I've plotted two polar curves.

PolarPlot[{3 Sin[t], 1 + Sin[t]}, {t, 0, 2 Pi}, PlotRange -> {-1, 3}]

enter image description here

I'd like to shade the region that lies inside the circle but outside of the cardioid. Then I'd like to find the area of the shaded region using Mathematica's Area command.

Is this possible using polar coordinates? Can someone share some suggestions?

Update: Thanks for posting some old questions I asked. Couldn't find them. Here is what I finally came up with:

Clear[r, t]
  PolarPlot[{3 Sin[t], 1 + Sin[t]}, {t, 0, 2 π}],
  ParametricPlot[{r Cos[t], r Sin[t]}, {t, π/6, 5 π/6}, {r, 1 + Sin[t], 3 Sin[t]}]]

enter image description here

Now, rather than using ImplicitRegion, I used:

Area[{r Cos[t], r Sin[t]}, {t, π/6, 5 π/6}, {r, 1 + Sin[t], 3 Sin[t]}]

Which returns an exact answer of $\pi$. Here is the timing.

Area[{r Cos[t], r Sin[t]}, {t, π/6, 5 π/6}, {r, 1 + Sin[t], 3 Sin[t]}] // 

{0.722002, π}


2 Answers 2


One approach is to use ImplicitRegion to represent the disk and cardioid regions by using your formulas as the maximum radius in polar coordinates and converting this to a cartesian representation that is easier to use with ImplicitRegion. Then we can get your desired region as the RegionDifference and plot it via DiscretizeRegion:

tocartesian = {t -> ArcTan[x, y], r -> Sqrt[x^2 + y^2]}
diskregion = ImplicitRegion[r < (3 Sin[t]) /. tocartesian // Simplify, {x, y}]
cardioidregion = ImplicitRegion[r < (1 + Sin[t]) /. tocartesian // Simplify, {x, y}]
crescentregion = RegionDifference[diskregion, cardioidregion]
DiscretizeRegion[crescentregion, PrecisionGoal->6]

Discretized crescent shaped region

(* Out[1]= 3.14159 *)

The area seems to be Pi. In theory we should be able to verify this symbolically with


but Mathematica seems to take a bit longer to solve the integral for this region (i.e. it didn't finish on my machine after a few minutes).

  • 1
    $\begingroup$ +1 - you could also use your replacement rule to transform the difference region, and use reg = r < (3 Sin[t]) && r > (1 + Sin[t]) /. tocartesian; {RegionPlot[reg, {x, -3/2, 3/2}, {y, 1/2, 3}], DiscretizeRegion@ImplicitRegion[reg, {x, y}]} $\endgroup$
    – Jason B.
    Commented Apr 27, 2017 at 16:06
  • $\begingroup$ @JasonB. Yeah, i was thinking the same, and trying out the same thing at the moment actually :) The thing that i'm trying to figure out his how to get the area of the region. It's easy and fast for the discretized version with RegionMeasure, but the implicit version seems to set off a nontrivial integral computation in Mathematica and doesn't seem to finish... $\endgroup$ Commented Apr 27, 2017 at 16:11
  • $\begingroup$ I did notice that Area[ ImplicitRegion[... was taking a long time, so I also didn't see a better way to get the area. $\endgroup$
    – Jason B.
    Commented Apr 27, 2017 at 17:06

The same as Thies Heidecke calculated it, only with other procedures.

reg = ImplicitRegion[TransformedField["Polar" -> "Cartesian", 
    r < 3 Sin[φ] && r > 1 + Sin[φ], {r, φ} -> {x, y}], {x, y}];
Area@DiscretizeRegion[reg, PrecisionGoal -> 6]



Another way with area = 1/2 Integrate[r^2(φ) dφ]:

r1 = 1 + Sin[φ];
r2 = 3 Sin[φ];
{φ1, φ2} = φ /. Solve[r1 == r2, φ] /. C[1] -> 0
{π/6, (5 π)/6}

area = 1/2 Integrate[r2^2 - r1^2, {φ, φ1, φ2}] //AbsoluteTiming
{0.156459, π}

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