How to compute this line integral

[jakey]

Hi guys, can anyone help me with evaluating this:
\int^{}_C |y| \,ds where C is the curve (x^2+y^2)^2=r^2(x^2-y^2)

any hints with the parametrization?

 

[HallsofIvy]

Are we to take it that r is some fixed constant? I'm going to call it "a" because I want to change to polar coordinates.

Would you believe the path is a straight line?

With x= r \cos(\theta) and y= r \sin(\theta), the equation of the curve becomes r^2= a^2(r^2 cos(2\theta)) or \cos(2\theta)= 1/a^2.

If a^2< 1, then that is impossible- there is no such curve. If a^2\ge 1, that is a straight line through the origin making angle \theta= (1/2)cos^{-1}(1/a^2). In particular, for a= 1, the graph is the x-axis.

More generally, from
tan(\theta/2)= \sqrt{\frac{1- cos(\theta)}{1+ cos(\theta)}}
with cos(2\theta)= 1/a^2

tan(\theta)= \sqrt{\frac{1- \frac{1}{a^2}}{1+ \frac{1}{a^2}}}= \sqrt{\frac{a^2- 1}{a^2+ 1}}

So the path is a straight line, through the origin, with slope
\sqrt{\frac{a^2- 1}{a^2+ 1}}.

 

[jakey]

Are we to take it that r is some fixed constant? I'm going to call it "a" because I want to change to polar coordinates.

Would you believe the path is a straight line?

With x= r \cos(\theta) and y= r \sin(\theta), the equation of the curve becomes r^2= a^2(r^2 cos(2\theta)) or \cos(2\theta)= 1/a^2.

If a^2< 1, then that is impossible- there is no such curve. If a^2\ge 1, that is a straight line through the origin making angle \theta= (1/2)cos^{-1}(1/a^2). In particular, for a= 1, the graph is the x-axis.

More generally, from
tan(\theta/2)= \sqrt{\frac{1- cos(\theta)}{1+ cos(\theta)}}
with cos(2\theta)= 1/a^2

tan(\theta)= \sqrt{\frac{1- \frac{1}{a^2}}{1+ \frac{1}{a^2}}}= \sqrt{\frac{a^2- 1}{a^2+ 1}}

So the path is a straight line, through the origin, with slope
\sqrt{\frac{a^2- 1}{a^2+ 1}}.

Hi hallsofivy, thanks for the reply.

but it seems that substituting x=t and y = t\sqrt{\frac{a^2-1}{a^2+1}} doesn't satisfy the curve above...

 

[Mute]

Hi hallsofivy, thanks for the reply.

but it seems that substituting x=t and y = t\sqrt{\frac{a^2-1}{a^2+1}} doesn't satisfy the curve above...

The problem is the line in HallsofIvy's derivation below:

With x= r \cos(\theta) and y= r \sin(\theta), the equation of the curve becomes r^2= a^2(r^2 cos(2\theta))

HallsofIvy accidentally missed that the right hand side was raised to another power of two, meaning the actual curve in polar coordinates is paramatrized by

r^4 = a^2 r^2 \cos(2\theta)
or
r^2= a^2 \cos(2\theta)

The curve is thus not as pleasent as a simple straight line through the origin.

 

[HallsofIvy]

Ah, well, if only life were so simple. I knew it was too good to be true! Thanks for the correction, Mute.