Is this integral set up correctly?

  • Thread starter Thread starter G01
  • Start date Start date
  • Tags Tags
    Integral Set
G01
Science Advisor
Gold Member
Messages
2,704
Reaction score
19
Find the area of the region within both circles;

r=cos(theta) and
r=sin(theta)

using a double integral.

I made the following:

2\int^{\pi/2}_{\pi/4} \int^{\cos\theta}_0 r dr d\theta

I multiplied by 2 because the area I have is only half of the total area to be found. Is this correct or am i doing something stupid?
 
Physics news on Phys.org
I'm not sure that it's a particularly good example for using multiple integrals, but you're effectively using Green's Theorem on something that can be readily handled by:
A = \frac{1}{2} \int r^2 d\theta
or, for those who prefer not to deal with polar integrals:
A=2 \int_0^{\frac{1}{2}} \left(\sqrt{(\frac{1}{2})^2-(x-\frac{1}{2})^2}-x \right)dx

I'm fairily sure that your expression will give a numerically correct result.

http://mathworld.wolfram.com/GreensTheorem.html
 
Last edited:
Thanks Nate, I agree that its probably easier to do this with
1/2 \int r^2 d\theta but our assignment was to do it using a double integral. Thanks for the help. I'll keep working on it and see if i get the right answer.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top