Choosing Coordinate Systems - Multivariable Calculus

A region is defined by $1 \leq x^2 + y^2 + z^2 \leq 4$ and $x^2 + y^2 \leq z^2$ with $z \geq 0$ . When setting up $\iiint_E f(x,y,z) \, dV$ in spherical coordinates, what is the correct range for the angle $\phi$ and why might this be counterintuitive?
Answer: $0 \leq \phi \leq \pi/4$ , which might seem backwards since the cone opens upward The condition $x^2 + y^2 \leq z^2$ with $z \geq 0$ describes the region inside an upward-opening cone. In spherical coordinates, this becomes $\rho^2 \sin^2 \phi \leq \rho^2 \cos^2 \phi$ , which simplifies to $\tan \phi \leq 1$ , or $\phi \leq \pi/4$ . This might seem counterintuitive because we think of the cone as 'opening upward,' but small values of $\phi$ correspond to points near the positive $z$ -axis. Choice B incorrectly interprets the region as outside the cone. Choice C ignores the cone constraint entirely. Choice D incorrectly extends into the lower hemisphere despite $z \geq 0$ .
A torus is generated by rotating a circle of radius $a$ in the $xz$ -plane, centered at $(b, 0, 0)$ with $b > a$ , around the $z$ -axis. To calculate the surface area of this torus, which standard coordinate system provides the most natural parameterization based on the object's fundamental symmetry?
Answer: Cylindrical coordinates, since the torus has rotational symmetry about the $z$ -axis. When choosing coordinate systems for surface area calculations, you should always look for the system that naturally captures the object's inherent symmetry. This makes parameterization simpler and computations more manageable. A torus has clear rotational symmetry about the z-axis since it's created by rotating a circle around that axis. Cylindrical coordinates $(r, \theta, z)$ are specifically designed to exploit this type of symmetry. You can parameterize the torus using two natural parameters: the angle $\theta$ around the z-axis (0 to $2\pi$ ) and the angle $\phi$ around the generating circle (also 0 to $2\pi$ ). This gives you $\vec{r}(\theta, \phi) = ((b + a\cos\phi)\cos\theta, (b + a\cos\phi)\sin\theta, a\sin\phi)$ , which leads to straightforward surface area calculations. Option A uses the implicit equation, but implicit forms make surface area calculations much more complex since you need to work with gradients and level surfaces rather than direct parameterization. Option B incorrectly suggests spherical coordinates—while a torus is indeed a closed 3D surface, spherical coordinates are suited for radial symmetry from a point, not rotational symmetry about an axis. Option C mentions rectangular coordinates with substitution, which doesn't leverage the torus's natural symmetry and unnecessarily complicates the problem. Study tip: Always match coordinate systems to geometric symmetry. Cylindrical for rotation about an axis, spherical for radial symmetry from a point, and Cartesian when no clear symmetry exists or when working with simple rectangular regions.
Consider the 'ice cream cone' solid bounded below by the cone $z=\sqrt{3(x^2+y^2)}$ and above by the sphere $x^2+y^2+z^2=4$ . To find the centroid of this solid, which coordinate system allows the necessary integrals to be set up without splitting the domain of integration?
Answer: Spherical, because both boundary surfaces correspond to constant values of a coordinate. The region is bounded by a cone and a sphere, a classic setup for spherical coordinates. The sphere is $\rho=2$ . The cone $z=\sqrt{3(x^2+y^2)}$ becomes $\rho\cos\phi = \sqrt{3}\rho\sin\phi$ , which simplifies to $\tan\phi = 1/\sqrt{3}$ , or $\phi=\pi/6$ . Thus, the entire solid can be described by $0 \le \rho \le 2$ , $0 \le \phi \le \pi/6$ , and $0 \le \theta \le 2\pi$ . Since all limits are constant, any integral over this domain is a single, unified setup. (A) Splitting the integral is unnecessary. (B) While a single cylindrical integral is possible as described, the limits for $z$ are functions of $r$ . The spherical system is superior because all its limits are constants, simplifying the setup significantly. (D) The definition of the centroid doesn't dictate the coordinate system for calculation; the geometry of the domain should.
A triple integral is set up to find the mass of a solid with density $\rho(x,y,z) = \sqrt{x^2 + y^2 + z^2}$ over the region where $x^2 + y^2 \leq z^2$ , $z \geq 0$ , and $x^2 + y^2 + z^2 \leq 9$ . Which coordinate system choice represents the best balance between boundary simplicity and integrand simplicity?
