $\oint_C \vec{F} \cdot d\vec{r}$ where $\vec{F} = \langle ye^{xy}, xe^{xy} + \cos y \rangle$ and $C$ is the boundary of the square $[0,\pi] \times [0,\pi]$ oriented counterclockwise, which method would minimize the computational work?
   Answer: Green's Theorem, since $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = e^{xy}(1 + xy) - e^{xy}(1 + xy) = 0$ Let's compute the partial derivatives for Green's Theorem. We have $P = ye^{xy}$ and $Q = xe^{xy} + \cos y$. Then $\frac{\partial P}{\partial y} = e^{xy} + y \cdot xe^{xy} = e^{xy}(1 + xy)$ and $\frac{\partial Q}{\partial x} = e^{xy} + x \cdot ye^{xy} = e^{xy}(1 + xy)$. Therefore $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = e^{xy}(1 + xy) - e^{xy}(1 + xy) = 0$. By Green's Theorem, the circulation is zero. This is much easier than parameterizing four separate line segments. Choice A would require substantial computation. Choice B has the correct conclusion but shows incorrect intermediate steps. Choice D would also work (the field is indeed conservative) but finding the potential function explicitly would take longer than applying Green's Theorem.
4. Consider the line integral $\int_C (x^2y \, dx + xy^2 \, dy)$ where $C$ is the boundary of the region $R = \{(x,y): 0 \leq x \leq 1, x^2 \leq y \leq \sqrt{x}\}$, traversed counterclockwise. Which approach would be most computationally efficient?
   Answer: Apply Green's Theorem: $\iint_R \left(\frac{\partial(xy^2)}{\partial x} - \frac{\partial(x^2y)}{\partial y}\right) dA = \iint_R (y^2 - x^2) \, dA$ When you encounter a line integral around a closed curve, you have three main tools: direct parameterization, checking for conservativeness, or applying Green's Theorem. The key is choosing the most efficient approach based on the vector field's properties. For $\vec{F} = \langle x^2y, xy^2 \rangle$, let's first check if it's conservative by testing if $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$. We have $\frac{\partial(x^2y)}{\partial y} = x^2$ and $\frac{\partial(xy^2)}{\partial x} = y^2$. Since $x^2 \neq y^2$ in general, the field is not conservative, eliminating option A. Green's Theorem converts the line integral to $\iint_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA$. Computing the partial derivatives: $\frac{\partial(xy^2)}{\partial x} = y^2$ and $\frac{\partial(x^2y)}{\partial y} = x^2$. Therefore, the integrand is $y^2 - x^2$, making option C correct. Option B would require parameterizing four separate curves (the boundary has segments along $y = x^2$, $y = \sqrt{x}$, and vertical segments), which is computationally intensive. Option D makes an algebraic error, incorrectly computing the partial derivatives as $2xy$ each, likely confusing mixed partials or chain rule applications. Study tip: Always check if a vector field is conservative first (quickest when it works), but when it's not conservative and you have a nice region, Green's Theorem typically beats direct parameterization. Double-check your partial derivative calculations—they're the most common source of errors.
5. To evaluate the triple integral $\iiint_E (x^2 + y^2 + z^2) \, dV$ where $E$ is the region bounded by $x^2 + y^2 + z^2 \leq 4$, $z \geq 0$, and $x^2 + y^2 \leq z^2$, which coordinate system and approach would be most efficient?
   Answer: Spherical coordinates with bounds $0 \leq \rho \leq 2$, $0 \leq \theta \leq 2\pi$, $0 \leq \phi \leq \pi/4$ The region E is the intersection of a hemisphere (upper half of sphere radius 2) with the interior of a circular cone ($x^2 + y^2 \leq z^2$ with $z \geq 0$). In spherical coordinates, the cone $x^2 + y^2 = z^2$ becomes $\rho^2\sin^2\phi = \rho^2\cos^2\phi$, which simplifies to $\tan\phi = 1$ or $\phi = \pi/4$. Since we want $x^2 + y^2 \leq z^2$ (inside the cone), we need $0 \leq \phi \leq \pi/4$. The integrand $x^2 + y^2 + z^2 = \rho^2$ is also simple in spherical coordinates. Choice A has incorrect bounds for a cone in cylindrical coordinates. Choice C uses Cartesian coordinates which would be computationally intensive. Choice D has the wrong range for $\phi$ (outside the cone rather than inside).
6. To compute $\iiint_E z \sqrt{x^2 + y^2} \, dV$ where $E$ is the region inside the cone $z^2 = x^2 + y^2$ and below the plane $z = 4$, which coordinate transformation would be optimal?
