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For a>0 we define

$\sum_n(a):=\big\{(x_1,...,x_n)\in R^n|x_1\geq 0,...,x_n\geq 0, \sum_{k=1}^nx_k\leq a\big\}$

1.1.: Show that the n-dimensional volume $v_n(\sum_n(a))=\frac{a^n}{n!}$

1.2.: Compute the volume of an object K $\in R^3$, which is enclosed by the planes x=1, y=0, x=y, z=0 and z=$e^{-x^2}$.

1.3.: As the center of gravity of a measurable set $K\in R^n$ with positive volume V we define the point $S=(s_1,...,s_n)$ with $s_k:=\frac{1}{V}\cdot \int_Kx_kd^nx,~~k=1,...,n.$ Compute the center of gravity of the standard simplex $\sum_3(1)\in R^3$ and the hemisphere $\{(x,y,z)|x^2+y^2+z^2 \leq 1, z\geq 0 \} \in R^3$.

Here are my ideas so far:

1.1.: I thought about proving that by induction.

For n=1 I get $x_1\leq a$ and $v_n$ would be a. But I don't see how I should approach n+1. I will try some more later.

1.2.: I tried drawing the space enclosed by those planes but I couldn't get anything out of it. There seems to be no space totally enclosed by it.

1.3.: I am totally clueless here.

Konstantin
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1 Answers1

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  1. Just set up the regular integral: $$\int_0^a\int_0^{a-x_1}...\int_0^{a-x_1-x_2-...-x_{n-1}}\,dx_ndx_{n-1}...dx_1\\ =\int_0^a\int_0^{a-x_1}...\int_0^{a-x_1-x_2-...-x_{n-2}}(a-x_1-...x_{n-1})\,dx_{n-1}...dx_1\\ \int_0^a\int_0^{a-x_1}...\int_0^{a-x_1-x_2-...-x_{n-3}}\frac{1}{2}(a-x_1-...x_{n-2})^2\,dx_{n-2}...dx_1\\$$

Can you try from here?

  1. It is a bounded object as below:

enter image description here

Notice that in the $xy$-plane, it is a triangle.

  1. For the simplex, you want to find $(s_1,s_2,s_3)$, where $$s_1=\frac{\iiint_K x\,dzdydx}{V}, s_2=\frac{\iiint_K y\,dzdydx}{V}, s_3=\frac{\iiint_K z\,dzdydx}{V}$$ where $V$ is the volume in $3$D you found from part 1. The integration region $K$ could be set up the same way as in 1, but in $3D$.

For the semisphere, you just need to change the $V$ and $K$ accordingly. Can you try from here?

KittyL
  • 16,965
  • One question though, how did you come up with the integral boundaries in 1? – Konstantin May 17 '15 at 12:03
  • @Konstantin: Think about the three dimensional case, the equation is $x+y+z\leq a$. The boundary is a plane that intersects $x,y,z$ axis all at the point $a$. In the first octant it is easy to imagine. When projected to $x,y$ plane, it is a line $x+y=a$, which intersects $x,y$ axis at $a$. – KittyL May 17 '15 at 17:39
  • Oh, okay, now I get it. But how come the upper boundary of the first integral is 1 and not a? – Konstantin May 17 '15 at 19:59
  • @Konstantin: Sorry my mistake. – KittyL May 17 '15 at 21:50
  • I am kind of having trouble getting the boundaries in 2 and 3. Been trying to find proper boundaries for a couple of hours now and can't get anything proper. I thought about $\int_0^1\int_0^x\int_y^\sqrt{-ln(z)}dxdydz$. Is that right? – Konstantin May 19 '15 at 15:11
  • @Konstantin: It's much easier setting it up in the order of $\iiint dz dy dx$ – KittyL May 19 '15 at 15:26
  • True, I did it this way and it worked. Thanks for all your help. – Konstantin May 19 '15 at 16:33