Area between polar curves help

[degreeplus]

Well I missed a day in class of my calculus for a dental appointment and now I'm completely lost. I looked at the notes and tried figuring these types of problems out several ways but I just can't see it.

I know to use the formula 1/2 definite integral of (r1)^2-(r2)^2 dthadum

** could someone teach me also how to make definite integral signs and input thadum (I have no clue on how to spell thadum either)?

Here's an example problem I'm frustrated with:

Find the area shared by the cirlces r=2cos(thadum) and r=2sin(thadum)

Here's some of my work:

I set sin= cos and thought that sin and cos are equal at pie/4 and 5pie/4
this doesnt make sense with the graph. Do I look at where it the interesect on the graph or not because Im confused on how to get the limits.

and then I just plugged those and my upper and lower limits of integration and plugged in r1=2cos and r2=2sin to the above formula.

I need help understanding on how to get upper and lower limits. I think that is my main problem. Thanks for whoever helps me out.

 

[soroban]

Hello, degreeplus!

Find the area shared by the cirlces \(\displaystyle \,r\:=\:2\cdot\cos(\theta)\,\) and \(\displaystyle \,r\:=\:2\cdot\sin(\theta)\)

                |
              * * *
          *     |     *
        *       |       *
       *        |        *
                |             ¼π
      *         |         * /
      *         +       * * *
      *         |   *:::::*     *
                |  *:::::         *
       *        |*:::::::*         *
        *       |:::::::*
          *     *:::::*             *
    ----------*-*-*-------+---------*--
                *                   *
                |
                |*                 *
                | *               *
                |   *           *
                |       * * *
                |

The area is comprised of two integrals . . .

. . \(\displaystyle \L A_1 \;=\;\frac{1}{2}\int^{\;\;\;\frac{\pi}{4}}_0\left(2\cdot\sin\theta)^2d\theta\)

. . \(\displaystyle \L A_2 \;=\;\frac{1}{2}\int^{\;\;\;\frac{\pi}{2}}_{\frac{\pi}{4}}\left(2\cdot\cos\theta)^2d\theta\)

Got it?

 

[degreeplus]

I was actually thinking about this and was wondering what the area between polar curves formula was for. So i guess use the formula some times not all of the time. Thanks for your help.

 

[degreeplus]

actually I still need help understanding.
I have this example here:
Find the area of the region that lies inside the circle r=1 and outside the cardioid r=1-cos

Is the method soroban used the same as the way you would approach this problem?

Here is how this problem is solved A= 1/2 definite integral from -pie/2 to pie /2 of (1^2-(1-cos)^2)dthadum.

The way soroban did the problem did not include the formula of area between polar curves. Is it just me or do they just have seperate approaches because I need to understand how to do the problem in a certain method not just solving the problem. Overall right now im still confused.

 

[soroban]

Hello, degreeplus!

Find the area of the region that lies inside the circle \(\displaystyle r\,=\,1\)
and outside the cardioid \(\displaystyle r\:=\:1\,-\,\cos\theta\)

Is the method soroban used the same as the way you would approach this problem? . no

Here is how this problem is solved: \(\displaystyle \:A\:=\:\frac{1}{2}\L\int^{\;\;\;\frac{\pi}{2}}_{-\frac{\pi}{2}}\,\left[1^2\,-\,(1\,-\,\cos\theta)^2\right]\,d\theta\;\;\) Right!

The way soroban did the problem did not include the formula of area between polar curves.
Do they just have seperate approaches? .yes

We need to make a fairly accurate sketch and carefully examine the region.


In the first problem, the region was not between two curves.

. . From \(\displaystyle 0\) to \(\displaystyle \frac{\pi}{4}\), the region was inside the circle \(\displaystyle r\,=\,2\sin\theta\)

. . From \(\displaystyle \frac{\pi}{4}\) to \(\displaystyle \frac{\pi}{2}\), the region was inside the circle \(\displaystyle r\,=\,2\cos\theta\)


With this problem, the region looks like this:

                            |
                  * * *     |
              *           * |
           *              * * *
         *            *     | *:::*
        *           *       |  *::::*
                   *        |  *:::::*
                            |  *::::::
       *          *         | *:::::::*
    ---*----------*---------*:-:-:-:-:*----
       *          *         | *:::::::*
                            |  *::::::
                   *        |  *:::::*
        *           *       |  *::::*
         *            *     | *:::*
           *              * * *
              *           * |
                  * * *     |
                            |

We find that the curves intersect at \(\displaystyle \theta\,=\,-\frac{\pi}{2},\,\frac{\pi}{2}\)
. . and that the region is between the two curves.

So we find the area inside the circle on that interval: \(\displaystyle \:\frac{1}{2}\L\int\)\(\displaystyle 1^2\,d\theta\)
. . and subtract the area inside the cardioid on that interval: \(\displaystyle \:\frac{1}{2}\L\int\)\(\displaystyle (1\,-\,\cos\theta)^2\,d\theta\)

And that's how they got that area formula.

 

[degreeplus]

Thanks that clears up a lot of uncertainty.