Answer: Spherical coordinates, because both the density function and sphere become very simple In spherical coordinates, the density becomes simply $\rho(\rho, \phi, \theta) = \rho$ , the sphere becomes $\rho = 3$ , and the cone becomes $\phi = \pi/4$ . This dramatic simplification of both the integrand and boundaries outweighs other considerations. Choice A ignores the clear spherical and rotational symmetries. Choice B recognizes rotational symmetry but leaves the density function complicated as $\sqrt{r^2 + z^2}$ . Choice D has the same issue as B, missing the optimal simplification available in spherical coordinates.
We wish to compute the flux of the vector field $\mathbf{F}(x,y,z) = \langle x, y, 1 \rangle$ across the portion of the inverted paraboloid $z = 8 - 2x^2 - 2y^2$ that lies above the plane $z=0$ . Which coordinate system provides the most straightforward parameterization of the surface for this calculation?
Answer: Cylindrical coordinates, because the surface is a paraboloid of revolution. When calculating flux through a surface, choosing the right coordinate system can dramatically simplify your parameterization. The key is to match the coordinate system to the natural symmetry of the surface. The surface $z = 8 - 2x^2 - 2y^2$ above $z = 0$ is a paraboloid of revolution about the z-axis. Since it has circular cross-sections in planes parallel to the xy-plane, cylindrical coordinates $(r, \theta, z)$ naturally capture this symmetry. You can parameterize using $x = r\cos\theta$ , $y = r\sin\theta$ , and $z = 8 - 2r^2$ , where $0 \leq r \leq 2$ and $0 \leq \theta \leq 2\pi$ . This gives you a clean parameterization with easily computable partial derivatives. Choice A, while mathematically correct, treats this as a graph $z = f(x,y)$ . This works but doesn't exploit the rotational symmetry, potentially making the flux integral more complex. Choice B suggests vertex-shifted spherical coordinates centered at $(0,0,8)$ . This is unnecessarily complicated since the surface isn't spherical and the shift doesn't align with any natural symmetry of the paraboloid. Choice C mentions spherical coordinates for the curved dome, but standard spherical coordinates don't naturally fit a paraboloid. The surface has circular, not spherical, symmetry. Remember: always choose coordinates that match your surface's natural symmetry. Surfaces of revolution call for cylindrical coordinates, spheres suggest spherical coordinates, and rectangular regions work well with Cartesian coordinates. This choice will minimize computational complexity in your flux calculations.
A solid is bounded by $y = \sqrt{x^2 + z^2}$ , $x^2 + z^2 = 4$ , and $y = 3$ . This region has rotational symmetry about the $y$ -axis. What is the most efficient coordinate system for integration, and what is the key insight for choosing it?
Answer: Modified cylindrical coordinates with $r^2 = x^2 + z^2$ , $y$ unchanged, recognizing the non-standard axis of symmetry The key insight is that the rotational symmetry is about the $y$ -axis, not the $z$ -axis. This requires modified cylindrical coordinates where $r^2 = x^2 + z^2$ and $y$ remains unchanged. In this system, the cone becomes $y = r$ , the cylinder becomes $r = 2$ , and the plane remains $y = 3$ . Choice A is conceptually correct but imprecisely stated. Choice B incorrectly applies standard cylindrical coordinates when the axis of symmetry is not the $z$ -axis. Choice C ignores the clear rotational symmetry that should be exploited. Choice D correctly identifies both the need for cylindrical-type coordinates and the non-standard axis orientation.
A solid object is created by drilling a cylindrical hole of radius $a$ through the center of a solid sphere of radius $2a$ . The axis of the cylinder coincides with a diameter of the sphere. To calculate the volume of the remaining object (a 'bead'), an integral is set up. Which coordinate system leads to an integral with the simplest integrand and limits?
