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plane polar coordinates

How does graph sketching work for plane polar coordinates?

In plane polar coordinates (r, φ) where the inequality r2 < r < r1, sketch r1 = 1 + cos φ and r2 = 3/2.
Starting with the easier one R2 = 3/2 is the locus of a point at a fixed distance from the origin - in other words a circle.

For the othe one you should draw up a table with R2 for various angles between 0 and 2pi in increments of pi/4, then join the dots (smoothly)
PS: meant to add that, depending on what exactly the question is asking for, you may need to shade the area corresponding to the given inequality.
Reply 3
Avatar for TSR George
TSR George
OP
17
Thanks, I've now got that! The next part of the question was to calculate the value of the area D.

(The area of integration, D, is defined in plane polar coordinates (r, φ) by the inequality r2 < r < r1, where r1 = 1 + cos φ and r2 = 3/2. Calculate the value of the area D. )

I'm having difficulty with setting up the integration, and working out the limits of the integral. Any help would be much appreciated!
1. Go to Desmos and graph the regions in question:

https://www.desmos.com/calculator

You want the area of a crescent.

2. Find θ\theta corresponding to the points of intersection of the curves.

Then it's either an application of A=∫r22dθA=\int \frac{r^2}{2} d \theta

or a double integral A=∬∣J∣drdθA=\iint |J| dr d\theta with appropriate limits which will end up as pretty much the same thing.
As atsruser has pointed out, area in polar coordinates is calculated as

A=∫12r2dθA=\int \frac{1}{2}r^2 d\theta

(You can sanity check this this for a circle, integrating between zero and 2pi).

For your question, the limits of integration are the intersections between the two curves, and the "shaded area" is the area of the cardioid shape less the area of the circle (both of them integrated between the angular limits just mentioned).
Reply 6
Avatar for TSR George
TSR George
OP
17
Original post by atsruser
1. Go to Desmos and graph the regions in question:

https://www.desmos.com/calculator

You want the area of a crescent.

2. Find θ\theta corresponding to the points of intersection of the curves.

Then it's either an application of A=∫r22dθA=\int \frac{r^2}{2} d \theta

or a double integral A=∬∣J∣drdθA=\iint |J| dr d\theta with appropriate limits which will end up as pretty much the same thing.


This is brilliant, thank you so much :biggrin:
Reply 7
Avatar for 3672N
3672N
10
Need a little help with this one plz
IMG_20201227_211455.jpg157.5KB

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