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?