Question

Let v be the region in R3 consisting of all points (x,y,z) such that:

0≤x, 0≤y, 0≤z,
x^2 + y^2 ≤ 1, x^2 + z^2 ≤ 1, y^2 + z^2 ≤ 1

Show that the volume of V equals:

2 (integral Phi from 0 to Pi/4)(integral r from 0 to 1) of ((1-r^2cos^2(Phi))^.5) r dr dPhi

Answer 1

equation should look like this:\[2\int\limits_{0}^{Pi/4}\int\limits_{0}^{1}\sqrt{(1-r^2cos^2(\phi))}rdrd\phi\]

Answer 2

@Zarkon , little help

Answer 3

this looks like its the unit sphere located in the first octant

Answer 4

its not the unit sphere

Answer 5

\[\int_{0}^{pi/2}\int_{0}^{pi/2}\int_{0}^{1} dr\ d\theta \ d\phi\]

Answer 6

x^2+y^2 = 1 is a circle in the xy plane

x^2 + z^2 = 1 is a circle in the xz plane

z^2 + y^2 = 1 is a circle in the zy plane

Answer 7

imnot to confident in the spherical and cylindrical coordinates

Answer 8

so you dont know how id go about showing that?

Answer 9

i cent think of it right at the moment :) ill work at it and see what i can devise tho

Answer 10

i still think its an eighth of the volume of a unit sphere

4pi/8 = pi/2 ... but ill see what i can come up with

Answer 11

is your integrand given in terms of spherical or cylindrical ??

Answer 12

unit sphere is given by x^2 + y^2 + z^2 = 1 , not what is given in the problem. intergrand is in spherical terms

Answer 13

It is not the unit sphere. It is the intersection of the three cylinders described by the 3 inequalities in the first octant.

Answer 14

they're not cylinders either, but parabolas opening up toward (0,0,0) on each axis

Answer 15

They are cylinders. What is \[
x^2 + y^2 \le 1
\]
It is a cylinder centered around the z-axis.

Wait, I will graph it for you.

Answer 16

pi/4 would split it in half and it is symmetrical in the region

Answer 17

are you sure? if I set x = 1 then both y and z have to be equal to 0, which implies a parabola

Answer 18

This is the later area of the cylinder
\[
x^2 + y^2 \le 1
\]
Its parametrization is
p(r,z) = { cos t, sin t, z}}

Answer 19

i would think that the term "phi" would indicate spherical and not cylindrical ...

Answer 20

either way, i got no real idea how to "prove" it

Answer 21

Suppose you are in the plane z=k, then
\[x^2 + y^2 \le 1
\]
it the graph of a disk of center the origin and of radius one in that plane.
Take all the planes z=k, you get a cylinder

Answer 22

It can't be a cylinder though,

Suppose you set x = 1, by the given parameters, both y = 0 and z = 0. This is the vertex of a parabola, not a cylinder, since we cannot have x = 1 and y > 0 or z > 0

Answer 23

See
http://mathworld.wolfram.com/SteinmetzSolid.html
For a solution to your problem. You have to adjust some parameters to do that.

Answer 24

If x=1, then the point (1,0,0) is part of your solid. Why should it be a vertex of a parabola?

Answer 25

because if x = 1, y has to be 0 and z has to be 0. y and z cannot take any other values.

Answer 26

I agree, but that will give you one point (1,0,0) in your solid. What is wrong with that?

Answer 27

The problem is that to make the disk at the top of the cylinder, we need the point (1,0,0) and all points (1,y,z) such that y^2 + z^2 = r^2 where r is the radius of the cylinder.

Answer 28

Read (18)(19) and (20) of
http://mathworld.wolfram.com/SteinmetzSolid.html
take r=1 and divide by 8. There are 8 octants.
You obtain the formula that you want.