Questions: Parametric Surfaces

5 questions to test your understanding

Score: 0 / 5
Question 1 Multiple Choice

You compute r_u × r_v for a parametric surface and get a nonzero vector. You want to use it as the surface normal in a flux integral. What is wrong with using this vector directly without modification?

ANothing — r_u × r_v is already a unit normal vector suitable for any calculation
BIt points inward; you must always reverse it to point outward
CIts magnitude is not 1; it encodes area scaling and must not be normalized before multiplying by dA
DIt only gives the normal at the origin, not at arbitrary surface points
Question 2 Multiple Choice

A surface is parametrized with r(u, v). You accidentally compute r_v × r_u instead of r_u × r_v. What is the effect on the resulting normal vector?

AThe normal vector is unchanged — cross product is commutative for perpendicular vectors
BThe magnitude changes but the direction stays the same
CThe direction reverses (inward vs outward flip) but the magnitude is the same
DThe result is no longer perpendicular to the surface
Question 3 True / False

For the parametrization r(x, y) = ⟨x, y, f(x,y)⟩ of a graph surface, the surface area element dS equals √(f_x² + f_y² + 1) dx dy.

TTrue
FFalse
Question 4 True / False

At a point where r_u × r_v = 0, the parametric surface has a well-defined tangent plane with a degenerate (zero-length) normal.

TTrue
FFalse
Question 5 Short Answer

Explain why the magnitude |r_u × r_v| must appear in a surface integral, rather than simply integrating over the parameter domain with area element du dv.

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