Find the intersection of $z=1-x^2$ and $x=y^3$ in $\mathbb R^3$

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I have these 2 surfaces, given in their Cartesian form, and want to find the resulting surface of intersection.

My approach is to find parametric representations. But I have two approaches, and not sure if both approaches are proper, or if one or the other is only proper.

Here are the equations:

$$\tag{1}z = 1 - x^2$$

$$\tag{2}x = y^3$$

Approach 1:

Substitute (2) into (1), we get the cartesian form of the intersection surface: $z = 1 - x^6$, then try to parameterize this, but I run into $2$ independent parameters with this approach.

Approach 2: Direct parameterization:

Just looking at both equations I see that '$x$' is the common variable so I set $x= t$. Then just plugging in '$t$' into the equations, and put them together in $(x,y,z)$, I get: $$(x,y,z) = (t, t^{1/3}, 1-t^2)$$

So not sure if either of these 2 approaches above give the correct parameterization.

Not sure if there is a systematic procedure to do parameterization of surfaces of intersection in $\mathbb R^3$, or if this is a creative type of process, and hence there is no systematic procedure to follow.

Hope someone can explain how to think about this kind of problem.

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Two quick comments. First, you do not want to introduce $t^{1/3}$, as then the parametrization will not be differentiable — let alone smooth — at the origin.

Second, when you eliminate a variable as you did in approach 1, you are not done; rather, you've obtained the projection of the intersection into the $xz$-plane in this case. You still need to give the $y$ value for each point on that curve in the $xz$-plane. To convince yourself, consider the system of linear equations $x+y=0$, $2x-y=0$. If you eliminate $y$, you get $3x=0$, so $x=0$. Clearly the solution set is not the entire $y$-axis.

Oh, and glad you enjoy the book. There are also lectures on YouTube, linked in my profile.