Cartesian to Cylindrical - Multivariable Calculus

A region $R$ in the $xy$ -plane is defined by $1 \leq x^2 + y^2 \leq 9$ and $y \geq 0$ . When this region is described using cylindrical coordinates, which of the following correctly gives the bounds?
Answer: $1 \leq r \leq 3$ , $0 \leq \theta \leq \pi$ , $z$ arbitrary The constraint $1 \leq x^2 + y^2 \leq 9$ becomes $1 \leq r^2 \leq 9$ in cylindrical coordinates, which simplifies to $1 \leq r \leq 3$ (since $r \geq 0$ ). The constraint $y \geq 0$ means $r\sin(\theta) \geq 0$ . Since $r > 0$ , this requires $\sin(\theta) \geq 0$ , which occurs when $0 \leq \theta \leq \pi$ . Option B incorrectly keeps $r^2$ in the bound instead of taking the square root. Option C uses the full range $0 \leq \theta \leq 2\pi$ , which would include the region where $y < 0$ . Option D restricts $\theta$ to $[0, \frac{\pi}{2}]$ , which only covers the first quadrant and misses the second quadrant where $y > 0$ and $x < 0$ .
A solid region is defined by the Cartesian inequalities $x^2 + y^2 \le 9$ , $y \ge |x|$ , and $0 \le z \le 2$ . Which of the following sets of inequalities describes this same region in cylindrical coordinates?
Answer: $0 \le r \le 3$ , $\pi/4 \le \theta \le 3\pi/4$ , $0 \le z \le 2$ We convert each inequality to cylindrical coordinates. The inequality $x^2 + y^2 \le 9$ becomes $r^2 \le 9$ , which simplifies to $0 \le r \le 3$ since $r$ must be non-negative. The inequality $0 \le z \le 2$ remains unchanged. The inequality $y \ge |x|$ describes the region in the xy-plane that is on or above the graphs of $y=x$ (for $x \ge 0$ ) and $y=-x$ (for $x < 0$ ). The line $y=x$ in the first quadrant corresponds to $\theta = \pi/4$ . The line $y=-x$ in the second quadrant corresponds to $\theta = 3\pi/4$ . The region between these lines in the upper half-plane ( $y>0$ ) corresponds to the angular range $\pi/4 \le \theta \le 3\pi/4$ . Combining these gives the description: $0 \le r \le 3$ , $\pi/4 \le \theta \le 3\pi/4$ , $0 \le z \le 2$ .
A sphere is given by $x^2+y^2+(z-2)^2=4$ and a cylinder by $x^2+y^2=3$ . Which of the following describes their curve(s) of intersection in cylindrical coordinates?
Answer: Two circles given by $r=\sqrt{3}, z=1$ and $r=\sqrt{3}, z=3$ First, convert the equations of the surfaces to cylindrical coordinates. The equation of the sphere $x^2+y^2+(z-2)^2=4$ becomes $r^2+(z-2)^2=4$ . The equation of the cylinder $x^2+y^2=3$ becomes $r^2=3$ , which means $r=\sqrt{3}$ (since $r \ge 0$ ). To find the intersection, substitute the value of $r^2$ from the cylinder equation into the sphere equation: $3 + (z-2)^2 = 4$ . Solving for $z$ : $(z-2)^2 = 1$ , which gives $z-2 = \pm 1$ . The solutions are $z = 2+1=3$ and $z=2-1=1$ . Therefore, the intersection consists of two curves, both with radius $r=\sqrt{3}$ : one circle in the plane $z=1$ and another circle in the plane $z=3$ .
A point has Cartesian coordinates $(-\sqrt{3}, -1, 5)$ . Which of the following are its cylindrical coordinates $(r, \theta, z)$ , where the angle $\theta$ is in the interval $[0, 2\pi)$ ?
Answer: $(2, 7\pi/6, 5)$ Given Cartesian coordinates $(x,y,z) = (-\sqrt{3}, -1, 5)$ . The z-coordinate is the same in both systems, so $z=5$ . The radial coordinate is $r = \sqrt{x^2+y^2} = \sqrt{(-\sqrt{3})^2 + (-1)^2} = \sqrt{3+1} = \sqrt{4} = 2$ . To find the angle $\theta$ , we use $\tan\theta = y/x = -1/(-\sqrt{3}) = 1/\sqrt{3}$ . The reference angle for which $\tan\theta_{ref} = 1/\sqrt{3}$ is $\theta_{ref} = \pi/6$ . Since both $x$ and $y$ are negative, the point lies in the third quadrant. To find the angle in the third quadrant, we add $\pi$ to the reference angle: $\theta = \pi + \pi/6 = 7\pi/6$ . The cylindrical coordinates are $(2, 7\pi/6, 5)$ .
