To find the intersection of the surface $z^2 = x^4 + y^2$ and the plane $z = 0$, we need to substitute $z = 0$ into the equation of the surface.

Starting with the equation of the surface: $z^2 = x^4 + y^2$ Substitute $z = 0$ into the equation: $0^2 = x^4 + y^2$ This simplifies to: $x^4 + y^2 = 0$ Now, since $x^4$ and $y^2$ are both non-negative for all real values of $x$ and $y$, the only way the sum can equal zero is if both $x^4 = 0$ and $y^2 = 0$, which occurs when: $x = 0 \quad \text{and} \quad y = 0$

Thus, the only point of intersection is at the origin $(0, 0, 0)$.

Conclusion:

The intersection of the surface $z^2 = x^4 + y^2$ and the plane $z = 0$ is a single point at the origin. Therefore, the intersection resembles a point.

---

Would you like any further details or clarifications?

Here are some related questions to explore further:

1. How does the shape of the surface $z^2 = x^4 + y^2$ change for different values of $z$?
2. What would the intersection look like if the plane had a different value, such as $z = 1$?
3. How would the equation change if the surface were $z^2 = x^2 + y^2$ instead of $z^2 = x^4 + y^2$?
4. Can you find the critical points of the surface $z^2 = x^4 + y^2$?
5. What is the significance of the higher-order term $x^4$ in the surface equation?

Tip: When analyzing surfaces and their intersections with planes, always check if substitution simplifies the problem to a basic geometric figure like a point, line, or curve.