The given equation is:

$z = 4 - y^2$

Step 1: Identifying the Surface Type

Rewriting the equation:

$z + y^2 = 4$

This represents a parabolic cylinder because:

• The equation involves a squared term in $y$ but no squared term in $x$.
• The equation can be rewritten as $z = 4 - y^2$, showing a parabolic shape in the $zy$-plane.
• The surface extends infinitely along the $x$-direction, indicating a cylinder rather than a paraboloid.

Thus, the surface is a parabolic cylinder.

Step 2: Cross-Section at $z = 0$

To find the cross-section at $z = 0$:

$0 = 4 - y^2$

Solving for $y^2$:

$y^2 = 4$

Taking the square root:

$y = \pm 2$

Thus, the cross-section equation is:

$y = \pm 2$

which represents two horizontal lines in the $xy$-plane.

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Step 3: Sketching the Surface

• The base curve in the $zy$-plane is a parabola opening downward with vertex at $(0,4)$.
• This parabola is extruded infinitely along the $x$-axis, forming a parabolic cylinder.

Would you like a 3D graph of this surface? Let me know!

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Follow-up Questions:

1. How does changing the equation to $z = 4 - x^2$ affect the surface?
2. What would the cross-section be for $x = 0$?
3. How does the graph change if we replace $y^2$ with $x^2 + y^2$?
4. What is the domain and range of the function $z = 4 - y^2$?
5. Can a parabolic cylinder be rotated? What equation would describe that?

Tip: A parabolic cylinder extends infinitely along one axis, unlike an elliptic paraboloid, which curves in both directions!