The Cartesian equation $z = \arctan(y/x)$ represents a surface, where the standard principal value of arctan in $(-\pi/2, \pi/2)$ is used. What is the equation of this surface in cylindrical coordinates and what is its geometric shape?
Answer: $z = \theta$ , representing a portion of a helicoid When converting between coordinate systems in multivariable calculus, you need to understand how each variable transforms and recognize the geometric meaning of the resulting equation. To convert $z = \arctan(y/x)$ to cylindrical coordinates, recall that $x = r\cos\theta$ , $y = r\sin\theta$ , and $z = z$ . The key insight is that $\frac{y}{x} = \frac{r\sin\theta}{r\cos\theta} = \frac{\sin\theta}{\cos\theta} = \tan\theta$ . Therefore, $z = \arctan(\tan\theta) = \theta$ (within the principal domain). The surface $z = \theta$ represents a helicoid—specifically, a portion of one. As you move around the z-axis (increasing $\theta$ ), the height $z$ increases linearly. This creates a twisted, ramp-like surface that spirals upward. Since we're using the principal value of arctan, we only get the portion where $\theta \in (-\pi/2, \pi/2)$ , making this a partial helicoid. Choice A incorrectly identifies this as a plane—while $z = \theta$ is linear in $\theta$ , it's not a plane because it spirals around the z-axis. Choice B gives $z = \tan(r)$ , which would result from misapplying the coordinate transformation. Choice C suggests $r = \tan(z)$ , which represents a completely different surface where radius depends on height, not the angular relationship we have here. Remember: when converting coordinate equations, always substitute the transformation relationships carefully and consider what the geometric shape actually looks like by thinking about how the surface behaves as each coordinate varies.
A point moves along a helical path described in cylindrical coordinates by $r = 2$ , $\theta = t$ , $z = 3t$ where $t$ is a parameter. At what value of $t$ (where $0 \leq t \leq 4\pi$ ) is the $x$ -coordinate of the point's position equal to $-\sqrt{3}$ ?
Answer: $t = \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{17\pi}{6}, \frac{19\pi}{6}$ The $x$ -coordinate is $x = r\cos(\theta) = 2\cos(t)$ . Setting $2\cos(t) = -\sqrt{3}$ gives $\cos(t) = -\frac{\sqrt{3}}{2}$ . In $[0, 2\pi]$ , this occurs at $t = \frac{5\pi}{6}$ and $t = \frac{7\pi}{6}$ . In $[2\pi, 4\pi]$ , adding $2\pi$ to each solution gives $t = \frac{17\pi}{6}$ and $t = \frac{19\pi}{6}$ . Since $\frac{19\pi}{6} < \frac{24\pi}{6} = 4\pi$ , all four values are valid.
A point $P$ has cylindrical coordinates $(r, \theta, z) = (4, \frac{2\pi}{3}, -2)$ . If this point is rotated about the $z$ -axis by an angle of $\frac{\pi}{6}$ in the positive direction, what are the Cartesian coordinates of the resulting point?
Answer: $(-2\sqrt{3}, 2, -2)$ After rotation, the new cylindrical coordinates are $(4, \frac{2\pi}{3} + \frac{\pi}{6}, -2) = (4, \frac{5\pi}{6}, -2)$ . Converting to Cartesian: $x = 4\cos(\frac{5\pi}{6}) = 4 \cdot (-\frac{\sqrt{3}}{2}) = -2\sqrt{3}$ , $y = 4\sin(\frac{5\pi}{6}) = 4 \cdot \frac{1}{2} = 2$ , $z = -2$ . Since $\frac{5\pi}{6}$ is in the second quadrant where cosine is negative and sine is positive, we get $(-2\sqrt{3}, 2, -2)$ .
A particle's position in Cartesian coordinates at time $t$ is given by $P(t) = (t^2, t, 5)$ . What is the rate of change of its radial cylindrical coordinate, $dr/dt$ , at $t=2$ ?
