# Length around intersection of polar curves

#### calcboi

##### New member
Sketch the 2 polar curves r = -6cos(theta), r = 2 - 2cos(theta).
a. Find the area of the bounded region that is common to both curves.
b. Find the length around the intersection of both curves.
I got a, but I don't know what to do for b because in my calculus book it only shows how to find the length of a single polar curve, not two. Please help!

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Sketch the 2 polar curves r = -6cos(theta), r = 2 - 2cos(theta).
a. Find the area of the bounded region that is common to both curves.
b. Find the length around the intersection of both curves.
I got a, but I don't know what to do for b because in my calculus book it only shows how to find the length of a single polar curve, not two. Please help!
Welcome to MHB, calcboi! Did you find the intersection points of the curves?

Suppose you divide the length around the intersection in a set of 3 curves.
Can you find each of those curves?
And then calculate the length of each of them?

#### calcboi

##### New member
Welcome to MHB, calcboi! Did you find the intersection points of the curves?

Suppose you divide the length around the intersection in a set of 3 curves.
Can you find each of those curves?
And then calculate the length of each of them?
I found the intersection points to be 2pi/3 to 4pi/3.
If I divided the length around the intersection into 3 curves, I could find the area from the curve to the origin if I knew the equation of each one. I don't understand how to come up with one polar equation to solve for the intersection or is there a way to do it for both polar curves.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
I found the intersection points to be 2pi/3 to 4pi/3.
If I divided the length around the intersection into 3 curves, I could find the area from the curve to the origin if I knew the equation of each one. I don't understand how to come up with one polar equation to solve for the intersection or is there a way to do it for both polar curves.
Good!

But I'm unclear now on what you want to calculate.
Is it the area contained by both curves?
Or is is the arc length around this intersection area?

Anyway, let's call your curves $r_1$ and $r_2$.
Curve $r_1$ is a circle to the left of the y-axis.
Curve $r_2$ intersects $r_1$ at $\theta=\frac \pi 2$, $\theta=\frac 2 3 \pi$, and in the origin.

To trace through the area enclosed by both curves, you would first start on curve $r_1$ with an angle $\theta=\frac \pi 2$ up to $\theta=\frac 2 3 \pi$.
Then trace curve $r_2$ from $\theta=\frac 2 3 \pi$ up to $\frac 4 3 \pi$.
And finally trace curve $r_1$ again from $\theta=\frac 4 3 \pi$ up to $\frac 3 2 \pi$.

For the area, you need the integral formula for the area of a polar curve.
For the arc length, you need the arc length formula for polar curves.

Do you have those?

#### calcboi

##### New member
Yes, I have the formulas and I got the area already. I have the polar arc length formula but it only works when there is one curve, just one r. I have r1 and r2, two different curves, and I am unsure how to calculate the arc length with two polar curves.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Yes, I have the formulas and I got the area already. I have the polar arc length formula but it only works when there is one curve, just one r. I have r1 and r2, two different curves, and I am unsure how to calculate the arc length with two polar curves.
Can you show us the formula you've got?

#### calcboi

##### New member
Can you show us the formula you've got?
Length of a Polar Curve

L = integral from a to b of square rt(r^2 + (dr/dtheta)^2) dtheta

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Length of a Polar Curve

L = integral from a to b of square rt(r^2 + (dr/dtheta)^2) dtheta
Good!

Use the following curve:
$$r = \left\{\begin{array}{ll} -6 \cos \theta & \qquad \text{ if } \frac \pi 2 \le \theta < \frac 2 3 \pi \\ 2-2 \cos \theta & \qquad \text{ if } \frac 2 3 \pi \le \theta < \frac 4 3 \pi \\ -6 \cos \theta & \qquad \text{ if } \frac 4 3 \pi \le \theta \le \frac 3 2 \pi \\ \text{undefined} & \qquad \text{ otherwise} \end{array}\right.$$
And take the integral from $a=\frac \pi 2$ to $b=\frac 3 2 \pi$.

#### calcboi

##### New member
Good!

Use the following curve:
$$r = \left\{\begin{array}{ll} -6 \cos \theta & \qquad \text{ if } \frac \pi 2 \le \theta < \frac 2 3 \pi \\ 2-2 \cos \theta & \qquad \text{ if } \frac 2 3 \pi \le \theta < \frac 4 3 \pi \\ -6 \cos \theta & \qquad \text{ if } \frac 4 3 \pi \le \theta \le \frac 3 2 \pi \\ \text{undefined} & \qquad \text{ otherwise} \end{array}\right.$$
And take the integral from $a=\frac \pi 2$ to $b=\frac 3 2 \pi$.
So do I use 3 different curves then since you gave me a piecewise function?
Like for one I use 6cos(theta) and so on.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
So do I use 3 different curves then since you gave me a piecewise function?
Like for one I use 6cos(theta) and so on.
It's the same thing.
You can use 3 different curves and add their lengths.
Or you can use 1 curve that has a piecewise definition.
To calculate its length you still have to split the integral into 3 integrals.