Polar coordinates locate a point by a distance r from the origin and an angle \theta from the positive x-axis, instead of by horizontal and vertical displacement. Nothing about that idea changes in this course. It resurfaces almost immediately as the angular half of cylindrical coordinates, the natural system for anything built around an axis of symmetry.
Picture a searchlight anchored at the origin. It sweeps through angle \theta and its beam reaches out a distance r. That's the whole system: point the light, then dial in how far to shine it. A negative r just flips the beam to the opposite ray, still at the same angle. It never means shining below the ground. The light can also spin all the way around and come back to where it started. So the very same point can be lit up by more than one angle-and-distance combination.
Polar coordinates are a way to locate a point by a distance r from the origin and an angle \theta from the positive x-axis, instead of by horizontal and vertical displacement. A given point has infinitely many valid polar representations. (r,\theta) and (r,\theta+360°) name the same point, and so does (-r,\theta+180°). A negative r is plotted by walking the OPPOSITE ray, not below the axis the way a negative height would suggest on an ordinary graph.
This is the single biggest source of polar-plotting errors. It only gets more important once triple integrals in cylindrical or spherical coordinates enter the picture. Those integrals depend on tracking which range of r, \theta, or a third spherical angle covers a solid region, without double-counting or missing any of it.
Formula
Polar to Rectangular Conversion
Recalled from Calculus II — Integral Calculus & Infinite Series · Converting Between Polar and Rectangular — first taught there.
For a point with polar coordinates (r, \theta), x = r\cos\theta, \qquad y = r\sin\theta.
This is nothing but the unit-circle definitions of cosine and sine — \cos\theta and \sin\theta are the x- and y-coordinates on a circle of radius 1 — stretched from radius 1 out to radius r.
Formula
Rectangular to Polar Conversion
Recalled from Calculus II — Integral Calculus & Infinite Series · Converting Between Polar and Rectangular — first taught there.
For a point with rectangular coordinates (x, y), not the pole, r = \sqrt{x^2 + y^2}, \qquad \tan\theta = \frac{y}{x}.
r = \sqrt{x^2+y^2} is just the distance formula from the origin to (x,y). Dividing the two conversion equations above, \frac{y}{x} = \frac{r\sin\theta}{r\cos\theta} = \tan\theta, whenever x \neq 0.
Constant r, moving \theta
Drag the \theta scrubber — the point stays exactly 3 units from the origin the entire time, tracing a circle.
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Watch r turn negative
Scrub past where the strip chart below dips under 0 — the point jumps onto the opposite ray, not below the axis.
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Worked example
Converting a polar point to rectangular
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1
Convert the polar point \left(5,\ \dfrac{2\pi}{3}\right) to rectangular coordinates.
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x = r\cos\theta = 5\cos\!\left(\dfrac{2\pi}{3}\right) = 5\!\left(-\dfrac12\right) = -2.5
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y = r\sin\theta = 5\sin\!\left(\dfrac{2\pi}{3}\right) = 5\!\left(\dfrac{\sqrt3}{2}\right) \approx 4.33
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\boxed{(x,y) \approx (-2.5,\ 4.33)}. This has negative x and positive y, consistent with \frac{2\pi}{3} sitting in the second quadrant.
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Where this goes next
Chapter 8: Triple integrals and coordinate systems
Cylindrical coordinates are (r,\theta) with a z appended, unchanged. Spherical coordinates reuse \theta again, alongside a new angle measured down from the z-axis. Every conversion formula reviewed here is reused directly, not replaced, once integrals move into three dimensions.
Quick review
Practice from earlier coursework, pulled in for this topic.
From College Algebra
Write \left(-8, \frac{5\pi}{6}\right) in standard form (r\ge0, 0\le\theta<2\pi).
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
The angle can't stay the same once you flip r's sign — you must also add \pi to \theta, giving \frac{5\pi}{6}+\pi=\frac{11\pi}{6}.
r must become positive, not stay negative — standard form requires r\ge0.
Adding \pi to \frac{5\pi}{6} gives \frac{5\pi}{6}+\frac{6\pi}{6}=\frac{11\pi}{6}, not \frac{5\pi}{3}.
Why
Since r=-8 is negative, apply the sign-flip rule: negate r and add \pi to \theta. \left(-8,\frac{5\pi}{6}\right) = \left(8, \frac{5\pi}{6}+\pi\right) = \left(8,\frac{11\pi}{6}\right), which is already in [0,2\pi).
Since r is negative, apply the sign-flip rule first (negate r, add \pi to \theta), then check whether the resulting angle is already in [0,2\pi).
From Calculus II — Integral Calculus & Infinite Series
Give one value of k (a nonzero integer) such that (5,\theta) and (5,\theta+2\pi k) name the same point, for any theta.
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Why
Adding any integer multiple of 2*pi to theta is a whole number of extra full turns, landing back on the same ray.
Correct answer
Also accepted:
Why
Any nonzero integer works -- a full 2pik turn always returns to the same ray.
From College Algebra
How many degrees is \frac{\pi}{5} radians?
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Why
\frac{\pi}{5} \times \frac{180°}{\pi} = \frac{180}{5} = 36°.
Correct answer
Also accepted:
Why
Quick review
Practice from earlier coursework, pulled in for this topic.
From College Algebra
Convert \left(0, \frac{7\pi}{9}\right) to rectangular coordinates.
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The angle is completely irrelevant when r=0 — every angle points at the identical spot, the pole.
It does NOT depend on the angle — r=0 means the point is at the pole regardless of \theta.
That mistakes the angle for an x-coordinate. Substituting r=0 directly: x=0\cos\theta=0, y=0\sin\theta=0.
Why
With r=0, both x=r\cos\theta and y=r\sin\theta come out to 0 no matter what \theta is — the point is the pole, (0,0).
Substitute r=0 into x=r\cos\theta and y=r\sin\theta and see what happens regardless of \theta.
From College Algebra
Find r for the rectangular point (-8, 15), sitting in the second quadrant.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
r=\sqrt{x^2+y^2}=\sqrt{64+225}=\sqrt{289}=17.
Correct answer
Also accepted:
Why
Use the distance formula r=\sqrt{x^2+y^2}.
From College Algebra
What is the rectangular form of the polar equation r = 10, a two-digit radius?
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r=10 squares to r^2=100, not r^2=10 — square both sides before substituting r^2=x^2+y^2.
r=10 describes a circle, not a line — there's no direct substitution turning r alone into a sum x+y.
The Pythagorean-style substitution adds x^2 and y^2; it never subtracts them.
Why
Squaring both sides of r=10 gives r^2=100. Since r^2=x^2+y^2 always, this becomes x^2+y^2=100 — a circle of radius 10 centered at the origin.
Square both sides of r=10 first, then substitute r^2=x^2+y^2.