Compute area inside curve:

$$ r=1+\cos(\theta) $$

Attempt to solve

Plot of this curve in polar coordinates looks like this.

enter image description here

Now i would like to compute area inside this curve. Now one reason that this kind of task feels difficult to me is that i don't think i understand polar coordinates system very good. For starters i haven't ever computed anything in this coordinate system. However i do know how to compute various areas in euclidean plane / euclidean space

the equation isn't represented in polar form. So there needs to be some kind of conversion between these two coordinate systems.

enter image description here

So i would like to have it in form where $(x,y) \mapsto (r\cos \theta,\ r\sin \theta)$. I have formula expressed in terms of $r$ which gives me length of hypotenuse with various inputs (when we are in polar coordinate system). I could probably define this in parametric form.

$$ r(\theta)=\begin{bmatrix} x=r\cos \theta \\ y=r\sin\theta \end{bmatrix} $$

$$ r(\theta)=\begin{bmatrix} x=(1+\cos\theta)\cos\theta \\ y=(1+\cos\theta)\sin\theta \end{bmatrix} $$

Now if i create plot of this parametric equation in Matlab i should get same result as from WolframAlpha.

Plot with Matlab:

enter image description here

Now i have parametric equation which describes the $r=1+\cos\theta$ in polar coordinates. How do you compute the area from this ? It is visible from the image that there is symmetry between areas where

$$ \mathrm{area}([0,\pi]) = \mathrm{area}([\pi,2\pi]) $$ Maybe this can be used as to our advantage when computing area with integrals. I haven't computed any integrals with parametric equation's so maybe someone could help with that ?

If someone does that would be highly appreciated.

  • $\begingroup$ It's much easier to compute this as an integral in polar co-ordinates than in cartesian co-ordinates. $\endgroup$
    – John Doe
    Feb 4 '18 at 19:10
  • $\begingroup$ only problem is i don't understand the polar coordinate system that well that i could do this @JohnDoe $\endgroup$
    – Tuki
    Feb 4 '18 at 19:11
  • $\begingroup$ Have a look at this video $\endgroup$
    – John Doe
    Feb 4 '18 at 19:15

Area of a curve given in polar coordinate is given by $$\rm{A} = \frac 12\int_{\theta_1}^{\theta_2} r^2(\theta) \:\rm{d} \theta$$

We can obtain this formula by dividing the curve into infinitely many thin sectors. (each subtending angle $\rm d \theta$ at the origin)

Then use the formula of area of sector to obtain $$dA=\frac{1}{2} r^2 d \theta$$

Now integrate both the sides,

$$\int \rm dA=A= \frac 12\int_{\theta_1}^{\theta_2} r^2(\theta) \:\rm{d} \theta$$

  • $\begingroup$ do you mind to explain or give me some sort of reference that where does this come from ? $\endgroup$
    – Tuki
    Feb 4 '18 at 19:12
  • $\begingroup$ @Tuki Have a look at my edit, does it help you? $\endgroup$ Feb 4 '18 at 19:24
  • 1
    $\begingroup$ Yes this does help indeed. It makes sense now how it works and thanks for the extra detail. $\endgroup$
    – Tuki
    Feb 4 '18 at 19:34

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