Choosing Coordinate Systems - Calculus 3

1. Consider the region $W$ bounded by the surfaces $z = x^2 + y^2$ and $x^2 + y^2 + z^2 = 2$. For the integral $\iiint_W (x^2 + y^2 + z^2) \, dV$, which coordinate system choice requires the most careful analysis of the intersection curve?
   Answer: All coordinate systems handle the intersection equally well since it reduces to $r = 1$ or equivalent in each system The correct answer is D. The intersection occurs where $z = x^2 + y^2$ and $x^2 + y^2 + z^2 = 2$. Substituting: $x^2 + y^2 + (x^2 + y^2)^2 = 2$. Let $u = x^2 + y^2$, then $u + u^2 = 2$, so $u^2 + u - 2 = 0$, giving $u = 1$ (since $u \geq 0$). Thus $x^2 + y^2 = 1$, or $r = 1$ in cylindrical coordinates, or $\rho \sin \phi = 1$ in spherical coordinates. Choice A exaggerates the Cartesian difficulty. Choice B incorrectly calls $r^2 + r^4 = 2$ a quartic—it's quadratic in $r^2$. Choice C misrepresents the spherical intersection condition.
2. The integral $\iint_D \frac{xy}{(x^2 + y^2)^{3/2}} \, dA$ is to be evaluated over the region $D = \{(x,y) : 1 \leq x^2 + y^2 \leq 9, x \geq 0, y \geq x\}$. Three students propose different strategies for handling the angular constraint $y \geq x$ in polar coordinates. Which analysis correctly identifies the optimal approach?
   Answer: The constraint $y \geq x$ in the first quadrant corresponds to $\pi/4 \leq \theta \leq \pi/2$, and the integrand becomes $\frac{\sin \theta \cos \theta}{r}$, creating a separable integral When evaluating double integrals over regions with both radial and angular constraints, polar coordinates often simplify the problem by converting the integrand and making the bounds more manageable. Let's work through this systematically. The region D has $1 \leq x^2 + y^2 \leq 9$ (an annulus), $x \geq 0$, and $y \geq x$. In polar coordinates, $x^2 + y^2 = r^2$, so the radial bounds become $1 \leq r \leq 3$. For the angular constraint $y \geq x$ with $x \geq 0$, we need $r\sin\theta \geq r\cos\theta$. Since $r > 0$, this gives $\sin\theta \geq \cos\theta$, or $\tan\theta \geq 1$, which means $\theta \geq \pi/4$. Combined with $x \geq 0$ (first and fourth quadrants), we get $\pi/4 \leq \theta \leq \pi/2$. The integrand transforms beautifully: $\frac{xy}{(x^2+y^2)^{3/2}} = \frac{r\cos\theta \cdot r\sin\theta}{(r^2)^{3/2}} = \frac{r^2\sin\theta\cos\theta}{r^3} = \frac{\sin\theta\cos\theta}{r}$. This creates a separable integral: $\int_{\pi/4}^{\pi/2} \sin\theta\cos\theta \, d\theta \int_1^3 \frac{1}{r} \, dr$. Answer C correctly identifies this approach. Answer A stops at finding the angular bounds without completing the analysis. Answer B incorrectly claims the origin singularity matters, but our region excludes the origin since $r \geq 1$. Answer D unnecessarily complicates the problem by suggesting case splits when the constraint gives straightforward bounds. Study tip: When converting rectangular constraints to polar coordinates, always check if the integrand becomes separable—this often signals the optimal approach.
3. Consider the integral $\iiint_T \frac{z^2}{x^2 + y^2 + z^2} \, dV$ where $T$ is the region inside both $x^2 + y^2 + z^2 \leq 4$ and $x^2 + y^2 + z^2 \geq 2z$. The second constraint can be rewritten as $x^2 + y^2 + (z-1)^2 \geq 1$. Which coordinate system best exploits the geometric relationship between these constraints?
   Answer: Standard spherical coordinates, because the inner constraint $\rho \geq 2\cos \phi$ creates a natural relationship with the outer bound $\rho \leq 2$ The correct answer is D. In standard spherical coordinates, $x^2 + y^2 + z^2 \geq 2z$ becomes $\rho^2 \geq 2\rho\cos\phi$, or $\rho \geq 2\cos\phi$ (for $\rho > 0$). Combined with $\rho \leq 2$, this creates clean bounds: $2\cos\phi \leq \rho \leq 2$ for $0 \leq \phi \leq \pi/2$. The integrand $\frac{z^2}{x^2+y^2+z^2} = \cos^2\phi$. Choice A misses the key insight about the inner boundary. Choice B complicates the outer boundary unnecessarily. Choice C doesn't exploit the integrand's natural form in spherical coordinates.
4. A solid region $D$ is defined by $x^2 + y^2 + z^2 \leq 9$, $z \geq 0$, and $x^2 + y^2 \geq z^2$. To evaluate $\iiint_D z^2 \sqrt{x^2 + y^2 + z^2} \, dV$, a student considers three coordinate systems. Which analysis correctly identifies the most suitable choice?
   Answer: Spherical coordinates are optimal because the integrand becomes $\rho^4 \cos^2 \phi$ and the region is $0 \leq \rho \leq 3$, $0 \leq \phi \leq \pi/4$, $0 \leq \theta \leq 2\pi$ The correct answer is C. In spherical coordinates: the integrand $z^2\sqrt{x^2+y^2+z^2}$ becomes $(\rho\cos\phi)^2 \cdot \rho = \rho^4\cos^2\phi$, the sphere gives $\rho \leq 3$, $z \geq 0$ gives $\phi \leq \pi/2$, and $x^2+y^2 \geq z^2$ becomes $\rho^2\sin^2\phi \geq \rho^2\cos^2\phi$, so $\tan\phi \geq 1$, giving $\phi \geq \pi/4$. Thus $\pi/4 \leq \phi \leq \pi/2$. Choice A has wrong integrand and constraint analysis. Choice B incorrectly claims cylindrical is better when spherical handles all constraints cleanly. Choice D ignores the complexity of Cartesian limits.
