# Thread: triple integrations

1. ## triple integrations

1. consider the solid tetrahedron determined by (0,0,0), (0,0,1),
(0,root13,0), (root2, root13, 0). SET UP the triple integrations so that its volume = triple integral of dy*dx*dz, and triple integral of dz*dy*dx.
is asking for the limit boundaries.
please explain step by step, i have the answers, but i dont' know how to get there.

2. find the center of mass of the solid above z = root(x^2+y^2) and below x^2 + y^2 + z^2 =4, and the density at a point equals the z-coordinate of the point.
i believe the center of mass = moment/mass.
with this problem please also explain step by step, and state all the equations you used.

THX!!!

2. Originally Posted by Econ_time
1. consider the solid tetrahedron determined by (0,0,0), (0,0,1),
(0,root13,0), (root2, root13, 0). SET UP the triple integrations so that its volume = triple integral of dy*dx*dz, and triple integral of dz*dy*dx.
is asking for the limit boundaries.
please explain step by step, i have the answers, but i dont' know how to get there.
Look at hand drawn picture below.

As you know the volume is,
$\int\int_S\int dV$
Where, $S$ is the set of points determined by this tetrahedron.
By Fubini's theorem it can be computed as,
$\int_A \int \int_{u(x,y)}^{v(x,y)} dz\, dA$
Where,
$u(x,y)$ is the lower surface bound on the tetrahedron in this case $u(x,y)=0$ and $v(x,y)$ is the equation of the plane above. To find the equation of that plane we need to find the equation of the plane passing through the points:
$(0,0,1)$, $(0,\sqrt{13},0)$, and $(\sqrt{3},\sqrt{13},0)$
I am not going to find the equation of the plane that is for thee to find.

Next, you need to express the region as Type I and Type II:
In that case,
$\int_0^{\sqrt{3}} \int_{\sqrt{13/3}x}^{\sqrt{13}} \int_0^{v(x,y)} dz\, dy\, dx$
And,
$\int_0^{\sqrt{13}} \int_0^{\sqrt{3/13}y} \int_0^{v(x,y)} dz\, dx\, dy$

Again where,
$v(x,y)$
Is the equation of plane expressed in terms of $x,y$