# Using triple integrals to find centre of gravity of an object with varying density

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• Oct 6th 2010, 02:49 AM
ben1479
Using triple integrals to find centre of gravity of an object with varying density
i have a really hard homework question where i have spent hours trying to figure it out.
the question is;
find the centre of gravity of a solid bound by $z=1-y^2$ (for y>=0), z=0, y=0, x=-1 and x=1 which has a mass density of $rho(x,y,z)=yz$ grams/m^3.
i have done some research on the web and found that the centre of gravity of an object with varying density is cg * W = g * SSS x * rho(x,y,z) dx dy dz where cg is center of gravity, W is weight(which i dont have), g is gravity(which is also not specified), SSS indicates a triple integral with respect to dx,dy,dz and rho(x,y,z) is the object density.

i asked my lecturer about it and this is what she replied;
You need an integral to find the mass (this will be just a number, not a function). You then need an integral to find the x-coordinate of the centre of gravity (this will also be just a number, not a function). Then one for the y-component and then one for the z-component. So in total, you should do 4 triple integrals.
please help as i cannot figure it out.
• Oct 6th 2010, 03:41 AM
HallsofIvy
This is very strange. Since this is homework, you must be taking a class in which you are being taught this, yet you used the internet to find a formula that should be in your text book- and you appear not to know how to use an integral to find the weight of an object with given density. All of those should have been covered in class before you were given homework like this.

One of the very first things you should have learned about triple integrals is that they can be used to find the volume of a three-dimensional region. And, in particular, if you are given the region as bounded by x= a, x= b, y= c, y= d, z= e, z= f(x,y) (where f(x,y) is a function of x and y) then its volume is given by
$\int_{x= a}^b \int_{y= c}^d\int_{z= e}^{f(x,y)} dzdydx$

And if the density is given by $\rho(x,y,z)$, you just include that instead of "1" as integrand to find the mass:
$\int_{x= a}^b \int_{y= c}^d\int_{z= e}^{f(x,y)} \rho(x, y, z) dzdydx$

"Weight" is the density times g but you don't really need g- it will cancel out. The "center of mass" is exactly the same as the "center of gravity".

The x- coordinate of the center of mass is given by
$\int_{x= a}^b \int_{y= c}^d\int_{z= e}^{f(x,y)} x\rho(x,y,z) dzdydx$
divided by the mass. The y-coordinate is given by
$\int_{x= a}^b \int_{y= c}^d\int_{z= e}^{f(x,y)} y\rho(x,y,z) dzdydx$
and the z- coordinate is given by
$\int_{x= a}^b \int_{y= c}^d\int_{z= e}^{f(x,y)} z\rho(x,y,z) dzdydx$

The mass integral (which has no "x", "y", or "z" multiplying $\rho(x,y,z)$) and those three integrals are the four integrals your lecturer wsa refering to. In your problem, $f(x,y)= 1- y^2$, $\rho(x,y,z)= yz$, a= -1, b= 1, (x between -1 and 1), c= 0 ( $y\ge 0$, and e= 0 (z= 0), but you do not give an upper bound on y. An upper bound for y, either as a number or as a function of x, must be given in order for this region to be properly defined.
• Oct 6th 2010, 04:21 AM
ben1479
thanks very much for that, it cleared up a lot.

with the y upper bound (y=d), am i right in thinking that it can be found by using z=1-y^2 where z=0? this would give me an upper limit of y=1 (d=1).