Question 1

Why is the surface area formula for z = f(x, y) based on partial derivatives?

Accepted Answer

The surface area formula uses partial derivatives because they measure the steepness of the surface at every point. By adding the squares of the partial derivatives ∂z/∂x and ∂z/∂y to one and taking the square root, we obtain the magnitude of the cross product of tangent vectors, which represents the differential surface area element dS on the curved surface.

Question 2

How do tangent vectors relate to calculating surface area?

Accepted Answer

Tangent vectors r_x and r_y are obtained by partial differentiation of the parametric representation r(x,y) = ⟨x, y, f(x,y)⟩. Their cross product r_x × r_y yields a vector perpendicular to the surface, and its magnitude gives the area of an infinitesimal parallelogram on the surface, which is integrated to find total surface area.

Question 3

Why are polar coordinates used for circular domains in surface area calculations?

Accepted Answer

Polar coordinates simplify integration over circular domains by transforming x = r cos θ and y = r sin θ, with dA = r dr dθ. For a circular boundary of radius R, this reduces the limits to r ∈ [0, R] and θ ∈ [0, 2π], making the double integral more manageable than rectangular coordinates.

Question 4

What is the practical purpose of calculating satellite dish surface area?

Accepted Answer

Engineers calculate the exact curved surface area of a parabolic satellite dish to estimate protective coating requirements, determine fabrication costs, and ensure structural specifications are met. The precise surface area enables accurate material planning and cost projections for manufacturing and maintenance.

Question 5

How does the cross product determine the surface area element?

Accepted Answer

The cross product r_x × r_y produces the vector ⟨-∂z/∂x, -∂z/∂y, 1⟩. Its magnitude equals √(1 + (∂z/∂x)² + (∂z/∂y)²), which defines the differential surface area element dS. Integrating dS over the domain D yields the total surface area of the curved surface.

Question 6

What role does the circular boundary play in surface area integration?

Accepted Answer

The circular boundary of the satellite dish rim defines the domain D over which the double integral is evaluated. This circular region in the xy-plane establishes the limits of integration and determines where the surface area calculation begins and ends, making it essential for setting up the correct integral bounds.

Question 7

How is the surface area formula constructed from parametric representation?

Accepted Answer

Starting with the parametric form r(x,y) = ⟨x, y, f(x,y)⟩, partial derivatives yield tangent vectors. Their cross product's magnitude produces the integrand √(1 + (∂z/∂x)² + (∂z/∂y)²). Integrating this expression over domain D gives the complete surface area formula A(S) = ∬_D √(1 + (∂z/∂x)² + (∂z/∂y)²) dA.

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