need help integrating

[toomers]

need help converting to polar equation

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


I would like to convert the above into a polar equation. Is it possible?

 

[lookagain]

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


I would like to convert the above into a polar equation. Is it possible?

\(\displaystyle y \ = \ \sqrt{\dfrac{[4(1 - b)^2x^2][x^2 + b^2(1 - x^2)]}{(1 - x^2)\bigg[\bigg(\dfrac{4bx^2}{1 - x^2} + 1\bigg)^2 + 1\bigg]} \ }\)


You need to know the parameters for b. \(\displaystyle \ \ \) I let b = 1/2 and graphed it [online]. For appropriate b-values, the graph of the function is the top half of a lemniscate * \(\displaystyle \ \) whose center is the origin. In your headline, you have the word "integrating," so I suspect you want the area under the curve but above the x-axis. Every \(\displaystyle x^2 \ \ \)term could be replaced with \(\displaystyle \ r^2cos^2\theta.\)

* A complete leminiscate is shown and discussed here: http://mathworld.wolfram.com/Lemniscate.html

 

[toomers]

I forgot to change the title. Actually I want it converted to polar coordinates first and then integrate from zero to pi/2. I don't know if it's possible to do it though. Replacing the x^2 as you say I doubt I can consolidate them to form a function with r(theta)=.... instead of the y=....

I must integrate it per d(theta) after it is in polar form.

 

[HallsofIvy]

The basic concept is pretty straightforward- though maybe very tedious. Replace each occurrence of "x" with \(\displaystyle r cos(\theta)\) and each occurrence of "y" with \(\displaystyle r sin(\theta)\). Then, depending on exactly how you want to do the integral, try to solve for r as one or more functions of \(\displaystyle \theta\).

 

[DrPhil]

I forgot to change the title. Actually I want it converted to polar coordinates first and then integrate from zero to pi/2. I don't know if it's possible to do it though. Replacing the x^2 as you say I doubt I can consolidate them to form a function with r(theta)=.... instead of the y=....

I must integrate it per d(theta) after it is in polar form.

Nothing quite makes sense to me - I have the feeling there was a lot more to this problem than you told us.

Was the function y(x) given in that form, or did you derive it from other information? I can't make sense of the dimensions, which I suspect should be the same for x and y (and thus r). The form you have can not be written in the form \(\displaystyle r = f(\theta)\).

Is the integral really \(\displaystyle \int_0^{\pi/2} r(\theta)\ d\theta\), which has dimensions of length? Or were you to find an area, in which the increment of integration would be \(\displaystyle (r\ d\theta)\) ?

 

[lookagain]

The form you have can not be written in the form \(\displaystyle r = f(\theta)\).

In polar coordinates, the equation of a lemniscate is \(\displaystyle \ r^2 \ = \ 2a^2\cos(2\theta). \ \ \) **

I claimed, that for particular b-values, the OP has given an equation for the graph of a lemniscate (in Cartesian coordinates).

Then, I think there is the potential to transform the given equation into a form of **, or ultimately, into the

\(\displaystyle r \ = \ f(\theta) \ \) form that you mentioned.








** Source:
http://en.wikipedia.org/wiki/Lemniscate_of_Bernoulli