What is the area of the region between two parametric curves?

[ineedhelpnow]

find the area of the region that lies inside the first curve and outside the second curve. $r=3cos(\theta)$, $r=1+cos(\theta)$

$3cos(\theta)=1+cos(\theta)$
$2cos(\theta)=1$
$cos(\theta)=\frac{1}{2}$
$\theta= \frac{\pi}{3}, \frac{5 \pi}{3}$$A=\frac{1}{2} \int_{\pi/3}^{5\pi/3} \ (3cos(\theta))^2 d \theta - \frac{1}{2} \int_{\pi/3}^{5\pi/3} \ (1+cos(\theta))^2 d \theta$

i did some by hand but it's super long. i put it into my calculator and got $4 \pi - \frac{9 \sqrt{3}}{4}$ is that right?

 

[MarkFL]

Take a look at the following plot, and see if you spot a problem with the integral you used (which incidentally, is not evaluated correctly).

View attachment 2824

 

[ineedhelpnow]

wrong interval?

 

[MarkFL]

wrong interval?

Correct...what interval should you use? Also, think about using some symmetry to your advantage...:D

 

[ineedhelpnow]

not sure. wasnt i just supposed to find the values of $\theta$?

 

[MarkFL]

You did find 2 valid values for $\theta$, but consider the periodicity of the cosine function:

\(\displaystyle \cos(\theta+2\pi k)=\cos(\theta)\)

You see, you were attempting to find the area outside the first curve and inside the second because of the values you chose. You were going from the first quadrant intersection to the fourth quadrant intersection, when you in fact want to go from the fourth to the first quadrant intersections.

So, how can you write the integral correctly?

 

[ineedhelpnow]

$2 \pi /3$ to $4 \pi /3$ ?

$- \pi/3$ to $\pi/3$

 

[MarkFL]

$2 \pi /3$ to $4 \pi /3$ ?

Let's use:

\(\displaystyle \cos(-x)=\cos(x)\)

And so we may write:

\(\displaystyle A=\frac{1}{2}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}f\left(\cos(\theta)\right)\,d\theta\)

Since the cosine function is even, we may simplify this as:

\(\displaystyle A=\int_{0}^{\frac{\pi}{3}}f\left(\cos(\theta)\right)\,d\theta\)

Look at the graph, do you see that the area above $r=0$ is equal to the area below because of symmetry?

 

[ineedhelpnow]

yeah

 

[MarkFL]

Do you also see how:

\(\displaystyle -\frac{\pi}{3}\le\theta\le\frac{\pi}{3}\)

is the interval we want?

 

[ineedhelpnow]

yes.

 

[MarkFL]

yes.

Cool...then commence to integrating...what do you get?

 

[ineedhelpnow]

i will commence to integrating right after i eat something (Tongueout)

 

[ineedhelpnow]

$2 \pi + \frac{9 \sqrt{3}}{4}$

 

[MarkFL]

$2 \pi + \frac{9 \sqrt{3}}{4}$

No, that's incorrect...can you post your work?

 

[ineedhelpnow]

here's the thing though. my calculator may have told me the answer. (Speechless)

is this the correct setup:

$\frac{1}{2} \int_{- \pi/3}^{\pi/3} \ 9cos^2(\theta)d\theta + \frac{1}{2} \int_{- \pi/3}^{\pi/3} (1+cos(\theta))^2d\theta$

 

[MarkFL]

You want to take the square of the outer curve minus the square of the inner curve...and I recommend using the symmetry I spoke of before. There is no need to break it up into two integrals, just square and combine terms...you should get (after applying a double-angle identity for cosine):

\(\displaystyle A=\int_0^{\frac{\pi}{3}} 4\cos(2\theta)-2\cos(\theta)+3\,d\theta\)

 

[ineedhelpnow]

$\pi$

 

[MarkFL]

$\pi$

Yep...:D

\(\displaystyle A=\left[2\sin(2\theta)-2\sin(\theta)+3\theta\right]_0^{\frac{\pi}{3}}=\sqrt{3}-\sqrt{3}+\pi-0=\pi\)