So I got this mass problem to solve: Find the mass of the wire formed by the intersection of two surfaces whose density is $\phi=x²$

$\underset{C}\int \phi ds $ along the curve:

$$ C:\left\{ \begin{array}{c} x²+y²+z²=1 \\ x+y+z=0 \\ \end{array} \right. $$

My question is not exactly in how to solve this - because I know how to solve it as my calculus teacher taught - but in how(if it can be solved that way) to solve it using linear algebra. It's just an effort I'm trying to do to integrate what I've learned in both courses in this semester so It might be just a dumb idea. I started by backsubstitution and got this equation of the intersection $$ 2x²+ 2xy + 2y² =1 $$ which I identify as a quadratic form, rewrote it in matrix form and diagonalized it. The elipsoid at the new "u v frame" is: $$ 3u² + v² = 1$$ whose parametrization is: $$ \gamma(t)=((1/\sqrt3)cos(t), sin(t)) $$ then I found the norm of the derivative as to substitute ds at the line integral $ ds=||\gamma'(t)||dt $

But now things started to get confusing and here's my problem I think

I don't know if I'm right but it seems I can't just compose $\phi(x)$ with $ \gamma(t) $ and proceed finding the primitive as $\phi=x²$ is in the "x y frame". At my humble opinion it seems that I should rotate somehow this density function as to find it's equation in the "u v frame".

  • Is it possible?
  • Is it practicle?
  • Am I proceeding right until now?

Sorry for any problems at my question, first time here;


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1 Answer 1


When you combine the two equations to find the intersection, you missed one variable $z$. So what you found is the projection of the intersection onto $xy$-plane. To make it really the intersection, you have to add an equation that represent the $z$-value. For example, you can add $z=-x-y$ and add that into your parametric equation.

After your represent the curve in parametric equation, you can represent $\phi=x^2$ using the $x$ component of the curve.

  • $\begingroup$ Thx! That seems right, missed that z coordinate, but the problem still remains =) I'll correct it! $\endgroup$
    – Janov
    May 18, 2015 at 18:18
  • $\begingroup$ @MarceloBroinizi: The problem is you cannot represent the intersection curve using one equation. For the representation of $\phi$, see my edited answer. $\endgroup$
    – KittyL
    May 18, 2015 at 18:30
  • $\begingroup$ Got that, but don't think u really understood what I asked: my parametrization is for the u v frame so I think I might have to find out how does density behaves in u direction instead of x direction. It doesn't seem to make sense composing the "x" of phi with a parametrization for - applying your correction - u, v, w plane. If it's really ok to do that it means the density is a invariant which is really strange for me. $\endgroup$
    – Janov
    May 18, 2015 at 19:41
  • $\begingroup$ @MarceloBroinizi: I see what you meant now. Yes, you will need to do a rotation to transform the ellipse. I think Stewart's book explains it clearly: stewartcalculus.com/data/ESSENTIAL%20CALCULUS/upfiles/topics/…. But I doubt whether it worth the extra effort to do that. Looks to me the polar coordinates would work well with $x,y$ values. $\endgroup$
    – KittyL
    May 18, 2015 at 19:48
  • $\begingroup$ Yes, you almost got now! ahaha! I did it, I found a good frame so the elipse now seems really nice to work, the problem is that I should also "rotate" somehow the x² by the same angle I rotated the elipse. But I don't know if I can in fact rotate a polynomial function in that way as it would become non-surjective; I'll try posting a picture. $\endgroup$
    – Janov
    May 18, 2015 at 20:02

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