Answer: $9/\sqrt{5}$ The radial cylindrical coordinate $r$ is given by $r = \sqrt{x^2+y^2}$ . Here, $x(t)=t^2$ and $y(t)=t$ . So, $r(t) = \sqrt{(t^2)^2 + t^2} = \sqrt{t^4+t^2}$ . To find $dr/dt$ , we differentiate $r(t)$ with respect to $t$ using the chain rule: $dr/dt = \frac{d}{dt}(t^4+t^2)^{1/2} = \frac{1}{2}(t^4+t^2)^{-1/2} \cdot (4t^3+2t) = \frac{4t^3+2t}{2\sqrt{t^4+t^2}} = \frac{2t^3+t}{\sqrt{t^4+t^2}}$ . Now, evaluate this derivative at $t=2$ : $dr/dt|_{t=2} = \frac{2(2^3)+2}{\sqrt{2^4+2^2}} = \frac{2(8)+2}{\sqrt{16+4}} = \frac{18}{\sqrt{20}}$ . To simplify, $\sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5}$ . So, $dr/dt = \frac{18}{2\sqrt{5}} = \frac{9}{\sqrt{5}}$ .
A curve in 3D space is defined parametrically by $x(t) = 3t$ , $y(t) = 4t$ , $z(t) = 5t$ for $t \geq 0$ . What is the equation of this curve when expressed in cylindrical coordinates $(r, \theta, z)$ ?
Answer: $r = z$ , $\theta = \arctan(\frac{4}{3})$ , $z = z$ First, convert the parametric equations to cylindrical coordinates. We have $r = \sqrt{x^2 + y^2} = \sqrt{(3t)^2 + (4t)^2} = \sqrt{9t^2 + 16t^2} = \sqrt{25t^2} = 5t$ . Also, $\theta = \arctan(\frac{y}{x}) = \arctan(\frac{4t}{3t}) = \arctan(\frac{4}{3})$ (constant). Since $z = 5t$ , we can express $t = \frac{z}{5}$ . Substituting: $r = 5 \cdot \frac{z}{5} = z$ . Therefore, the curve in cylindrical coordinates is $r = z$ , $\theta = \arctan(\frac{4}{3})$ . Option A incorrectly keeps the parameter $t$ . Option B has $r = 5z$ instead of $r = z$ . Option D has both incorrect $r$ and $\theta$ expressions.
Consider the region in 3D space described by the intersection of the cylinder $x^2 + y^2 = 4$ and the plane $z = x + y$ . If a point in this region has cylindrical coordinates $(r, \theta, z)$ , which equation correctly relates $z$ to $r$ and $\theta$ ?
Answer: $z = r(\cos(\theta) + \sin(\theta))$ When you encounter problems involving coordinate transformations in multivariable calculus, your goal is to express one coordinate system in terms of another using the fundamental conversion relationships. To find how $z$ relates to $r$ and $\theta$ , you need to use the cylindrical-to-Cartesian conversion formulas. In cylindrical coordinates, $x = r\cos(\theta)$ and $y = r\sin(\theta)$ . Since the region lies on the plane $z = x + y$ , you can substitute these expressions: $z = r\cos(\theta) + r\sin(\theta) = r(\cos(\theta) + \sin(\theta))$ . Let's examine why each answer choice works or fails: Choice A gives $z = 2(\cos(\theta) + \sin(\theta))$ . This incorrectly assumes $r = 2$ for all points in the region. While the cylinder has radius 2, points inside the cylinder have $r \leq 2$ , so this only describes points on the cylinder's surface. Choice B correctly gives $z = r(\cos(\theta) + \sin(\theta))$ . This properly accounts for all radial distances within the cylindrical region. Choice C shows $z = r\sqrt{\cos^2(\theta) + \sin^2(\theta)} = r \cdot 1 = r$ . This incorrectly simplifies the plane equation, ignoring that we need $x + y$ , not the distance formula. Choice D gives $z = r^2(\cos(\theta) + \sin(\theta))$ . This introduces an incorrect $r^2$ term that doesn't arise from the coordinate conversion. Study tip: Always write out the basic conversion formulas ( $x = r\cos\theta$ , $y = r\sin\theta$ ) first, then substitute directly into the given constraint equation. This systematic approach prevents algebraic errors.
A vector field is given in Cartesian coordinates by $\vec{F}(x,y,z) = \langle -y, x, z \rangle$ . Find the cylindrical component of the vector field in the direction of increasing $\theta$ , denoted $F_{\theta}$ , at the point with Cartesian coordinates $(1, \sqrt{3}, 4)$ .
