Line integral coordinates problem

[venrera]

Hello
I have been trying to solve this problem for a while now, but still cant quite figure out how to tackle it.
The problem is to solve the integral ∫ ydx+zdy+xdz, over a curve defined by these equations:
x^2+y^2+z^2=1 (unit ball)
x+y+z=0 (plane)
The resulting curve is a circle, but to progress further, I must first describe it by transformed coordinates, and therein lies my problem:
I have no idea what they should be. Since both equations contain the dimension z, I guess they should be spherical coordinates.
What i think is required, is to find trigonometric expressions to substitute for the coordinates, that satisfy both of these equations through trigonometric identities.
Problem is, my knowledge of these is just about enough for cases in 2D, thats why the z dimension poses such a problem for me.

To clarify: its not the eventual solution to the integral that i am searching for, its the transformed coordinates.
I´ll be glad for any help or advice.

 

[HallsofIvy]

We can write x+ y+ z= 0 as z= -x- y so that \(\displaystyle z^2= (-x- y)^2= (x+ y)^2= x^2+ 2xy+ y^2\). Then \(\displaystyle x^2+ y^2+ z^2= x^2+ y^2+ x^2+ 2xy+ y^2= 2x^2+ 2xy+ 2y^2= 1\) or \(\displaystyle x^2+ xy+ y^2= 1/2\).

That "xy" term means we will have to rotate the coordinate system. Let \(\displaystyle x= x'cos(\theta)+ y'sin(\theta)\) and \(\displaystyle y= -x'sin(\theta)+ y'cos(\theta)\).
So
\(\displaystyle x^2= x'^2 cos^2(\theta)+ 2x'y'cos(\theta)sin(\theta)+y'^2 sin^2(\theta)\)
\(\displaystyle xy= -x'^2 sin(\theta)cos(\theta)+ x'y'(cos^2(\theta)-sin^2(\theta)+ sin^2(\theta)+ y'^2sin(\theta)cos(\theta)\)
\(\displaystyle y^2= x'^2sin^2(\theta)- 2x'y'cos(\theta)sin(\theta)+ y'^2cos^2(\theta)\).

So \(\displaystyle x^2+ xy+ y^2= x'^2 cos^2(\theta)+ 2x'y'cos(\theta)sin(\theta)+y'^2 sin^2(\theta)-x'^2 sin(\theta)cos(\theta)+ x'y'(cos^2(\theta)-sin^2(\theta)+ sin^2(\theta)+ y'^2sin(\theta)cos(\theta)+ x'^2sin^2(\theta)- 2x'y'cos(\theta)sin(\theta)+ y'^2cos^2(\theta)\)
\(\displaystyle = x'^2(cos^2(x)- sin(\theta)cos(\theta)+ sin^2(\theta))+ x'y'(cos^2(\theta)+ 4cos(\theta)sin(\theta)- sin^2(\theta))+ y'^2(sin^2(\theta)+ sin(\theta)cos(\theta)+ cos^2(\theta)\).

We want to remove the "x'y'" term so we want \(\displaystyle cos^2(\theta)+ 4 cos(\theta)sin(\theta)+ sin^2(\theta)= 0\). We can solve that using the quadratic formula: \(\displaystyle sin(\theta)= \frac{-4cos(\theta)\pm\sqrt{16cos^2(\theta)- 4cos^2(\theta)}}{2}= cos(\theta)\frac{-4\pm\sqrt{12}}{2}\).

That is, \(\displaystyle tan(\theta)= \frac{-4\pm\sqrt{12}}{2}= -2\pm\sqrt{3}\)

 

[venrera]

Thank a bunch for the reply! Am going to be honest, and confess this is a bit beyond me. It looks to me like all the problems we have been solving so far were a lot less complex. Ill have the opporotunity to ask my teacher the same question, and hopefully, together with his reply, and this reply, ill be able to piece it together.
Many thanks for taking the time for writing it all down. At least Ill know in what direction i should be looking now.
This problem is also me basically doing ¨work in advance¨, so maybe i should return to it after getting some more experience.