Answer: Cylindrical, because the inner boundary is a cylinder, simplifying the bounds on $r$ . The object has a clear rotational symmetry about the z-axis, suggesting cylindrical or spherical coordinates. The boundaries are the cylinder $r=a$ and the sphere $r^2+z^2 = (2a)^2$ . In cylindrical coordinates, the limits are straightforward: $\theta$ from $0$ to $2\pi$ , $r$ from $a$ to $2a$ , and $z$ from $-\sqrt{4a^2-r^2}$ to $\sqrt{4a^2-r^2}$ . The volume element is $r\,dz\,dr\,d\theta$ . This is manageable. (A) In spherical coordinates, the outer boundary $\rho=2a$ is simple, but the inner boundary $r=a$ becomes $\rho\sin\phi = a$ , or $\rho=a\csc\phi$ . This makes the lower limit for $\rho$ a function of $\phi$ , which is more complicated than the cylindrical setup. (B) While the object has this symmetry, this reason alone is insufficient; cylindrical coordinates exploit this symmetry more effectively for this particular geometry. (D) The washer method is a valid technique from single-variable calculus, but it is not a 3D coordinate system. The question asks for the best coordinate system for a triple integral.
Let $\mathcal{W}$ be the wedge-shaped region in the first octant bounded by the cylinder $x^2+y^2=4$ , the plane $z=0$ , and the plane $z=y$ . When setting up the volume integral $\iiint_\mathcal{W} dV$ , which coordinate system results in the maximum number of integration variables with constant bounds?
Answer: Cylindrical coordinates, which yields two variables with constant bounds. The region's projection onto the $xy$ -plane is a quarter-circle of radius 2 in the first quadrant. This suggests cylindrical coordinates are a good choice. In cylindrical coordinates, the region is described by $0 \le \theta \le \pi/2$ , $0 \le r \le 2$ , and $0 \le z \le y = r\sin\theta$ . The iterated integral is $\int_0^{\pi/2} \int_0^2 \int_0^{r\sin\theta} r \,dz\,dr\,d\theta$ . The variables $\theta$ and $r$ have constant bounds. (B) In Cartesian coordinates, the integral is $\int_0^2 \int_0^{\sqrt{4-x^2}} \int_0^y \,dz\,dy\,dx$ . Only the outer variable $x$ has constant bounds if we integrate in this order. (C) In spherical coordinates, the cylinder $x^2+y^2=4$ is $\rho\sin\phi=2$ and the plane $z=y$ is $\cos\phi = \sin\phi\sin\theta$ . These boundaries are complex functions of the coordinates. (D) This is incorrect, as cylindrical coordinates provide constant bounds for two of the three variables.
Consider integrating over the region where $x^2 + y^2 + z^2 \leq 9$ , $x^2 + y^2 \geq 1$ , and $z \geq 0$ . A student chooses spherical coordinates and sets up the integral with limits $1/\sin\phi \leq \rho \leq 3$ , $0 \leq \phi \leq \pi/2$ , $0 \leq \theta \leq 2\pi$ . What is the primary error in this setup?
Answer: The $\rho$ limits $1/\sin\phi \leq \rho \leq 3$ are undefined when $\phi = 0$ , requiring split integration The constraint $x^2 + y^2 \geq 1$ becomes $\rho^2 \sin^2 \phi \geq 1$ , or $\rho \geq 1/\sin \phi$ . However, this is undefined when $\phi = 0$ (along the positive $z$ -axis), and for small $\phi$ , $1/\sin \phi$ can exceed 3, violating the outer sphere constraint. The integration must be split: for $\phi$ values where $1/\sin \phi \leq 3$ (i.e., $\sin \phi \geq 1/3$ ), use $1/\sin \phi \leq \rho \leq 3$ ; for smaller $\phi$ values, the cylindrical constraint is automatically satisfied. Choice A incorrectly suggests including the lower hemisphere. Choice B doesn't recognize that the constraint can be expressed in spherical coordinates. Choice D is false; the constraint can be expressed but creates integration complications.
A region is defined by $x^2 + y^2 \leq 1$ , $0 \leq z \leq x^2 + y^2 + 1$ . When computing $\iiint_E (x^2 + y^2)^{3/2} \, dV$ , a student argues that despite the clear cylindrical symmetry, rectangular coordinates might be preferable because the integrand becomes $r^3$ in cylindrical coordinates, which is 'more complicated' than the original form. How should this reasoning be evaluated?
