Recent content by Unemployed

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

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...

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

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

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

For r = 0 to √2 ?

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.

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...

i meant 2pi

0 to pi?

So phi is 0-pi/4 and theta is also pi/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

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)...

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

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)...