Thread: what does it mean when a solid object is symmetric about a plane?

Suppose a region in the first octant is bounded by the coordinate planes and the plane $\displaystyle x+y+z=2$
and the density of the solid is $\displaystyle \delta(x,y,z) = 2xy$

I am asked to find center of mass.

I know how to find center of mass and moment about planes.

But how is this solid symmetric about xy and xz plane and not about yz plane? cause book says moment about $\displaystyle xy$
and $\displaystyle xz$ are equal by symmetry. ($\displaystyle M_{xy}$ and $\displaystyle M_{xz}$ are equal because of symmetry)

Is it possible to explain a bit?


You will find Plotting graph of $\displaystyle x+y+z=2, x=0, y=0$ and $\displaystyle z=0$ below.

Maybe my last post wasn't so clear. Sorry about that. Let me rephrase it.

Let's say i have an object bounded by four planes $\displaystyle (x = 0, y = 0, z = 0$ and $\displaystyle x + y + z - 2 = 0)$


And density function of that object is $\displaystyle \delta(x,y,z) = 2x$ And asked to find first moment of that bounded object about three planes($\displaystyle xy,\,\, yz$ and $\displaystyle zx$ plane).


We know from the definition that the first moment of this 3d object about $\displaystyle xy,\,\, yz\,\,$ and $\displaystyle zx$ planes are respectively:

$\displaystyle
M_{xy} = \int_0^2\int_0^{2-x}\int_0^{2-x-y} 2\,\,x\,\,z\,\,dz\,\, dy\,\, dx
$

$\displaystyle
M_{yz} = \int_0^2\int_0^{2-x}\int_0^{2-x-y} 2\,\,x^2\,\, dz\,\, dy\,\, dx
$

$\displaystyle
M_{zx} = \int_0^2\int_0^{2-x}\int_0^{2-x-y} 2\,\,x\,\,y\,\, dz\,\, dy\,\, dx
$

The book without evaluating all three of this, evaluated only $\displaystyle M_{xy}$ and $\displaystyle M_{yz}$ and said that by symmetry $\displaystyle M_{xy} = M_{xz}$ The graph of this object is given below. My question is in what way this bounded object is symmetric? And how did they figure that out? Any explanation regarding this question is welcome. I've no idea how they deduced that..

The dark blue surface in the graph below is $\displaystyle x + y + z - 2 = 0$. And $\displaystyle x = 0, z = 0$ are wireframe planes in rainbow color. And finally $\displaystyle y = 0$ wireframe plane is transparent with wireframe visible. If my question isn't clear please let me know.

I found the solution to my problem. Those who are interested if you plug in any value for $\displaystyle x$ and $\displaystyle y$ into this
equation for plane $\displaystyle x + y + z - 2 = 0$ the value of $\displaystyle y$ and $\displaystyle z$ will be always the same.

So they are equidistant from $\displaystyle xz$ and $\displaystyle xy$ plane.

The same is true for $\displaystyle x$ and $\displaystyle z$. No matter what value you plug into within the domain of the solid $\displaystyle x + y + z - 2=0$ for $\displaystyle x$ and $\displaystyle z$ and
solve for $\displaystyle y$ the value of $\displaystyle y$ and $\displaystyle z$ will be always same.

So in that sense the enclosed solid is equidistant from $\displaystyle xz$ and $\displaystyle xy$ plane and the solid is symmetric about $\displaystyle xz$ and $\displaystyle xy$ plane. Solved.