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Equation of Graph in Polar Coordinates

  • Thread starter salman213
  • Start date
1. The question was find the area between the curves using DOUBLE Integrals

Area between:
r = sin theta
r = cos theta


well to draw them i made them into cartesian form by

r^2 = rsin theta
r^2 = rcos theta

so

x^2 + y^2 = y

x^2 + y^2 = x

completing square

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


these are two circles

their intersection or bounded region that we need to find the area is like a disc...

I know if i use the double integral

Integral of (Integral of 1) dA OVER D where D is the intersection bounded area


i get the area i need....


but I dont know how to define that region

if someone can help me define the region it will be helpful

I know THETA will change from 0 to pi/2

but R will change from 0 to some equation of that area but i dont know how to find that equation!!

HELP!!
 

lanedance

Homework Helper
3,304
2
Hi Salman

If you look at r = sin theta in polar coordinates, I assume you mean the the polar equation is:

r = (sin(theta), theta)

which is defintely not a circle, similar for the cos
 
so

in polar coordinates
r = sin (theta)

is NOT equal to
r^2 = rsin (theta) = x^2 + y^2 = y


??
 

lanedance

Homework Helper
3,304
2
Hi salman, sorry jumped the mark on that one, think I know what you are doing now...

so you have 2 circles radius 1/2 at centres (0,1/2) and (1/2,0), and need to calculate the area between them

If you draw a picture where do the circles intersect? at the orgin, and at a point along y = x (or equivalently theta = pi/4) due to the symmetry of the situation

think about the bounds for r in each half of the area, both upper & lower will correspond to one of the circles,

I would try and write it as sum of 2 integrals, one from theta 0 to pi/4, and theta pi/4 to pi/2, with appropriate r bounds. So do the r integral first & define lower & up bounds in terms of theta for each segment...
 
so in other words i can just do


theta limits 0 to pi/4
r limits r = sin (theta) to the intersection point


add that to


theta limits pi/4 to pi/2
r limits intersection point to r = cos (theta)


would that be the proper limits?
 

lanedance

Homework Helper
3,304
2
lookin good - so the first half is

[tex]\int^{\pi/4}_0 d\theta \int^{sin{\theta}}_{0}dr (?)[/tex]

what is your integrand for area? do you know what a jacobian is?
 
yea got it ty...it was also on my quiz yesterday so it was great i asked here :)

r*dr*d(theta) :)
 

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