Question

Let F(x, y, z) = (x+y, z, z−x) and let S be the boundary surface of the solid region between the paraboloid z = 9 − x² − y² and the xy-plane. Evaluate ZZ (s) F ·

Answer 1

The question is about finding the surface integral of a vector field over a solid region bounded by a paraboloid and xy-plane, analogous to calculating electric flux in physics.

Step-by-step explanation:

The student is asking about the evaluation of a surface integral over the boundary surface of a solid region using a vector field F(x, y, z). Specifically, the surface integral ∫∫ F · dS over the surface S, which is the boundary surface of the solid region between the paraboloid z = 9 - x² - y² and the xy-plane, is sought. To solve this, one would typically parameterize the surface, compute the surface normal vector, and integrate using the dot product of F and the differential area element dS.

An analogous concept in physics is the electric flux through a surface for a given electric field, as it also involves a surface integral. The electric flux is calculated by taking the dot product of the electric field vector and the area vector over the surface.

Answer 2

The double integral of F · dS over the surface S is ∫∫_S (z, z, z−x) · dS = ∫∫_S (z+z+z−x) dA, where dA represents the area element on the xy-plane.

To evaluate the double integral of F · dS over the surface S, we first need to parameterize the surface S. The given paraboloid equation z = 9 − x² − y² implies that z is a function of x and y. We can use a parameterization of the form (x, y, g(x, y)) to represent points on the surface S.

Next, we need to find the outward unit normal vector dS. Since S is the boundary surface of the solid region between the paraboloid and the xy-plane, the normal vector points outward from the region. In this case, the outward unit normal vector is (−∂g/∂x, −∂g/∂y, 1).

Now, we can calculate F · dS, which is the dot product of the vector field F and the outward unit normal vector dS. This gives us (x+y) * (-∂g/∂x) + z * (-∂g/∂y) + (z−x) * 1.

After obtaining this expression, we integrate it over the parameter domain of S. Since S lies in the xy-plane, we integrate with respect to x and y, and the area element dA is dx dy.