Recent content by Unemployed

  1. U

    Integrate ln(cube root (2+y^3)) dydx

    Homework Statement integrate ln(cube root (2+y^3)) dydx with limits of y =sqrt (x) to 1 of x from 1 to 0 Homework Equations integral of ln x= x ln x. This ends up being very convoluted, haven't even begun to insert limits The Attempt at a Solution
  2. U

    Change to cartesian double integral to polar coordinates and evaluate

    Homework Statement integrate 1/((1+x^2+y^2)^2) dx dy Both x and y going from 0 to infinity Homework Equations x^2+y^2 =r The Attempt at a Solution After that I get 1/(1+r^2) ^2 Cannot visualize the function, do not know what the limits are. If I could have any help it...
  3. U

    Volume bound by rho=2+2cos phi

    just to check answer: I got 653.33 Pi. Can anyone verify if that is correct?
  4. U

    Volume bound by rho=2+2cos phi

    ok typo: dV= rho squared sin phi drho dphi d theta I cannot visualize this object to find the volume bound by rho= 5+2cos(phi) I want to know the limits of dtheta and dphi Integrating rho squared is easy I am guessing the limits are 0-pi/2 for phi and 0-2pi for theta
  5. U

    Volume bound by rho=2+2cos phi

    Homework Statement Find the volume bounded rho=5+2cosphi Homework Equations dV=rho squared drho d phi d theta The Attempt at a Solution I am guessing this is some cylindrical shape. Theta should be 0-2pi and phi=0 pi/2
  6. U

    CLYINDRICAL coordinates of volume bound by z=r and z^2+y^2+x^2=4

    For r = 0 to √2 ?
  7. U

    Spherical coordinates: volume bound by z=r andz^2+y^2+x^2=4

    I did draw a picture. It is still very hard to visualize. I am guessing 2pi because the cone is centered on the origin and it is the positive x-z plane. Please help me.
  8. U

    CLYINDRICAL coordinates of volume bound by z=r and z^2+y^2+x^2=4

    Homework Statement Find the smaller volume bound by cone z=r and sphere z^2+y^2+x^2=4 using cylindrcal coordinates Homework Equations dV=r-dr d-theta dz The Attempt at a Solution Limits on r: z to sqrt (4-z^2) limits on theta: 2pi to 0 limits on z: 2-0 Did this and got 8...
  9. U

    Spherical coordinates: volume bound by z=r andz^2+y^2+x^2=4

    i meant 2pi
  10. U

    Spherical coordinates: volume bound by z=r andz^2+y^2+x^2=4

    0 to pi?
  11. U

    Spherical coordinates: volume bound by z=r andz^2+y^2+x^2=4

    So phi is 0-pi/4 and theta is also pi/4?
  12. U

    Spherical coordinates: volume bound by z=r andz^2+y^2+x^2=4

    Thanks. So the general formula is (rho^2 sin phi drho d-phi d-theta) z=rho-cos-phi r=rho-sin-phi rho cos phi=rho-sin-phi tan phi=1 So now with the limits rho from 0-2 phi from 0-1 or is it 0-pi/4? and theta from 0-pi/2 I got 8/3 *(cos 0-cos 1)*pi/2 1.296
  13. U

    Spherical coordinates: volume bound by z=r andz^2+y^2+x^2=4

    Homework Statement Using spherical coordinares, find the smaller volume bounded by the cone z=r and the sphere z^2+y^2+x^2=4 Homework Equations x^2+y^2+z^2=4 ; rho=2, z=rhocosphi The Attempt at a Solution Shot in the dark: Tried function integrating (rho squared - rhocosphi)...
  14. U

    Find points on a plane at(0,0,0) 2x+3y+yz=12

    it still does say find point on the plane 2x+3y+4z+12 and the point (0,0,0) But I am guessing he meant find the point closest to 0,0,0 The prior example x+y+z=3 at point (1,2,3) does show use of the perpendicular. Thanks for the help deciphering
  15. U

    Find points on a plane at(0,0,0) 2x+3y+yz=12

    These are problems given in notes as HW made up as the professor goes along rather than straight out of the book. You are right in that I am not sure in what the hell he is asking. I assume the plane is 2x+3y+4z=12 Then it says: Find the point on the plane and the point (0,0,0)...