Polar Integral Conversion Correct?

  • Thread starter Thread starter bigedd1227
  • Start date Start date
  • Tags Tags
    Integral Polar
bigedd1227
Messages
1
Reaction score
0
I'm not sure if I'm doing this conversion correctly. I have to convert the following double integral into polar. The integration part I can do, I just want to make sure I converted correctly. And sorry about the formatting bc I don't kno how to do the math formatting.


Int(Int((x^2+y^2))dy)dx
the bounds on the outer integral are -1 to 1
the bounds on the inner integral are -sqrt(1-y^2) to sqrt (1-y^2)

now here is what I got converting it to polar

Int(Int(r^3)dr)dtheta
outer bounds = 0 to 2pi
inner bounds = 0 to 1

Did I do the conversion correctly?
Thanks for the help
 
Physics news on Phys.org
Lookin' good, but in the first integral the bounds on the inner integral should be -sqrt(1-x^2) to sqrt (1-x^2).
 
Last edited:
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