Math Is Fun Forum

White_Owl

Member

Registered: 2010-03-03

Posts: 106

Size of the region, defined in polar

Find the area of the region inside

and outside

I understand that this should be calculated by a double integral of

and

. And I see that the area I am looking for can be calculated by halves (one half above x-axes and one below).
So my current solution is:

But I do not understand what are the limits of

?

White_Owl

Member

Registered: 2010-03-03

Posts: 106

Re: Size of the region, defined in polar

After sleeping on it, I realize that the area is bounded by

on the inside and

on the outside.
In other words:

while

Therefore:

Ok.... If the result is negative, I took the wrong order on of the limits... Not sure which one.

White_Owl

Member

Registered: 2010-03-03

Posts: 106

Re: Size of the region, defined in polar

(Note that the upper limits of integration are different.) Now calculate the half-area outside the first curve and inside the second curve:

Why??? And I do not understand where did pi/2 come from?

The half-area inside the first curve and outside the second is then given by

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Sorry, but I do not think so....

But if we take that approach, then we can try:

So which answer is a correct one???

White_Owl

Member

Registered: 2010-03-03

Posts: 106

Re: Size of the region, defined in polar

This is the image I am working with:

is a blue curve and

is a red one.
It is easy to see that they have three points of intersections: pi/3, pi, 2pi/3

White_Owl

Member

Registered: 2010-03-03

Posts: 106

Re: Size of the region, defined in polar

Ok, I found another approach.
As we can see, the blue curve (3*cos(theta)) exists in the first and fourth quadrants. But the red curve (1+cos(theta)) exists in all four quadrants. So we can divide the region in question into four segments:

And mirror these two into Q3 and Q4.

If I calculate it like this, then

And the total area:

Well... It certainly looks plausible. But am I correct or not?

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Size of the region, defined in polar

That is what I believe to be the correct answer.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

zetafunc

Moderator

Registered: 2014-05-21

Posts: 2,436

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Re: Size of the region, defined in polar

But if we take that approach, then we can try:

So which answer is a correct one???

No, you're integrating r with respect to theta. You instead need to integrate:

.

Your upper limit of integration for 3cosθ should be pi/2, not pi. (For which θ does 3cosθ vanish? Which values does r = 3cosθ take for θ between pi/2 and pi?)

Your method should then be fine, provided you use the square of r, and double your result at the end, noting the symmetry of both shapes.

Last edited by zetafunc (2015-06-02 21:21:58)