   Answer: Cylindrical coordinates with bounds $0 \leq r \leq 4$, $0 \leq \theta \leq 2\pi$, $r \leq z \leq 4$ The cone $z^2 = x^2 + y^2$ with $z \geq 0$ (since we're below the plane $z = 4$) becomes $z = \sqrt{x^2 + y^2} = r$ in cylindrical coordinates. The region $E$ is inside this cone and below $z = 4$, so for each fixed $r$ and $\theta$, $z$ ranges from the cone surface $z = r$ up to $z = 4$. At $z = 4$, the cone has radius $r = 4$, so $r$ ranges from $0$ to $4$. The integrand becomes $z \sqrt{x^2 + y^2} = zr$ with volume element $r \, dr \, d\theta \, dz$. Choice A has wrong $z$ bounds (doesn't account for cone constraint). Choice B has $r$ and $z$ bounds reversed. Choice C uses spherical coordinates unnecessarily and with incorrect bounds.
7. To find the volume of the region bounded by $z = x^2 + y^2$ and $z = 2 - x^2 - y^2$, which method and coordinate system combination would be most efficient?
   Answer: Cartesian coordinates with double integration: $V = \iint_R [(2-x^2-y^2) - (x^2+y^2)] \, dA$ where $R$ is $x^2+y^2 \leq 1$ The surfaces intersect where $x^2 + y^2 = 2 - x^2 - y^2$, giving $2(x^2 + y^2) = 2$, so $x^2 + y^2 = 1$. The paraboloid $z = x^2 + y^2$ opens upward, and $z = 2 - x^2 - y^2$ opens downward. For $x^2 + y^2 < 1$, we have $x^2 + y^2 < 2 - x^2 - y^2$, so the downward paraboloid is above the upward one. The volume is $V = \iint_R [(2-x^2-y^2) - (x^2+y^2)] \, dA = \iint_R (2-2x^2-2y^2) \, dA$ where $R: x^2+y^2 \leq 1$. Converting to polar: $V = \int_0^{2\pi} \int_0^1 (2-2r^2) r \, dr \, d\theta = 2\pi \int_0^1 (2r-2r^3) \, dr = 2\pi[r^2-r^4/2]_0^1 = 2\pi(1-1/2) = \pi$. Choice A uses triple integration unnecessarily. Choice C uses wrong coordinate system and bounds. Choice D has incorrect intersection radius.
8. For the surface integral $\iint_S \vec{F} \cdot d\vec{S}$ where $\vec{F} = \langle x^3, y^3, z^3 \rangle$ and $S$ is the sphere $x^2 + y^2 + z^2 = a^2$, which approach most effectively exploits the symmetry of the problem?
   Answer: Use the Divergence Theorem with $\nabla \cdot \vec{F} = 3(x^2 + y^2 + z^2)$ to get $\iiint_V 3(x^2 + y^2 + z^2) \, dV$ When evaluating flux integrals through closed surfaces, the Divergence Theorem is often your most powerful tool, especially when the surface has high symmetry like a sphere. The theorem converts the surface integral $\iint_S \vec{F} \cdot d\vec{S}$ into the volume integral $\iiint_V (\nabla \cdot \vec{F}) \, dV$, which can dramatically simplify calculations. For $\vec{F} = \langle x^3, y^3, z^3 \rangle$, the divergence is $\nabla \cdot \vec{F} = 3x^2 + 3y^2 + 3z^2 = 3(x^2 + y^2 + z^2)$. Using the Divergence Theorem transforms your problem into $\iiint_V 3(x^2 + y^2 + z^2) \, dV$, which is perfect for spherical coordinates where $x^2 + y^2 + z^2 = \rho^2$. Option A incorrectly assumes $\nabla \cdot \vec{F} = 3a^2$ everywhere on the surface, but the divergence varies with position—it only equals $3a^2$ specifically on the sphere surface, not throughout the volume. Option B misapplies symmetry arguments; while the components have certain symmetries, this vector field doesn't have the antisymmetric properties needed to make the integral zero. Option C gives the correct normal component formula but suggests direct parameterization, which leads to a much more complex calculation involving $x^4 + y^4 + z^4$ that doesn't simplify easily. The key insight: when you see a flux integral through a closed surface, always check if the Divergence Theorem simplifies your work. Spherical symmetry combined with the Divergence Theorem often transforms difficult surface integrals into manageable volume integrals.

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