Question

Let S denote the part of the conical surface x² + y² = z² that is between the planes z = 2 and z = 3, oriented so that the top of S points away from the z axis. Calculate the surface area of S.

Answer 1

To calculate the surface area of the part of the conical surface between two planes, parameterize the surface using spherical coordinates and integrate the norm of the cross product of the partial derivatives of the parameterization. Evaluate the integral to find the surface area of S, which is 8*pi.

Step-by-step explanation:

To calculate the surface area of the part of the conical surface that is between the planes z=2 and z=3, we first need to parameterize the surface. Let's use spherical coordinates, where z = p*cos(theta), r = p*sin(theta), and phi ranges from 0 to 2*pi. The surface area is then given by the integral of the norm of the cross product of the partial derivatives of the parameterization, integrated over the range of phi.

The norm of the cross product is sqrt((2*sin(theta))^2 + (2*p*sin(theta)*cos(theta))^2 + (1)^2). Integrate this expression over phi from 0 to 2*pi and theta from 0 to pi/4. The result will give you the surface area of S.

By evaluating the integral, the surface area of S is found to be 8*pi.