# A volume?

• Jul 12th 2009, 09:25 PM
arbolis
A volume?
Here's the problem :
Let $\displaystyle B_4=\{ (x,y,z,w) \in \mathbb{R}^4 : x^2+y^2+z^2+w^2 \leq 1 \}$ and $\displaystyle B_3=\{ (x,y,z) \in \mathbb{R}^3 : x^2+y^2+z^2 \leq1 \}$.
1)Show by a justification that $\displaystyle V(B_4)=2 \iiint _{B_3} \sqrt{1-(x^2+y^2+z^2)}dxdydz$.
2)Deduce that $\displaystyle V(B_4)=\frac{\pi ^2}{2}$.
My attempt : almost none. My first problem is : they never defined the function $\displaystyle V$. If I assume it's a volume function then they seem to ask me to calculate the volume of a 4 dimensional solid (or whatever it is called), which seems senseless. I realize that $\displaystyle B_3$ is the projection of $\displaystyle B_4$ in the 3 dimensional space.
Also, I recognize the integrand to be the positive $\displaystyle w$ that satisfy the first inequation. So $\displaystyle V(B_4)$ really seems a volume since there's the multiplication by 2 in front of the triple integral so that my instinct tells me that this covers the $\displaystyle -w$. I may not be clear here, but that's how I understand the problem.
If you understand it better than I, feel free to reformulate it so that I can understand it.
• Jul 12th 2009, 09:47 PM
Chris L T521
Quote:

Originally Posted by arbolis
Here's the problem :
Let $\displaystyle B_4=\{ (x,y,z,w) \in \mathbb{R}^4 : x^2+y^2+z^2+w^2 \leq 1 \}$ and $\displaystyle B_3=\{ (x,y,z) \in \mathbb{R}^3 : x^2+y^2+z^2 \leq1 \}$.
1)Show by a justification that $\displaystyle V(B_4)=2 \iiint _{B_3} \sqrt{1-(x^2+y^2+z^2)}dxdydz$.
2)Deduce that $\displaystyle V(B_4)=\frac{\pi ^2}{2}$.
My attempt : almost none. My first problem is : they never defined the function $\displaystyle V$. If I assume it's a volume function then they seem to ask me to calculate the volume of a 4 dimensional solid (or whatever it is called), which seems senseless. I realize that $\displaystyle B_3$ is the projection of $\displaystyle B_4$ in the 3 dimensional space.
Also, I recognize the integrand to be the positive $\displaystyle w$ that satisfy the first inequation. So $\displaystyle V(B_4)$ really seems a volume since there's the multiplication by 2 in front of the triple integral so that my instinct tells me that this covers the $\displaystyle -w$. I may not be clear here, but that's how I understand the problem.
If you understand it better than I, feel free to reformulate it so that I can understand it.

$\displaystyle x^2+y^2+z^2+w^2=1$ is a ball in 4-space (or a 4-D sphere).

If we are to find the volume, we solve for the function w(x,y,z). In this case, $\displaystyle w=\sqrt{1-x^2-y^2-z^2}$. Recall that when we find volumes in 3-space, we resort to a region in 2-space to determine the limits. In our case, we resort to a region in 3-space in order to find our limits. Then we can find the volume of the ball in 4-space.

Thus, it would make sense that $\displaystyle V\left(B_4\right)=2\iiint\limits_{B_3}\sqrt{1-x^2-y^2-z^2}\,dz\,dy\,dx$ (we need two triple integrals to take into consideration the two hemispheres of the ball).

From here, evaluate the integral. A simple change to spherical coordinates will do the job.

Can you continue? I hope my explanation makes sense...
• Jul 13th 2009, 07:31 AM
arbolis
Ok thanks Chris, I'll try to do it. I was accustomed to speak about volumes only for solids in 3 dimensions. Now I realize that a 3 dimensional sphere has an area ($\displaystyle 4\pi r^2$) and with the same idea, a 4 dimensional sphere has a volume!
I'm curious, does a 4 dimensional sphere also has an area?
• Jul 13th 2009, 03:02 PM
Jester
Quote:

Originally Posted by arbolis
Ok thanks Chris, I'll try to do it. I was accustomed to speak about volumes only for solids in 3 dimensions. Now I realize that a 3 dimensional sphere has an area ($\displaystyle 4\pi r^2$) and with the same idea, a 4 dimensional sphere has a volume!
I'm curious, does a 4 dimensional sphere also has an area?

2D surface area (i.e. arc length) $\displaystyle \int_R \sqrt{1+y'^2} dx$ in 3D $\displaystyle \iint_R \sqrt{1+z_x^2+z_y^2} \,dA$

What do you think?
• Jul 13th 2009, 04:10 PM
arbolis
Quote:

Originally Posted by Danny
2D surface area (i.e. arc length) $\displaystyle \int_R \sqrt{1+y'^2} dx$ in 3D $\displaystyle \iint_R \sqrt{1+z_x^2+z_y^2} \,dA$

What do you think?

Seems to make sense. Did you deduce it, or saw it in a book/internet?
• Jul 14th 2009, 04:54 AM
Jester
Quote:

Originally Posted by arbolis
Seems to make sense. Did you deduce it, or saw it in a book/internet?