Question

Find the volume under the surface (z=16-x²-y²) with (xgeq 2).

a) Triple integral of (√16-x²-y²) with respect to (x), (y), and (z) in the given bounds.

b) Double integral of (√16-x²-y²) with respect to (x) and (y) in the given bounds.

c) Integral of (√16-x²-y²) with respect to (z) from 0 to 16.

d) None of the above

Answer 1

Double integral of (√16-x²-y²) with respect to (x) and (y) in the given bounds. Hence the correct option is b.

To find the volume under the surface z=16−x^2 −y^2 with x≥2, we can use a double integral over the specified region in the xy-plane.

The given surface is a paraboloid opening downward, centered at the origin, with its vertex at (0,0,16). The condition x≥2 restricts the surface to the region where x is greater than or equal to 2.

The double integral involves integrating the function
`√(16-x^2-y^2)` over this region. This essentially means finding the volume of the solid below the surface within the specified bounds.

The integral is set up as:

∬
`√(16-x^2-y^2)`

Here, D is the region in the xy-plane defined by x≥2. Evaluating this double integral provides the volume under the surface. This approach is appropriate because the region of integration corresponds to the given constraint x≥2. Hence the correct option is b.