Answer: Incorrect, because $r^3$ is actually simpler to integrate than $(x^2 + y^2)^{3/2}$ and the boundaries are much simpler The student's reasoning is flawed. In cylindrical coordinates, $(x^2 + y^2)^{3/2} = r^3$ , and with the Jacobian $r$ , we integrate $r^4$ , which is straightforward. More importantly, the boundaries become $0 \leq r \leq 1$ , $0 \leq \theta \leq 2\pi$ , $0 \leq z \leq r^2 + 1$ , which are much simpler than the rectangular form. The complexity reduction in boundaries far outweighs any perceived complexity in the integrand. Choice A makes an incorrect generalization about polynomial vs. power expressions. Choice C incorrectly suggests the coordinate transformation adds significant complexity. Choice D contains an error about the Jacobian cancellation.
The region $\mathcal{R}$ in the $xy$ -plane is bounded by the cardioid $r = 1 + \cos\theta$ . This region is used as the base of a solid whose height at any point $(x,y)$ is given by $h(x,y) = \sqrt{x^2+y^2}$ . Which coordinate system is most appropriate for calculating the volume of this solid?
Answer: Cylindrical coordinates, because the base is given in polar coordinates and height is a function of $r$ . The solid is described by its base in the $xy$ -plane and its height. This is a setup for a double or triple integral. The base is given in polar coordinates ( $r = 1 + \cos\theta$ ), and the height function is $h(x,y) = \sqrt{x^2+y^2} = r$ . This structure strongly points to cylindrical coordinates. The volume is given by the triple integral $\int_0^{2\pi} \int_0^{1+\cos\theta} \int_0^r r \,dz\,dr\,d\theta$ , or more simply by the double integral $\iint_\mathcal{R} h(x,y) dA = \int_0^{2\pi} \int_0^{1+\cos\theta} r \cdot r \,dr\,d\theta$ . In either case, the polar/cylindrical setup is the most direct. (A) Spherical coordinates are for 3D radial symmetry, which is not present here. (C) Converting the cardioid equation $r=1+\cos\theta$ to Cartesian coordinates gives $\sqrt{x^2+y^2}=1+x/\sqrt{x^2+y^2}$ , which is very complex. (D) The standard polar/cylindrical system already places the origin at the cusp of the cardioid, which is the natural center for the given equation.
You want to find the center of mass of a solid with density $\delta(x,y,z) = x^2 + y^2 + z^2$ over the region $x^2 + y^2 + z^2 \leq 4$ , $z \geq \sqrt{x^2 + y^2}$ . The calculation requires finding $\bar{x} = \frac{\iiint_E x \cdot \delta \, dV}{\iiint_E \delta \, dV}$ . Which coordinate system should be chosen and why?
Answer: Spherical coordinates, because both the density and the sphere boundary are simplified Despite the numerator containing $x = \rho \sin \phi \cos \theta$ , spherical coordinates are still optimal because: (1) the density becomes $\delta = \rho^2$ , (2) the sphere becomes $\rho = 2$ , (3) the cone becomes $\phi = \pi/4$ , and (4) the $x$ integral $\int_0^{2\pi} \cos \theta \, d\theta = 0$ by symmetry, making $\bar{x} = 0$ immediately apparent. Choice B focuses on one boundary while ignoring the density simplification. Choice C overemphasizes the complexity of $x$ in spherical coordinates. Choice D is unnecessarily complicated and not a standard approach.
Consider a region where you need to integrate $f(x,y,z) = x^2 + y^2$ over the solid bounded by $x^2 + y^2 = 4$ , $z = 0$ , and $z = \sqrt{9 - x^2 - y^2}$ . A student claims that cylindrical coordinates are optimal because the integrand contains $x^2 + y^2$ . What is the most accurate assessment of this reasoning?
Answer: Correct, because $x^2 + y^2 = r^2$ simplifies the integrand and the cylinder $x^2 + y^2 = 4$ becomes $r = 2$ The student's reasoning is correct. The cylindrical boundary $x^2 + y^2 = 4$ becomes $r = 2$ , the integrand $x^2 + y^2$ becomes $r^2$ , and the upper boundary becomes $z = \sqrt{9 - r^2}$ , which is still manageable. While the upper boundary is hemispherical, the combination of the cylindrical lateral boundary and the simplified integrand makes cylindrical coordinates optimal. Choice B overemphasizes the spherical upper boundary. Choice C incorrectly suggests equal efficiency. Choice D makes a false general claim about polynomial integrands.
A solid region $\mathcal{E}$ is described by the inequalities $1 \le x^2 + y^2 + z^2 \le 4$ and $z \ge 0$ . To compute the integral $\iiint_\mathcal{E} (x^2+y^2) dV$ , which coordinate system transforms the domain $\mathcal{E}$ into a rectangular box in its corresponding coordinate space?