5. Consider the integral $\iiint_E (x^2 + y^2)^{3/2} \, dV$ where $E$ is the region bounded by the cone $z = \sqrt{x^2 + y^2}$ and the sphere $x^2 + y^2 + z^2 = 16$. Which coordinate system would be most efficient for evaluating this integral, and what is the primary reason for this choice?
   Answer: Cylindrical coordinates, because both the integrand and the boundary surfaces exhibit rotational symmetry about the z-axis The correct answer is B. The integrand $(x^2 + y^2)^{3/2}$ becomes $r^3$ in cylindrical coordinates, and both boundary surfaces (cone: $z = r$, sphere: $r^2 + z^2 = 16$) have simple expressions. The rotational symmetry about the z-axis makes cylindrical coordinates optimal. Choice A ignores the complexity of the limits in Cartesian form. Choice C is incorrect because while spherical coordinates work for the sphere, the integrand becomes $\rho^3 \sin^3 \phi$ (not $\sin^4 \phi$) and the cone creates awkward limits. Choice D gives the right coordinate system but oversimplifies the reasoning by focusing only on one boundary.
6. Consider evaluating $\iiint_E e^{\sqrt{x^2+y^2+z^2}} \, dV$ where $E$ is the region $\{(x,y,z) : 1 \leq x^2 + y^2 + z^2 \leq 4, z \geq \sqrt{x^2 + y^2}\}$. A student argues that while spherical coordinates simplify the integrand and outer boundary, the constraint $z \geq \sqrt{x^2 + y^2}$ creates complications. Which assessment is most accurate?
   Answer: The constraint becomes $\tan \phi \leq 1$, so $0 \leq \phi \leq \pi/4$, and combined with $1 \leq \rho \leq 2$, this creates straightforward rectangular-type bounds The correct answer is B. In spherical coordinates, $z = \rho\cos\phi$ and $\sqrt{x^2+y^2} = \rho\sin\phi$. The constraint $z \geq \sqrt{x^2+y^2}$ becomes $\rho\cos\phi \geq \rho\sin\phi$, which simplifies to $\cos\phi \geq \sin\phi$ (for $\rho > 0$), or $\tan\phi \leq 1$, giving $0 \leq \phi \leq \pi/4$. This creates simple bounds. The integrand $e^{\sqrt{x^2+y^2+z^2}}$ becomes $e^\rho$. Choice A has the wrong inequality direction. Choice C incorrectly claims the constraint can't be simplified. Choice D undervalues how well spherical coordinates handle both the integrand and all boundaries simultaneously.
7. Consider evaluating $\iint_R \frac{x + y}{\sqrt{x^2 + y^2}} \, dA$ where $R$ is the region $\{(x,y) : 1 \leq x^2 + y^2 \leq 4, x \geq 0, y \geq 0\}$. A student claims that polar coordinates will simplify this integral significantly. Which statement best evaluates this claim?
   Answer: Correct, because the integrand becomes $\cos \theta + \sin \theta$ and the region becomes $1 \leq r \leq 2$, $0 \leq \theta \leq \pi/2$ The correct answer is A. In polar coordinates, $\frac{x+y}{\sqrt{x^2+y^2}} = \frac{r\cos\theta + r\sin\theta}{r} = \cos\theta + \sin\theta$, and the region becomes $1 \leq r \leq 2$, $0 \leq \theta \leq \pi/2$. The integral becomes $\int_0^{\pi/2} \int_1^2 (\cos\theta + \sin\theta) \cdot r \, dr \, d\theta$, which separates cleanly. Choice B is wrong because no integration by parts is needed. Choice C incorrectly suggests the simplification loses information—it's actually the key advantage. Choice D is wrong because polar coordinates handle $x,y \geq 0$ perfectly as $0 \leq \theta \leq \pi/2$.
8. The integral $\iiint_E xyz \, dV$ is to be evaluated over the region $E = \{(x,y,z) : x^2 + y^2 + z^2 \leq 1, x \geq 0, y \geq 0, z \geq 0\}$. Two students propose different approaches: Student A uses spherical coordinates, Student B uses Cartesian coordinates with the order $dz \, dy \, dx$. Which analysis is most accurate?
   Answer: Student A's approach is superior because the integrand becomes $\rho^4 \sin^3 \phi \cos \phi \sin \theta \cos \theta$ which separates into a product of single-variable integrals The correct answer is A. In spherical coordinates, $xyz = (\rho\sin\phi\cos\theta)(\rho\sin\phi\sin\theta)(\rho\cos\phi) = \rho^3\sin^2\phi\cos\phi\sin\theta\cos\theta$. With the Jacobian $\rho^2\sin\phi$, the integrand becomes $\rho^5\sin^3\phi\cos\phi\sin\theta\cos\theta$. The integral separates: $\int_0^1 \rho^5 d\rho \cdot \int_0^{\pi/2} \sin^3\phi\cos\phi d\phi \cdot \int_0^{\pi/2} \sin\theta\cos\theta d\theta$. Choice B ignores that Cartesian coordinates require complex limits. Choice C is wrong—the integral is positive, not zero. Choice D understates the advantage of separability in spherical coordinates.

Start this quiz