calc:image below please.

Question

anonymous · Answer

attachment

anonymous · Answer

Actually, drawing this is a lot harder than I expected... I'll be back with a screen shot.

anonymous · Answer

attachment

anonymous · Answer

does y range from 2 to 3 or does y range from 0 to 2?

anonymous · Answer

One way to define $E$ is by
$$E:=\left\{(x,y,z)~:~2-y\le x\le 6-2y,~~\color{red}{0\le y\le2},~~0\le z\le \sqrt{4-y^2}\right\}$$
The red part should answer that question

anonymous · Answer

probably 0 to 2 because of z= something something something

anonymous · Answer

yeah i got it now, thanks. I was having a debate between y from 0 to 2 from yz planes or 2 to three from xy planes.

anonymous · Answer

Right, so the integral is
$$\int\int\int_E~\frac{z}{4-y}~dV\
\int_0^2\int_{2-y}^{6-2y}\int_0^\sqrt{4-y^2} \frac{z}{4-y}~dz~dx~dy$$

anonymous · Answer

No, I am gonna be uber and use spherical coordinates.

anonymous · Answer

Just kidding. I dun think its gonna be possible with spherical coordinates.

anonymous · Answer

Yeah, I'm not sure about that... It looks like it can be done with the Cartesian coordinates, though, so that's fortunate.