Question

help please
find the area of the region that lies inside the first curve and outside the second curve

Answer 1

\[r=3\cos (\theta)\]
\[r=1+\cos(0)\]

Answer 2

so how do i know which area is it ?

Answer 3

@zepdrix

Answer 4

Do you mean that?
\[r=1+\cos(0)\]
or that?
\[r=1+\cos(\theta)\]

Answer 5

there are two functions :O

Answer 6

I mean the second one?

Answer 7

cos(o)=1
and 1 + cos(o)=2
so you mean r=2??

Answer 8

@marcelie

Do you follow?

Answer 9

\(\color{black}{\displaystyle r_1(\theta)=3\cos(\theta) }\)
\(\color{black}{\displaystyle r_2(\theta)=\cos(\theta)+1 }\)

The area
\(\color{black}{\displaystyle \int_{\pi/3}^{-\pi/3}\frac{1}{2}\left\{\left[3\cos(\theta)\right]^2-\left[\cos(\theta)+1\right]^2\right\}d\theta }\)

Answer 10

(Let me recheck the setup)

Answer 11

oh okay . there are two functions so i graphed it

Answer 12

https://www.desmos.com/calculator/y5apxfqlsc

Answer 13

oh okay so which part would be shaded ?

Answer 14

This little banana-like part on the left (the top and bottom).

Answer 15

Oh my bad.

Answer 16

I did inside the second curve, and outside the curve.
For that my setup above would be right.

Answer 17

For inside the first, and outside the second, you are taking the other part of the circle (not from \(\pi/3\) to \(-\pi/3\), but), from \(-\pi/3\) to \(\pi/3\).

\(\color{black}{\displaystyle \int_{-\pi/3}^{\pi/3}\frac{1}{2}\left\{\left[3\cos(\theta)\right]^2-\left[\cos(\theta)+1\right]^2\right\}d\theta }\)

Answer 18

Alternatively, if you want to note the symmetry between the top and bottom parts of the area considered, you may also say

\(\color{black}{\displaystyle 2\int_{0}^{\pi/3}\frac{1}{2}\left\{\left[3\cos(\theta)\right]^2-\left[\cos(\theta)+1\right]^2\right\}d\theta }\)

Answer 19

```
btw, correct to my previous comment:
I did inside the second curve, and outside the first curve.
For that my setup above would be right.
```

Answer 20

More info ... http://tutorial.math.lamar.edu/Classes/CalcII/PolarArea.aspx
(It suits most students for most purposes.)

Answer 21

ohh.. so then its outside the curve - inside the curve ?

Answer 22

The setup above is inside the first curve, and outside the second.

Answer 23

Or, if you considered set theory ... \(r_1-r_2\).
(All elements in \(r_1\) that are not in \(r_2\))

Answer 24

I'm just trying to give better analogies of material you might have previously learned ... ignore if you didn't ..

Answer 25

yeah it kinda seems familiar

Answer 26

So, do you see why I set up the integral that way? (If you see either of the way - whether you like to use symmetry or not, it's fine, as long as you have one sufficient way of doing it0

Answer 27

If you understand the setup we can proceed to integrate.

Answer 28

(If not, you will need to let me know about any problems you have with it.)

Answer 29

oh okay i think i get it but how did you get the bounds?

Answer 30

Oh, I looked at the graph.

Answer 31

You can click the lines at which the desired area starts or ends, and you will see one of the equations of y selected. In this equation, you will see the angle of boundary.

Answer 32

(I suck at explaining how and where to click even in most critical moment.)

Answer 33

lol its okay but is there an algebraic way to solve it ?
in order to find the bounds

Answer 34

You can set the curves equal to each other, and solve.

Answer 35

\(r_1=r_2\), and then, of course, choose the solutions over the interval \([0,2\pi]\).

Answer 36

i got theta = pi/3

Answer 37

And another solution is -pi/3.

Answer 38

Your solution is in a form \(\theta =2\pi k\pm\frac{2}{3}\pi\) ...

Answer 39

In any case, we know where the boundaries are coming from, yes?

Answer 40

oh how come ? theres another 5pi/3 right ?

Answer 41

No

Answer 42

Over \([0,2\pi]\), only the aforementioned two solutions.

Answer 43

If we can go on with integration, let me know, if not I am willing to address more about the boundaries ...

(Not rushing, just asking what you would like to do now)

Answer 44

I guess let's integrate?

Answer 45

I actually prefer to use the symmetry to eliminate the 1/2.

\(\color{black}{\displaystyle 2\int_{0}^{\pi/3}\frac{1}{2}\left\{\left[3\cos(\theta)\right]^2-\left[\cos(\theta)+1\right]^2\right\}d\theta }\)

\(\color{black}{\displaystyle \int_{0}^{\pi/3}\left\{8\cos^2(\theta)-2\cos\theta-1\right\}d\theta }\)

\(\color{black}{\displaystyle \int_{0}^{\pi/3}\left\{8\left(\frac{1}{2}\cos(2\theta)+\frac{1}{2}\right)-2\cos\theta-1\right\}d\theta }\)

\(\color{black}{\displaystyle \int_{0}^{\pi/3}\left\{4\cos(2\theta)-2\cos\theta+3\right\}d\theta }\)

Answer 46

and from there simple integration, etc, although I would normally just use a calculator for this if I had to perform a quick calculation.

Answer 47

Anyway, if any questions, ask.
(I'll do my best to address them if I'm online)

Answer 48

oh okay it make sense now :) tysm

Answer 49

yw