Spherical Coordinates - Calculus 3

1. A point has Cartesian coordinates $(-2\sqrt{3}, 2, 4)$. When expressed in spherical coordinates with $0 \leq \theta < 2\pi$ and $0 \leq \phi \leq \pi$, what is the value of $\theta$?
   Answer: $\frac{5\pi}{6}$ When converting from Cartesian to spherical coordinates, you need to find three values: $\rho$ (distance from origin), $\phi$ (angle from positive z-axis), and $\theta$ (azimuthal angle in the xy-plane). The key insight is that $\theta$ depends only on the x and y coordinates, not z. To find $\theta$, you use the relationship $\tan \theta = \frac{y}{x}$. With the point $(-2\sqrt{3}, 2, 4)$, we have $x = -2\sqrt{3}$ and $y = 2$, so: $\tan \theta = \frac{2}{-2\sqrt{3}} = -\frac{1}{\sqrt{3}}$ Since the point lies in the second quadrant (negative x, positive y), you need an angle between $\frac{\pi}{2}$ and $\pi$. The reference angle where $\tan = \frac{1}{\sqrt{3}}$ is $\frac{\pi}{6}$, so in the second quadrant: $\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$. Let's examine why the other answers are wrong. Choice A ($\frac{2\pi}{3}$) gives $\tan \theta = -\sqrt{3}$, not $-\frac{1}{\sqrt{3}}$. Choice B ($\frac{7\pi}{6}$) is in the third quadrant where both sine and cosine are negative, but our point has positive y. Choice D ($\frac{4\pi}{3}$) is also in the third quadrant and gives $\tan \theta = \sqrt{3}$. The correct answer is C: $\frac{5\pi}{6}$. Study tip: Always check which quadrant your point lies in first—this immediately eliminates half the answer choices and prevents sign errors in spherical coordinate conversions.
2. Consider the region in spherical coordinates defined by $2 \leq \rho \leq 4$, $\frac{\pi}{6} \leq \theta \leq \frac{\pi}{3}$, and $\frac{\pi}{4} \leq \phi \leq \frac{\pi}{2}$. What is the $z$-coordinate range of this region?
   Answer: $0 \leq z \leq 2\sqrt{2}$ In spherical coordinates, $z = \rho\cos\phi$. The minimum $z$ occurs when $\rho$ is minimum and $\phi$ is maximum: $z_{min} = 2\cos(\frac{\pi}{2}) = 0$. The maximum $z$ occurs when $\rho$ is maximum and $\phi$ is minimum: $z_{max} = 4\cos(\frac{\pi}{4}) = 4 \cdot \frac{\sqrt{2}}{2} = 2\sqrt{2}$. Choice B incorrectly excludes $z = 0$. Choice C uses $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ instead of $\cos(\frac{\pi}{4})$. Choice D combines both errors.
3. In spherical coordinates, the equation $\phi = \frac{\pi}{4}$ represents a specific geometric surface. What is the Cartesian equation of this surface?
   Answer: $z^2 = x^2 + y^2$ When converting between spherical and Cartesian coordinates, you need to understand what each spherical variable represents. The angle $\phi$ measures the angle from the positive z-axis down to the point, while $\rho$ is the distance from origin and $\theta$ is the azimuthal angle. The key conversion relationships are: $x = \rho \sin \phi \cos \theta$, $y = \rho \sin \phi \sin \theta$, and $z = \rho \cos \phi$. When $\phi = \frac{\pi}{4}$, we have $\cos \phi = \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$ and $\sin \phi = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$. From $z = \rho \cos \phi$, we get $z = \rho \cdot \frac{\sqrt{2}}{2}$, so $\rho = \frac{z\sqrt{2}}{\sqrt{2}} = z\sqrt{2}$. Also, $x^2 + y^2 = \rho^2 \sin^2 \phi = \rho^2 \cdot \frac{1}{2}$. Substituting $\rho = z\sqrt{2}$: $x^2 + y^2 = (z\sqrt{2})^2 \cdot \frac{1}{2} = 2z^2 \cdot \frac{1}{2} = z^2$. Therefore, $z^2 = x^2 + y^2$, which is answer D. Choice A gives $z = \sqrt{x^2 + y^2}$, which only represents the upper half of our cone ($z \geq 0$). Choice B incorrectly includes a factor of 2. Choice C represents a sphere, not a cone, and would correspond to a completely different spherical equation. Remember: constant $\phi$ in spherical coordinates always creates a cone with vertex at the origin. When $\phi = \frac{\pi}{4}$, this cone makes a 45° angle with the z-axis, giving equal relationships between vertical and horizontal distances.
4. A curve in space is parameterized in spherical coordinates by $\rho(t) = 2t$, $\theta(t) = t^2$, $\phi(t) = \frac{\pi}{2} - \frac{t}{2}$ for $0 \leq t \leq 2$. What is the $z$-coordinate when $t = 1$?
   Answer: $2\sin(\frac{1}{2})$ At $t = 1$: $\rho = 2$, $\theta = 1$, $\phi = \frac{\pi}{2} - \frac{1}{2}$. The $z$-coordinate is $z = \rho\cos\phi = 2\cos(\frac{\pi}{2} - \frac{1}{2}) = 2\sin(\frac{1}{2})$, using the identity $\cos(\frac{\pi}{2} - x) = \sin(x)$. Choice A incorrectly expands $\sin(\frac{1}{2})$. Choice C uses $\cos(\frac{1}{2})$ instead of $\sin(\frac{1}{2})$. Choice D has an incorrect factor of $\sqrt{2}$.
5. In spherical coordinates, consider the intersection of the surfaces $\rho = 4$ and $\phi = \frac{\pi}{3}$. This intersection forms a circle. What is the radius of this circle in Cartesian coordinates?
   Answer: $2\sqrt{3}$ The intersection of $\rho = 4$ (sphere of radius 4) and $\phi = \frac{\pi}{3}$ (cone) is a circle on the sphere at a fixed distance from the $z$-axis. The distance from the $z$-axis is $r = \rho\sin\phi = 4\sin(\frac{\pi}{3}) = 4 \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3}$. This is the radius of the circle. Choice A uses $\cos$ instead of $\sin$. Choice C leaves the expression unsimplified. Choice D uses the wrong trigonometric function.

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