# Thread: Need help with the boundaries of a 3D solid

1. ## Need help with the boundaries of a 3D solid

How do you find the boundaries of a volume defined by
$-1\leq x+y+3z\leq 1$
$1\leq 2y-z\leq 7$
$-1\leq x+y\leq 1$

to compute $\int\int\int 9z^2 dxdydz$

2. Various ways. Projection is one such way.

Simply eliminate all the 'z's and see wht it looks like from the point of view of the x-y plane.

Do the same for x-z and y-z.

Are you sure you must use dxdydz? Often there is one way that is easier than another. Maybe dydzdx would be more straight-forward. You must keep your eyes open.

Exploit symmetries and redundancies. I don't see much in this one except that the first and third constraints are a little odd. If z > 0, does the third one mean anything? Subtract 3z in the first and compare it to the third. What does that mean?

3. Originally Posted by TKHunny
Various ways. Projection is one such way.

Simply eliminate all the 'z's and see wht it looks like from the point of view of the x-y plane.

Do the same for x-z and y-z.

Are you sure you must use dxdydz? Often there is one way that is easier than another. Maybe dydzdx would be more straight-forward. You must keep your eyes open.

Exploit symmetries and redundancies. I don't see much in this one except that the first and third constraints are a little odd. If z > 0, does the third one mean anything? Subtract 3z in the first and compare it to the third. What does that mean?
I used the constraints to my advantage and this is what I came up with. Can anybody confirm whether the following is correct?

$\int_{\frac{-15}{6}}^{\frac{5}{6}} \int_{\frac{3+2x}{2}}^{\frac{5+2x}{2}}\int_{\frac{-1-2x-2z}{4}}^{\frac{9-2x-2z}{4}} 9z^2 dydzdx$