Find the area bounded by the cardioid

  • Thread starter STEMucator
  • Start date
  • #1
STEMucator
Homework Helper
2,075
140

Homework Statement



Find the area bounded by the cardioid [itex]x^2 + y^2 = (x^2+y^2)^{1/2} - y[/itex]

Homework Equations



Area of R = [itex]\int \int_R dxdy = \int \int_{R'} |J| dudv[/itex]

J Is the Jacobian.

The Attempt at a Solution



Switching to polars, x=rcosθ and y=rsinθ our region becomes [itex]r^2 = r(1-sinθ) → r = 1-sinθ[/itex]
where 0 ≤ θ ≤ 2π.

Also, the Jacobian of polars is just r.

So our integral becomes :

[itex]\int \int_R dxdy = \int \int_{R'} |J| dudv = \int_{0}^{2π} \int_{0}^{1-sinθ} r \space drdθ[/itex]

and using the identity [itex]sin^2θ = (1/2)(1-cos(2θ))[/itex], we can effectively evaluate it.

I have two concerns. The first concern is did I set this up right. My second concern which is more of a worry is how do I KNOW that 0 ≤ θ ≤2π without analytically showing it? It's leaving a sour taste that I'm not justifying it.
 
Last edited:

Answers and Replies

  • #2
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,559
770
It looks properly set up. Nice job. To answer your last question, I would draw the graph in polar coordinates to be sure. That is presumably no different than what you would do in any area problem in either rectangular or polar coordinates to check the limits and shape of the region. Polar graphs can surprise you by looping inside them selves or covering themselves more than once.
 

Related Threads on Find the area bounded by the cardioid

Replies
1
Views
2K
  • Last Post
Replies
2
Views
728
  • Last Post
Replies
4
Views
817
  • Last Post
Replies
8
Views
3K
  • Last Post
Replies
1
Views
2K
Replies
3
Views
747
  • Last Post
Replies
3
Views
671
  • Last Post
Replies
4
Views
12K
Top