# Finding area between two curves Polar Coordinates

## Homework Statement

Find the area inside the circle r = 3sinθ and outside the carotid r = 1 + sinθ

## The Attempt at a Solution

Alright so I graphed it and found that they intersect at ∏/6 and 5∏/6.
I can't think of a good way to approach the problem. The carotid has some of it's area beneath the x-axis otherwise I would take the area of the 3sinθ - the area of the other one. Will it work if set the limits of integration to pi/6 and 5pi/6 and take the integral of (3sinθ - (1+sinθ)? I'm a bit lost

STEMucator
Homework Helper

## Homework Statement

Find the area inside the circle r = 3sinθ and outside the carotid r = 1 + sinθ

## The Attempt at a Solution

Alright so I graphed it and found that they intersect at ∏/6 and 5∏/6.
I can't think of a good way to approach the problem. The carotid has some of it's area beneath the x-axis otherwise I would take the area of the 3sinθ - the area of the other one. Will it work if set the limits of integration to pi/6 and 5pi/6 and take the integral of (3sinθ - (1+sinθ)? I'm a bit lost

Yes, that all seems reasonable. Just compute your integral and you're done.

• 1 person
I got 1.369705. Is there any online integral doer I can use to check my work?

STEMucator
Homework Helper
I got 1.369705. Is there any online integral doer I can use to check my work?

Yes, that's the correct answer. Wolfram alpha is good for checking your work afterwards.

• 1 person
LCKurtz
Homework Helper
Gold Member

## Homework Statement

Find the area inside the circle r = 3sinθ and outside the carotid r = 1 + sinθ

## The Attempt at a Solution

Alright so I graphed it and found that they intersect at ∏/6 and 5∏/6.
I can't think of a good way to approach the problem. The carotid has some of it's area beneath the x-axis otherwise I would take the area of the 3sinθ - the area of the other one. Will it work if set the limits of integration to pi/6 and 5pi/6 and take the integral of (3sinθ - (1+sinθ)? I'm a bit lost

NO. That's the wrong integrand. Look up the formula for polar area.