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## Calculus 2

### Unit 5: Lesson 8

Area: polar regions (two curves)

# Worked example: Area between two polar graphs

AP.CALC:
CHA‑5 (EU)
,
CHA‑5.D (LO)
,
CHA‑5.D.1 (EK)
,
CHA‑5.D.2 (EK)
More practice with the area encloised by polar functions, this time with two graphs.

## Want to join the conversation?

• At , why does he neglect the 1/2 shown in the previous equation. He jumps from 1/2 (3sintheta)^2 to simply 9 (sintheta)^2. Why would it not be 9/2....
• So first he sets up two different equations for the two different regions but then he discusses that both the regions have the same area hence he only uses one equation and multiplies it by 2 which cancels out the 1/2 in front and hence he just takes out the 9 as is and as there is no 1/2 out front, it remains as 9 and not 9/2. Hope that helps.
• at the exam we have to make the graph by ourselves, then how can we calculate the integration intervals? Like without a graph how can you calculate it's from 0 to pi/4 and fro pi/4 to pi/2?
great vids btw!
• In my course we were given the following steps to graph a polar function: 1) recognize what kind of graph you are dealing with first. The general forms of polar graphs are good to know. For example, r = asin𝛉 and r = acos𝛉 are circles, r = cos(n𝛉) is a rose curve, r = a + bcos𝛉 where a=b is a cardioid, r = a + bcos𝛉 where a<b is a limaçon, and r^2 = a^2sin(2𝛉) and r^2 = a^2cos(2𝛉) are lemniscates. Knowing what the generic graph looks like will help you make sure that your graph is correct. 2) Set the polar expression equal to zero and solve for theta. This will give you the various "pitch lines." These pitch lines and the coordinate axes will also tell you your limits of integration, depending on what curve you are trying to integrate. For example, the rose curve cos(2𝛉) equals zero when theta is equal to π/4, 3π/4, 5π/4, and 7π/4. If you look at the graph of cos(2𝛉), you will see that each of the petals of the rose curve are nestled between these lines. If this is too much to graph at once for you, then try just graphing the first quadrant and going on from there. 3) set dr/d𝛉 equal to zero. This will give you candidates for where the curve is at a maximum. In the case of cos(2𝛉), it will give you the endpoints of each petal. In some cases you may also need to find where the tangent line is undefined. 4) sketch the pitch lines found in step two. Then plot the points found in step 3. Remember to keep in mind what the generic form of your graph looks like, and then draw your graph! I hope this helps some! :)
• why is it that for rcos(theta) the radius is 0 at pie, when clearly cos(pie) is -1 and so the radius would have to be -3? where is the graph from when theta equals to pie/2 to 3pie/2?
• In the polar coordinate system, r denotes the distance of the point from the origin. Having -a for r means going a distance of a in the opposite direction. Suppose that at an angle of pi, r = -3. This means that you have to go 3 in the direction. This is just (in fact exactly) the same as going -3 along the number line in the negative direction ( -(-3) = 3 ). In general, mathematically:
( -r , theta ) = ( r , theta - 0.5pi )
https://en.wikibooks.org/wiki/Calculus/Polar_Introduction
I apologize for the fact that much or all of the literature of the page is in degrees.
• where does he get the limits pi/4 to pi/2 for the second integral?
• If you look at the rays starting from the center, you can see that they represent various values of the angle theta. You can even split the whole polar plane into even more rays. All of those rays from 0 to pi/2 intersect the shaded region, so it means that theta goes from 0 to pi/2 (a total of pi/2 degrees) for our shaded region. Since we evaluate one integral from 0 to pi/4 (a total of pi/4 degrees), we evaluate the other integral from pi/4 to pi/2 (another total of pi/4 degrees). The total interval has a degree measure of pi/4 + pi/4, so we get our pi/2 total, for the whole shaded region.
• at , why did he choose pi/4?
• Because the angle pi/4 is when the area stops being bounded by the first circle (r = 3 sin theta) and starts being bounded by the second circle (r = 3 cos theta). With the first integral, he is trying to measure the red area, which is bounded by the first circle (r = 3 sin theta) from angle 0 to pi/4.
• At , why are the integrals multiplied by 1/2?
• Sal, you forgot to double the answer! In case you forgot, it is because you only solved for one half of the area and it is symmetric.
(1 vote)
• No, he didn't forget to double the answer. He did, actually double the answer at the really beginning at . That is why, that 1/2 at the front is gone.
• What would the limits of the integration be if we were to use double integration?
• You can use the symmetry as in the video:
The integral would look like ∫∫r dr dθ
The bounds of the outer integral would be 0 to π/4
The bounds of the inner integral would be 0 to 3cosθ.
Evaluating the inner integral will give you a result that looks EXACTLY like the integral he writes at in the video where he has 1/2 ∫(3cosθ)^2 dθ, with 0 and π/4 being the limits.
Hope that helped!