Find the mass of a wire whose shape is that of a curve of intersection of the sphere $x^2 + y^2 + z^2 = 1$ and the plane $x + y + z = 0$ if the density of the wire is $x^2$.

I know that this problem is just a simple computation with line integrals wrt to arc length, but the issue is that I can't find a parametrization. This very question is answered here, but I'm not entirely sure how to get the parametrization given (not enough sleep last night. . . )

Could someone help me derive that parametrization?


I found this question in the Apostol's Calculus book and I couldn't find the answser by myself. Fortunately, I understood the parameterization made by muzzlator.

As muzzlator said, you should find two orthogonal vectors such that they are on the plane and on the sphere. Thus, their components should satisfy the equations $$x^2 + y^2 + z^2 = 1 \space \space \space \space(I)$$ and $$x+y+z=0 \space \space \space \space(II)$$

Your first vector can be arbitrary. For find the second, you can do the dot product with the first e use the fact that it should be unitary. For the first vector, muzzlator imposed $x=0$, which results $y=-z$ by equation II. And if use the equation I, you gonna find $y=\pm\frac{1}{\sqrt2}$. You can choose the negative or the positive, it doesn't make difference. If you choose $y=\frac{1}{\sqrt2}$, you will find the first vector:$$\vec v_1=\frac{1}{\sqrt2}(0,1,-1)$$

Now you take the second vector as $\vec v_2= (x,y,z)$, do the dot product with $\vec v_1$ and use equation (I). Thus, you will find the second vector: $$\vec v_2=\frac{1}{\sqrt6}(-2,1,1)$$

Note that you would have two solutions for $\vec v_2$, but what matters is that they are orthogonal and satisfy I and II. So, there isn't any problem if you choose the other solution. Now, to get the parameterization, you sholud note that the curve of intersection is a circle with radius one (make a drawing), in which are the vectors $\vec v_1$ and $\vec v_2$. Therefore, imagine a cartesian plane, where the orthogonal axes are $\vec v_1$ and $\vec v_2$; the parameterization of this circle is $$\mathbf{x}(t) = \vec v_1\cos t + \vec v_2\sin t$$ $$\mathbf{x}(t) = \frac{1}{\sqrt{2}} (0,1, -1) \cos t + \frac{1}{\sqrt{6}}(-2, 1, 1) \sin t$$


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