MHB Triple Integral Bounded by Planes

[erin]

Hi All,

Question: Consider the tetrahedron, T, bounded by planes x=2, y=0, z=0 and 3x-6y-2z=0. Determine the integral \iiintyDV which is the y coordinate of the centre of mass.

I am getting a negative area which leads me to believe I'm doing something wrong. Working is attached.

View attachment 4305

Help would be greatly appreciated.

Thanks :)

 

[Opalg]

The only thing that is wrong is the limit of integration for $y$. You have $y$ going from $0$ to $1$. But in fact $y$ can be at most $\frac12x$, otherwise $z$ would have to be negative. Replace the upper limit for the $y$ integral by $\frac12x$ and I think you will get the right answer.

 

[erin]

Thanks. I re-did the integral and got -3/8. The integral is for the y-coordinate for the centre of mass so I don't understand why I am getting a negative number? Do you have any further insights? Thanks for your help.

 

[Opalg]

I re-did the integral and got -3/8. The integral is for the y-coordinate for the centre of mass so I don't understand why I am getting a negative number?

You already did the $z$-integral, so taking it from there, we get $$\int_0^2\int_0^{x/2}\bigl(\tfrac32xy - 3y^2\bigr)\,dy\,dx = \int_0^2\Bigl[\tfrac34xy^2 - y^3\Bigr]_0^{x/2}dx = \int_0^2\bigl(\tfrac3{16}x^3 - \tfrac18x^3\bigr)\,dx = \Bigl[\tfrac1{64}x^4\Bigr]_0^2 = \frac14.$$