Question

a cone where the angle of vertex is 2pi/3 can someone help me fill in the red blanks

Answer 1

try b = 2pi/3

Answer 2

lol nope that didnt work out

Answer 3

lol can you show me the picture of cone above. lol its hard to tell thsi way.

Answer 4

i have another problem im working on thats slightly easier its in spherical coordinates

Answer 5

\[b = \frac{ 8\sec \theta }{ \sqrt{3}}\]

Answer 6

hmm
is that 8/secthetha ssqrt(3)

Answer 7

no 8sec theata / sqrt 8

Answer 8

f = pi/4

Answer 9

nope...umm are you taking into account the vertex angle cause i think that changes things

Answer 10

can you help me with this problem instead

Answer 11

the answers for the previous question are

Answer 12

The cone z=sqrt(x^2+y^2) has angle pi/2 at the vertex. To see how the equation changes when the angle is 2*pi/3 you can set x=0 and look in the yz-plane. Then the cone gives a triangle in this plane with opposite side y, adjacent side z and angle pi/3. So y/z=tan(pi/3)=sqrt(3). So y = sqrt(3)*z. You could do something similar for x. So the equation for the cone is z=sqrt(3)*sqrt(x^2+y^2). For your bounds, phi goes from 0 to pi/3, and because the cone has a flat top and z=rho*cos(phi) you get that z<=8/sqrt(3) so rho is between 0 and 8/(sqrt(3)*cos(phi)).

Answer 13

lol its tough i never liked these topics much..

Answer 14

try limits for second question
integral V dzdrdthetao
z-> 0->4(x^2+y^2)^0.5
r = 0->1
theta= 0->2pi

Answer 15

how would i do this using spherical co ordinates

Answer 16

Here is what I got