Triple Integral Example: Solving with the Order dydzdx and Correct Bounds

[PsychonautQQ]

Homework Statement

Triple Integral (x^6e^y)dV bounded by z=1-y^2 z=0 x=-1 x=1

The Attempt at a Solution

So I chose to try to integrate this in the order dydzdx
My bounds for the dy integral were from zero to (1-z)^(1/2)
my bounds for the dz integral were from 0 to 1
and my bounds for the dx integral were from -1 to 1.

I did the whole crazy integral and got 2/7 which my online thing is saying is wrong X_x doh...

Do these bounds look correct?

 

[vanhees71]

I'm not sure which integration region is meant here, because it sounds not very well defined. I guess, it's meant to integrate over a "paraboloic tunnel-like volume" between $x=-1$ and $x=1$ and from $z=0$ to $z=1-y^2 > 0$. This implies that also $y \in [-1,1]$. Then your integral would read
$$\int_{-1}^{1} \mathrm{d} x \int_{-1}^{1} \mathrm{d} y \int_0^{1-y^2} \mathrm{d} z x^6 \exp y.$$
Perhaps this gives the correct result?

 

[PsychonautQQ]

yes, I believe it does. What would the integral look like if I evaluated dydzdx? would y go from 0 to (1-z)^(1/2) and z go from 0 to 1?

 

[vanhees71]

Hm, if you want to do the $y$ integral first (which I'd consider a bad idea if I think about it ;-)), then for each $z \in [0,1]$ it should run in $y \in [-\sqrt{1-z},\sqrt{1-z}]$. Then the integral would be
$$\int_{-1}^{1} \mathrm{d} x \int_0^1 \mathrm{d} z \int_{-\sqrt{1-z}}^{\sqrt{1-z}} \mathrm{d} y \; x^6 \exp y.$$
Mathematica tells me that this gives the same result as it should be, but after the $y$ integral which is a bit easier in this way, the $z$ integral looks pretty cumbersome!