Answer: $2$ When converting vector fields between coordinate systems, you need to express the field components in terms of the new coordinate system's unit vectors. This question tests your ability to transform from Cartesian to cylindrical coordinates. To find $F_{\theta}$ , you need the unit vector in the direction of increasing $\theta$ . In cylindrical coordinates, $\hat{\theta} = \langle -\sin\theta, \cos\theta, 0 \rangle$ . At the point $(1, \sqrt{3}, 4)$ , first convert to cylindrical: $r = \sqrt{1^2 + (\sqrt{3})^2} = 2$ and $\theta = \arctan(\sqrt{3}/1) = \pi/3$ . Therefore, $\sin\theta = \sqrt{3}/2$ and $\cos\theta = 1/2$ . The unit vector becomes $\hat{\theta} = \langle -\sqrt{3}/2, 1/2, 0 \rangle$ . Now compute the dot product: $F_{\theta} = \vec{F} \cdot \hat{\theta} = \langle -\sqrt{3}, 1, 4 \rangle \cdot \langle -\sqrt{3}/2, 1/2, 0 \rangle = (-\sqrt{3})(-\sqrt{3}/2) + (1)(1/2) + (4)(0) = 3/2 + 1/2 = 2$ . Answer A) gives $1$ , which you'd get if you only calculated the $y$ -component contribution and forgot the $x$ -component. Answer C) gives $-\sqrt{3}$ , which is just the $x$ -component of $\vec{F}$ at this point—a common error when students confuse Cartesian components with cylindrical ones. Answer D) gives $0$ , which would result from incorrectly assuming the $z$ -component affects $F_{\theta}$ or making sign errors in the calculation. Remember: cylindrical components require dot products with the appropriate unit vectors. Always convert your point's coordinates first, then construct the unit vectors, then compute the dot product carefully.
A solid region is bounded by the surfaces $r = 1$ , $r = 3$ , $\theta = \frac{\pi}{6}$ , $\theta = \frac{\pi}{3}$ , $z = 0$ , and $z = r\cos(\theta)$ in cylindrical coordinates. At the point with cylindrical coordinates $(2, \frac{\pi}{4}, 1)$ , is this point inside, outside, or on the boundary of this region?
Answer: Inside the region When you encounter a region defined by multiple inequalities in cylindrical coordinates, you need to check whether a given point satisfies all the boundary conditions simultaneously. The region is defined by six surfaces creating boundaries: $1 \leq r \leq 3$ , $\frac{\pi}{6} \leq \theta \leq \frac{\pi}{3}$ , and $0 \leq z \leq r\cos(\theta)$ . To determine if point $(2, \frac{\pi}{4}, 1)$ lies inside this region, check each coordinate against its bounds. For the $r$ -coordinate: $1 \leq 2 \leq 3$ ✓ For the $\theta$ -coordinate: $\frac{\pi}{6} \leq \frac{\pi}{4} \leq \frac{\pi}{3}$ . Since $\frac{\pi}{6} = \frac{2\pi}{12}$ , $\frac{\pi}{4} = \frac{3\pi}{12}$ , and $\frac{\pi}{3} = \frac{4\pi}{12}$ , we have $\frac{2\pi}{12} \leq \frac{3\pi}{12} \leq \frac{4\pi}{12}$ ✓ For the $z$ -coordinate: We need $0 \leq 1 \leq r\cos(\theta) = 2\cos(\frac{\pi}{4}) = 2 \cdot \frac{\sqrt{2}}{2} = \sqrt{2} \approx 1.414$ . So $0 \leq 1 \leq 1.414$ ✓ Since all three conditions are satisfied with strict inequalities (not equalities), the point is inside the region, making (D) correct. (A) is wrong because we have complete information to make this determination. (B) is incorrect because all boundary conditions are satisfied. (C) is wrong because the point doesn't lie exactly on any of the six bounding surfaces—all inequalities are strict. Study tip: Always check each coordinate systematically against its bounds, and remember that "inside" means all inequalities are satisfied without equality.
The equation of a surface in Cartesian coordinates is $(x^2+y^2)^2 = 4(x^2-y^2)$ . Which of the following is the equation for this surface in cylindrical coordinates?
