Switching bounds in multiple integrals.

Question

sh3lsh · Answer

$$\int\limits_{0}^{3}\int\limits_{0}^{2z^2}\int\limits_{0}^{6-2x}8z dz dy dx$$

sh3lsh · Answer

So, I know I have to switch dz and dy, but I'm not too sure on how to go about that.

anonymous · Answer

The limits don't make sense... The first integral removes $z$, but the second integral's limits reintroduce it.

sh3lsh · Answer

Therefore, shouldn't I switch dz and dy?

anonymous · Answer

Oh sorry, didn't notice your second comment. It looked like you were given a 3d region, let's see those bounds again.

sh3lsh · Answer

This was the original question: 
Let a solid defined by the domain E have density f(x,y,z)=8z. Then the mass of the solid is
m=(triple integral) 8zdV

If E is the solid bounded in the x direction by x=0 and x=3, lying between the parabolic cylinder y=2z^2 and the plane y=0, and bounded in the z direction by z=0 and 2x+z=6, find the total mass of the solid.

sh3lsh · Answer

Wait, is easier than expected. 
I can just switch the bounds without worry. 

Thanks for taking your time though!