Answer: Spherical coordinates, because the boundaries are concentric hemispheres. When you encounter a region defined by spherical inequalities like $x^2 + y^2 + z^2$ , immediately consider which coordinate system will simplify both the boundaries and the computation. The region $\mathcal{E}$ is bounded by two concentric spheres (radii 1 and 2) with $z \geq 0$ , creating a thick hemispherical shell. In spherical coordinates, where $x^2 + y^2 + z^2 = \rho^2$ , this region transforms beautifully: $1 \leq \rho \leq 2$ , $0 \leq \phi \leq \pi/2$ (since $z \geq 0$ ), and $0 \leq \theta \leq 2\pi$ . This creates a perfect rectangular box: $[1,2] \times [0,\pi/2] \times [0,2\pi]$ in $(\rho,\phi,\theta)$ space. Answer D correctly identifies that spherical coordinates transform the concentric hemispheres into a rectangular domain with constant bounds. Answer A mentions that $x^2 + y^2 = r^2$ in cylindrical coordinates, which is true, but cylindrical coordinates don't simplify the spherical boundaries $x^2 + y^2 + z^2$ . You'd still have complicated limits involving $\sqrt{1-z^2}$ and $\sqrt{4-z^2}$ . Answer B is incorrect because spherical coordinates do create a rectangular box, as demonstrated above. Answer C fails because in Cartesian coordinates, the region isn't a rectangular box at all—it's a curved hemispherical shell with complex integration limits. Study tip: When you see $x^2 + y^2 + z^2$ in boundaries, immediately think spherical coordinates. When you see $x^2 + y^2$ in boundaries or integrands, consider cylindrical coordinates. Match the coordinate system to the natural symmetry of your problem.
Consider the solid cylinder defined by $(x-2)^2 + y^2 \le 4$ , bounded by the planes $z=0$ and $z=3$ . To compute the moment of inertia of this solid about the $z$ -axis, using the integral $\iiint_E (x^2+y^2) dV$ , which coordinate system setup offers the most practical path to a solution?
Answer: A standard cylindrical system, where the domain is $0 \le r \le 4\cos\theta$ and the integrand is $r^2$ . The cylinder is off-center. In a standard cylindrical system, the inequality $(x-2)^2 + y^2 \le 4$ becomes $(r\cos\theta-2)^2 + (r\sin\theta)^2 \le 4$ , which simplifies to $r^2 - 4r\cos\theta + 4 \le 4$ , or $r \le 4\cos\theta$ . The integrand for the moment of inertia, $x^2+y^2$ , becomes simply $r^2$ . The integral setup is $\int_{-\pi/2}^{\pi/2} \int_0^{4\cos\theta} \int_0^3 (r^2) r \,dz\,dr\,d\theta$ . This is a tractable integral. (B) If we use a translated system $x'=x-2, y'=y$ , the domain becomes simple ( $r' \le 2$ ), but the integrand becomes $x^2+y^2 = (x'+2)^2 + y'^2 = r'^2 + 4x' + 4 = r'^2 + 4r'\cos\theta' + 4$ . The more complex integrand makes this choice less practical. (C) In Cartesian coordinates, the limits for $x$ and $y$ would involve square roots, making integration difficult. (D) Spherical coordinates are not well-suited for cylindrical geometries.
A wire is shaped like the curve of intersection of the cylinder $x^2 + y^2 = 9$ and the plane $z = 1 + x$ . To find the mass of the wire, assuming a density function $\delta(x,y,z) = z$ , one must compute a line integral. Which coordinate system provides the most direct parameterization of the curve, allowing the line integral to be expressed in terms of a single variable?
Answer: Cylindrical coordinates, using the angle $\theta$ as the single parameter. The curve's projection onto the $xy$ -plane is a circle of radius 3, which is perfectly described in cylindrical coordinates. We can parameterize the curve using a single variable, $\theta$ , as $\vec{r}(\theta) = \langle 3\cos\theta, 3\sin\theta, 1+3\cos\theta \rangle$ for $\theta \in [0, 2\pi]$ . This is the most direct approach. (A) Cartesian coordinates require splitting the curve into two halves ( $y = \pm\sqrt{9-x^2}$ ) and leads to more complex derivatives for the arc length element. (C) The curve does not lie on a sphere; for example, at $\theta=0$ , the point is $(3,0,4)$ with squared distance 25, while at $\theta=\pi$ , the point is $(-3,0,-2)$ with squared distance 13. (D) Translating the coordinate system does not simplify the fundamental circular nature of the cylinder.