Answer: $r^2 = 4\cos(2\theta)$ When converting equations between coordinate systems, you need to systematically substitute the conversion formulas and simplify. For cylindrical coordinates, the key relationships are $x = r\cos\theta$ , $y = r\sin\theta$ , and $x^2 + y^2 = r^2$ . Starting with $(x^2+y^2)^2 = 4(x^2-y^2)$ , substitute the cylindrical relationships. The left side becomes $(r^2)^2 = r^4$ . For the right side, you need $x^2 - y^2 = r^2\cos^2\theta - r^2\sin^2\theta = r^2(\cos^2\theta - \sin^2\theta)$ . Using the double angle identity, $\cos^2\theta - \sin^2\theta = \cos(2\theta)$ , so the right side becomes $4r^2\cos(2\theta)$ . This gives you $r^4 = 4r^2\cos(2\theta)$ . Dividing both sides by $r^2$ (assuming $r \neq 0$ ), you get $r^2 = 4\cos(2\theta)$ , which is answer C. Let's examine why the other options are wrong: Answer A gives $r = 4\cos(2\theta)$ , which would result from incorrectly taking the square root of both sides without accounting for the $r^2$ factor. Answer B, $r^2 = 4\cos^2\theta$ , comes from failing to use the double angle identity and leaving $\cos^2\theta - \sin^2\theta$ as just $\cos^2\theta$ . Answer D, $r^4 = 4\cos(2\theta)$ , represents stopping before the final simplification step. Remember: coordinate conversions require careful algebraic manipulation and trigonometric identities. Always check that your final form makes sense by considering special angle values.
Point $P$ has Cartesian coordinates $(2, -2, 1)$ . Point $Q$ has cylindrical coordinates $(r, \theta, z) = (3, \pi, 5)$ . What is the square of the distance between $P$ and $Q$ ?
Answer: $45$ When you encounter problems involving points in different coordinate systems, you need to convert everything to the same system before calculating distances. Here, point P is given in Cartesian coordinates while point Q is in cylindrical coordinates. First, convert Q's cylindrical coordinates $(r, \theta, z) = (3, \pi, 5)$ to Cartesian using the conversion formulas: $x = r\cos\theta$ , $y = r\sin\theta$ , and $z = z$ . For point Q: $x = 3\cos(\pi) = 3(-1) = -3$ , $y = 3\sin(\pi) = 3(0) = 0$ , and $z = 5$ . So Q has Cartesian coordinates $(-3, 0, 5)$ . Now calculate the square of the distance between P $(2, -2, 1)$ and Q $(-3, 0, 5)$ using the distance formula: $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$ $d^2 = (-3 - 2)^2 + (0 - (-2))^2 + (5 - 1)^2 = (-5)^2 + (2)^2 + (4)^2 = 25 + 4 + 16 = 45$ Answer A (33) likely comes from incorrectly using $\cos(\pi) = 0$ and $\sin(\pi) = -1$ , swapping the trigonometric values. Answer B (21) might result from calculation errors in the distance formula. Answer D (41) could occur from sign errors or arithmetic mistakes in the squaring process. Study tip: Always double-check your trigonometric values when converting coordinates. Remember that $\cos(\pi) = -1$ and $\sin(\pi) = 0$ , and verify your coordinate conversion before applying the distance formula.
The equation of a surface in Cartesian coordinates is $x^2 + y^2 = 2ay$ , where $a > 0$ is a constant. This surface is a cylinder. Which choice correctly gives its equation in cylindrical coordinates and the minimal angular range required to trace its circular base exactly once?
Answer: $r = 2a \sin\theta$ , for $\theta \in [0, \pi]$ To convert the equation $x^2 + y^2 = 2ay$ to cylindrical coordinates, substitute $x^2+y^2=r^2$ and $y=r\sin\theta$ . This gives $r^2 = 2a(r\sin\theta)$ . This equation can be satisfied if $r=0$ (which is the z-axis) or if we divide by $r$ (assuming $r \ne 0$ ), which gives $r = 2a\sin\theta$ . The Cartesian equation can be rewritten as $x^2 + (y-a)^2 = a^2$ , which is a circle in the xy-plane centered at $(0, a)$ with radius $a$ . Since $a>0$ , this circle lies in the upper half-plane ( $y \ge 0$ ) where $\theta$ is in the interval $[0, \pi]$ . As $\theta$ varies from $0$ to $\pi$ , $r$ starts at $0$ , increases to $2a$ at $\theta=\pi/2$ , and returns to $0$ at $\theta=\pi$ , tracing the circle exactly once. The range $\theta \in [0, 2\pi)$ would trace the circle twice.
A line segment connects point $P_1 = (4, 0, 2)$ and point $P_2 = (-4, 4, 6)$ . What are the cylindrical coordinates $(r, \theta, z)$ of the midpoint of this segment, assuming $0 \le \theta < 2\pi$ ?
