Questions: Surface Area and Surface Integrals

5 questions to test your understanding

Score: 0 / 5
Question 1 Multiple Choice

What is the role of |r_u × r_v| in the surface area formula A = ∬_D |r_u × r_v| du dv?

AIt computes the normal vector at each point, which is needed to orient the surface
BIt measures how much the parametrization stretches a small parameter rectangle du dv into actual surface area, accounting for distortion
CIt equals the determinant of the Jacobian matrix and is needed to convert between coordinate systems
DIt ensures the integral is taken over the correct parameter domain D
Question 2 Multiple Choice

A surface S is the graph z = g(x, y) over domain D. Which formula gives the correct surface area?

A∬_D dA, since x and y directly parametrize the surface
B∬_D √(g_x² + g_y²) dA, the magnitude of the gradient of g
C∬_D √(1 + g_x² + g_y²) dA, where g_x and g_y are partial derivatives of g
D∬_D (1 + g_x + g_y) dA, a linear correction for the tilt of the surface
Question 3 True / False

The scalar surface integral ∬_S f dS equals the double integral of f over the parameter domain D, weighted by |r_u × r_v|.

TTrue
FFalse
Question 4 True / False

A different parametrization of the same surface will give a different value for the surface area integral.

TTrue
FFalse
Question 5 Short Answer

Explain why the formula for the surface area of a graph z = g(x, y) is ∬_D √(1 + g_x² + g_y²) dA rather than simply ∬_D dA, and what goes wrong if you omit the square root factor.

Think about your answer, then reveal below.