quiz8_sol - P = (2 , 1 , 3) lies on the surface z = x 2-y 2...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Date: Mar 23, 2011 MA3160-08 Quiz 8 No Calculators! Justify all answers! Name (print): Solutions 1. Write a triple integral for the volume above the cone z = p x 2 + y 2 and below the sphere of radius 4 centered at the origin (do not evaluate the integral) (a) (3 pts.) using cylindrical coordinates, Solution In cylindrical coordinates the cone is z = r and the sphere is r 2 + z 2 = 16. The two surfaces intersect when r 2 + r 2 = 16, that is, when r = 8 = 2 2. Volume = Z 2 π θ = 0 Z 2 2 r = 0 Z 16 - r 2 z = r r d z d r d θ (b) (2 pts.) using spherical coordinates. Solution In spherical coordinates the cone is φ = π/ 4 and the sphere is ρ = 4. Volume = Z 2 π θ = 0 Z π/ 4 φ = 0 Z 4 ρ = 0 ρ 2 sin φ d ρ d φ d θ 2. (5 pts.) Find the parametric equation of the line that is perpendicular to the surface z = x 2 - y 2 at the point P = (2 , 1 , 3). Solution Note that the point
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: P = (2 , 1 , 3) lies on the surface z = x 2-y 2 . The surface z = x 2-y 2 is the level surface x 2-y 2-z = 0 of the function of three variables f ( x , y , z ) = x 2-y 2-z . A vector perpendicular to this level surface at P = (2 , 1 , 3) is ~ u = grad f (2 , 1 , 3) = (2 x ~ i-2 y ~ j-~ k ) (2 , 1 , 3) = 4 ~ i-2 ~ j-~ k Therefore, the line through P ( P has position vector ~ r o = 2 ~ i + ~ j + 3 ~ k ) in the direction ~ u is ~ r ( t ) = ~ r o + t ~ u = (2 + 4 t ) ~ i + (1-2 t ) ~ j + (3-t ) ~ k or, in component form x ( t ) = 2 + 4 t y ( t ) = 1-2 t z ( t ) = 3-t...
View Full Document

This note was uploaded on 04/04/2011 for the course MA 3160 taught by Professor Staff during the Spring '08 term at Michigan Technological University.

Ask a homework question - tutors are online