Double integral in polar coordinate

  • Thread starter agro
  • Start date
  • #1
46
0

Homework Statement


With a > 0, b > 0, and D the area defined by

[tex]D: \frac{x^2}{a^2} + \frac{y^2}{b^2} \leq 1[/tex]

Change the integral expression below:

[tex]\iint\limits_D (x^2+y^2) dx\,dy[/tex]

by using x = a r cos θ, y = b r sin θ. After that evaluate the integral.

The Attempt at a Solution



First of all, I can already visualize that the area is an ellipse, and because of that if we want to express the outside boundary in polar coordinates, it is

[tex]r_1(\theta) = \frac{ab}{\sqrt{(b \, cos \theta)^2+(a \, \sin \theta)^2}}[/tex]

Therefore, since x^2 + y^2 = r^2, the integral can be expressed in terms of θ and r as:

[tex]\int_0^{2\pi} \int_0^{r_1(\theta)} \! (r^2)r \, dr\,d\theta[/tex]

I've also understood how this construct comes from the Riemann sums.

However, I don't understand the substitution hinted from the problem. For every random point (r, θ), clearly x = r cos θ and not x = a r cos θ! Can someone give a geometric interpretation/explanation, and from that show how one changes the integral expression?

Thanks a lot.
 

Answers and Replies

  • #2
193
2
x = r cos θ and y = r sin θ maps out a circle, not an ellipse.

The equation of an ellipse is:

x = a cos θ
y = b sin θ

Substitute these into your integral.
 

Related Threads on Double integral in polar coordinate

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
1
Views
977
  • Last Post
Replies
7
Views
11K
  • Last Post
Replies
12
Views
896
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
1K
Top