calculs iv

Hello, i need help in solving some calculus iv problems.

(10 pts) Consider the surface given by the equation is r 2 + z 2 = 8. (a – 5 pts) Demonstrate that (r, θ, z) = (2, π, 2) is a point on the surface. Solution. We just need to show that (2, π, 2) satisfies the equation r 2 +z 2 = 8, but this is clear because r 2 +z 2 = (2)2 + (2)2 does equal 8. (b – 5 pts) Is (x, y, z) = (1, √ 3, 2) a point on the surface? (Note that the point is given in rectangular coordinates). Solution. We need to change (1, √ 3, 2) into cylindrical coordinates, and then proceed as in part (a). So if x = 1, y = √ 3 and z = 2, then r = p x 2 + y 2 = q (1)2 + ( √ 3) 2 = √ 4 = 2, θ = tan−1 ( √ 3)) = π/3, and z = 2. Now the point (2, π/6, 2) is on the surface because (2, π/6, 2) satisfies the equation r 2+z 2 = 8, that is r 2+z 2 = 4+4 = 8. P