Answer: $(2, \pi/2, 4)$ First, find the Cartesian coordinates of the midpoint $M$ using the midpoint formula: $M = \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2} \right)$ . $M = \left( \frac{4+(-4)}{2}, \frac{0+4}{2}, \frac{2+6}{2} \right) = (0, 2, 4)$ . Next, convert these Cartesian coordinates $(x, y, z) = (0, 2, 4)$ to cylindrical coordinates $(r, \theta, z)$ . $z = 4$ . $r = \sqrt{x^2+y^2} = \sqrt{0^2+2^2} = 2$ . Since $x=0$ and $y > 0$ , the point lies on the positive y-axis, so $\theta = \pi/2$ . Thus, the cylindrical coordinates are $(2, \pi/2, 4)$ .
In cylindrical coordinates, the equation $r = 2\sin(\theta)$ represents a curve. What is the Cartesian equation of this same curve?
Answer: $x^2 + (y-1)^2 = 1$ Starting with $r = 2\sin(\theta)$ , multiply both sides by $r$ : $r^2 = 2r\sin(\theta)$ . Since $r^2 = x^2 + y^2$ and $r\sin(\theta) = y$ , we get $x^2 + y^2 = 2y$ . Rearranging: $x^2 + y^2 - 2y = 0$ . Completing the square: $x^2 + (y^2 - 2y + 1) = 1$ , which gives $x^2 + (y-1)^2 = 1$ . This is a circle with center $(0,1)$ and radius $1$ . Option A shows the intermediate step before completing the square. Option B represents a circle centered at $(1,0)$ , which would come from $r = 2\cos(\theta)$ . Option D would result from $r = 2\cos(\theta)$ before completing the square.
The cylindrical coordinate system uses the transformation $x = r\cos(\theta)$ , $y = r\sin(\theta)$ , $z = z$ . If a function $f(x,y,z) = x^2 + y^2 + z^2$ is expressed in cylindrical coordinates, what is the simplified form?
Answer: $f(r,\theta,z) = r^2 + z^2$ When converting functions from Cartesian to cylindrical coordinates, you're essentially substituting the coordinate relationships to express the same function in a different variable system. The key insight is recognizing when algebraic simplification can dramatically clean up your expression. Starting with $f(x,y,z) = x^2 + y^2 + z^2$ , you substitute the cylindrical coordinate transformations: $x = r\cos(\theta)$ and $y = r\sin(\theta)$ . This gives you: $f(r,\theta,z) = (r\cos(\theta))^2 + (r\sin(\theta))^2 + z^2 = r^2\cos^2(\theta) + r^2\sin^2(\theta) + z^2$ Now comes the crucial step: factor out $r^2$ from the first two terms to get $r^2(\cos^2(\theta) + \sin^2(\theta)) + z^2$ . Since the fundamental trigonometric identity tells us that $\cos^2(\theta) + \sin^2(\theta) = 1$ , this simplifies beautifully to $r^2 + z^2$ . Choice A shows the substitution but stops before applying the trigonometric identity—it's mathematically correct but not simplified. Choice C also represents the intermediate step where you've factored out $r^2$ but haven't used the identity $\cos^2(\theta) + \sin^2(\theta) = 1$ . Choice D incorrectly includes $\theta^2$ instead of recognizing that $\theta$ cancels out completely through the trigonometric identity. The correct answer is B: $f(r,\theta,z) = r^2 + z^2$ . Study tip: When converting to cylindrical coordinates, always look for opportunities to use $\cos^2(\theta) + \sin^2(\theta) = 1$ . This identity frequently appears and can dramatically simplify expressions involving $x^2 + y^2$ terms.
The point $P$ has Cartesian coordinates $(-3, 3\sqrt{3}, 7)$ . If we represent this point in cylindrical coordinates as $(r, \theta, z)$ where $0 \leq \theta < 2\pi$ , what is the value of $\theta$ ?
Answer: $\frac{2\pi}{3}$ To find $\theta$ , we use $\theta = \arctan(\frac{y}{x})$ with consideration for the quadrant. Here, $x = -3 < 0$ and $y = 3\sqrt{3} > 0$ , so the point is in the second quadrant. We have $\tan(\theta) = \frac{3\sqrt{3}}{-3} = -\sqrt{3}$ . The reference angle is $\arctan(\sqrt{3}) = \frac{\pi}{3}$ . Since we're in the second quadrant, $\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$ . Option B gives $\frac{\pi}{3}$ , which would correspond to the first quadrant. Option C gives $\frac{4\pi}{3}$ , which is in the third quadrant. Option D gives $\frac{5\pi}{3}$ , which is in the fourth quadrant.

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