# plane polar coordinates

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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.

In plane polar coordinates (r, φ) where the inequality r2 < r < r1, sketch r1 = 1 + cos φ and r2 = 3/2.

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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)

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)

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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.

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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!

(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!

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#5

(Original post by

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!

**pippabethan**)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!

https://www.desmos.com/calculator

You want the area of a crescent.

2. Find corresponding to the points of intersection of the curves.

Then it's either an application of

or a double integral with appropriate limits which will end up as pretty much the same thing.

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#6

**pippabethan**)

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!

(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).

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(Original post by

1. Go to Desmos and graph the regions in question:

https://www.desmos.com/calculator

You want the area of a crescent.

2. Find corresponding to the points of intersection of the curves.

Then it's either an application of

or a double integral with appropriate limits which will end up as pretty much the same thing.

**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 corresponding to the points of intersection of the curves.

Then it's either an application of

or a double integral with appropriate limits which will end up as pretty much the same thing.

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