equation for area in polar form?

Question

for those of you who are into polar form, if you have
r = 1-sin(theta) and r=sin(theta)
and you need to find the area of the region BETWEEN inner loops of them, how do you set up the equation?

the formula for area bounded between functions is:
the integral from alpha to beta of 1/2 *[(r2)^2-(r1)^2]d(theta)

alpha and beta would be pi/6 and 5pi/6 in this case, i think

which one is supposed to be used as the farther one(r2) and which one as the closer one(r1) since we have to find the are BETWEEN the two...??

Answer 1

[Edit: I just thought of a way to do this. Read the whole thing, but see the footnote at the end.]

Will come back later with more on this ... am still working on it, especially the limits. Have sketched it out ... First, it's that r must be positive in polar coordinates. That's a hangover from rectangular coordinates where r = sqrt(x^2 + y^2) must be positive. But the point (r, theta) where r = sin theta and theta = 7 pi/6 (210 degrees) is r = -1/2. When that happens, you of r. In rectangular coordinates, the point (r, theta), with r and theta defined as above, is located in the first quadrant where r is 1/2 and theta is pi/6 or 30 degrees. Rectangular coordinates are x = 1/2 cos pi/6 = sqrt(3)/4 and y = 1/2 sin pi/6 = 1/4. (x,y) = (0.433, 0.250).

The equation r = sin theta is a circle of radius 1/2 centered at (0,1/2) in the x-y plane.

The equation r = 1 - sin theta is a "cardioid" -- a symmetrical upright heart-shaped gizmo starting at the origin (where there's a discontinuous slope), with a shallow loop in the first quadrant, passing through (1,0) (rectangular), then bulging outward in the fourth quadrant, before coming to a point at (0,-2) at the bottom.

Since r = sin theta only exists in the first and second quadrants (top hemisphere), and since the left and right sides of this thing are symmetrical, you can get a solution only, then double the result to get your answer.

The shape of the (venn diagram) intersection is sort of like two circles that barely overlap each other -- sort of almond-shaped but pointy at the ends.

Algebraically, they intersect at the origin, and also where
sin theta = 1 - sin theta ==> sin theta = 1/2 ==> theta = pi/6.

When theta = pi/6 (30 degrees), r = 1/2 in both equations.

In rectangular coordinates, that's the same point (0.433, 0.250) we discussed above. So the area of interest (first quadrant only) extends between the origin and that point.

[Edit] This is a hard problem, but maybe I have a way to get the area. For convenience, let t be the angle instead of theta, and let p be pi. Let's limit both curves to the first quadrant (t between 0 and p/2). Let Curve 1 be the circle r = sin t, and let Curve 2 be r = 1 - sin t.

We want to find the area between Curves 1 and 2. The two curves intersect at the origin (point O) and where sin t = 1 - sin t ==> sin t = 1/2 ==> t = p/6. The intersection is (r,t) = (1/2, p/6). Call that point P, and get its x and y coordinates:

x = r cos t = 1/2 cos p/6 = sqrt(3)/4 = 0.433 (point A)
y = r sin t = 1/2 sin p/6 = 1/4 = 0.25 (point B)

Okay, here's the plan. In the region we care about [OA], Curve 2 is greater than Curve 1. On a piece of paper, make a blow-up of the first quadrant on x from 0 to 1/2 and on y from 0 to 1/2. Locate points A and B on the x and y axes respectively. Also locate the points O and P.

Draw OP (a 30 degree angle), and sketch in Curves 1 (concave upward) and Curve 2 (concave downward) between points O and P. If you integrate Curve 2 for t from p/6 to p/2 using the area formula 1/2 r^2 dt = 1/2 [1 - (sin t)^2] dt, that will give you the area of Curve 2 the line OP. Let's call that Area 2.

Now get the area of the triangle OPA = 1/2 [OA] [AP] = (1/2)[sqrt(3)/4](1/4) = sqrt(3)/32. Let's call that Area 3.

Finally we want the area Curve 1 between O and A. In a moment we'll come back to how to get that, but for right now, let's just label that as Area 1.

Now, looking at your diagram, the answer you're looking for is Area 2 + Area 3 - Area 1. That gives you the first-quadrant area between Curves 1 and 2. Using symmetry, just double that to get your final result.

To get Area 1, rectangular coordinates may be easiest. Curve 1 is a circle of radius 1/2 centered at (0,1/2). The equation of this circle is

x^2 + (y - 1/2)^2 = (1/2)^2 = 1/4
(y - 1/2)^2 = 1/4 - x^2 = (1 - 4x^2)/4
y - 1/2 = +/- sqrt(1 - 4x^2)/2
y = 1/2 - sqrt(1 - 4x^2)/2

Here, we don't care about the plus sign; we're only concerned with y values less than 1/2.

y = (1/2) [1 - sqrt(1 - 4x^2)]

To get Area 1, integrate y = f(x) dx on x from 0 to A = sqrt(3)/4.

This was not an easy problem, and I spent a lot of time on it. There may be a better way; I don't know. In any event, there's a plan here to get the answer. Good luck!

[Edit -- Footnote-- Better Idea!!] Instead of Areas 1 and 3 as described above, get Area 4 which is the integral of Curve 1, using the area formula, from t = 0 to pi/6. That gives the area between Curve 1 and the 30-degree line OP. Then all you have to do is add Areas 2 and 4 together.

Answer 2

The two curves intersect at 1-sin(t)=|sin(t)|=>sin(t)=1/2 or t=pi/6. and t=5pi/6 for sin(t)>0. For sin(t)<0, 1+|sin(t)|=|sin(t)| which has no solution. This breaks the problem up into two regions:
(a) 5pi/6>t>pi/6, |sin(t)|>1-sin(t)
(b) 5pi/6|sin(t)|

If you want this problem to be meaningful in a straightforward way, you need to define r=|sin(t)| in the lower half-plane where sin(t)<0, for otherwise one of your radii becomes negative--which isn't supposed to happen in polar coordinates.

Answer 3

SCROLL DOWN TO VERY BOTTOM OF THE SITE

http://groups.google.ca/group/alt.math.undergrad/browse_thread/thread/accf8e7966e695f5/869da9ae159e242a?lnk=st&q=equation+for+area+in+polar+form%3F&rnum=7&hl=en#869da9ae159e242a

Answer 4

if it is of any use, there is a graph of the curves at

http://qwpey.com/mathindex/gl/glx.html