The trajectory of a particle is given by the vector function $\vec{r}(t) = \langle e^{-t}\cos(t), e^{-t}\sin(t), \sqrt{1-e^{-2t}} \rangle$ for $t \ge 0$ . The particle's motion is constrained to a particular surface. Which coordinate system is most natural for describing this surface?
Answer: Spherical, because the path lies on the surface $x^2+y^2+z^2=1$ . To identify the surface, we check the relationship between the components. Let's calculate the squared magnitude of the position vector, which corresponds to $\rho^2$ in spherical coordinates: $x^2+y^2+z^2 = (e^{-t}\cos t)^2 + (e^{-t}\sin t)^2 + (\sqrt{1-e^{-2t}})^2 = e^{-2t}(\cos^2 t + \sin^2 t) + (1-e^{-2t}) = e^{-2t} + 1 - e^{-2t} = 1$ . Since $x^2+y^2+z^2=1$ , the particle is always on the surface of a sphere of radius 1 centered at the origin. Therefore, spherical coordinates are the most natural choice. (A) While the components suggest cylindrical coordinates, the relationship between $r=e^{-t}$ and $z=\sqrt{1-r^2}$ reveals the spherical nature. (B) The magnitude of the position vector is constant (it is 1), but this is the conclusion, not the premise. The choice should be justified by the calculation. (D) Sticking with Cartesian coordinates obscures the simple spherical geometry of the motion.
A surface $\mathcal{S}$ is defined by the Cartesian equation $(x^2 + y^2)^2 + z = 1$ , for $z \ge 0$ . To compute the surface area of $\mathcal{S}$ , which coordinate system is most appropriate for parameterizing the surface?
Answer: Cylindrical coordinates, because the expression $x^2+y^2$ appears in the equation. The presence of the term $x^2+y^2$ strongly suggests cylindrical coordinates. In this system, $x^2+y^2 = r^2$ , and the surface equation simplifies to $r^4 + z = 1$ , or $z = 1-r^4$ . This is a surface of revolution around the $z$ -axis, which is easily parameterized by $r$ and $\theta$ : $\vec{s}(r, \theta) = \langle r\cos\theta, r\sin\theta, 1-r^4 \rangle$ . This makes calculating the surface area element straightforward. (A) In spherical coordinates, the equation becomes $(\rho^2\sin^2\phi)^2 + \rho\cos\phi = 1$ , which is not a simplification. (C) While possible, the domain for $(x,y)$ is a disk, $x^2+y^2 \le 1$ , and evaluating the resulting integral usually requires switching to polar coordinates anyway, so cylindrical is more direct. (D) Changing axis orientation doesn't simplify the core algebraic structure.
Consider the solid region $\mathcal{E}$ enclosed above by the sphere $x^2 + y^2 + z^2 = 16$ and below by the cone $z = \sqrt{x^2 + y^2}$ . When setting up a triple integral for the volume of $\mathcal{E}$ , which coordinate system results in an iterated integral where all limits of integration are constants?
Answer: Spherical coordinates, because the boundaries correspond to constant values of $\rho$ and $\phi$ . In spherical coordinates, the sphere $x^2 + y^2 + z^2 = 16$ is simply $\rho = 4$ . The cone $z = \sqrt{x^2 + y^2}$ becomes $\rho\cos\phi = \sqrt{(\rho\sin\phi\cos\theta)^2 + (\rho\sin\phi\sin\theta)^2} = \rho\sin\phi$ , which simplifies to $\tan\phi = 1$ , or $\phi = \pi/4$ . The volume integral is then $\int_0^{2\pi} \int_0^{\pi/4} \int_0^4 \rho^2\sin\phi \,d\rho\,d\phi\,d\theta$ . All six limits of integration (0, $2\pi$ , 0, $\pi/4$ , 0, 4) are constants. (A) In cylindrical coordinates, the upper limit for $z$ is $\sqrt{16-r^2}$ , which is not constant. (B) In Cartesian coordinates, the limits for $z$ , $y$ , and $x$ would all involve variables. (D) This is incorrect, as spherical coordinates achieve this.

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