Candy
If you integrate e^(x+1) I'm sure you'd say e^(x+1) and you'd be right
When intergrating e^(x+y) wrt x treat y as a constant
When integratin e^(x+y) wrt y treat as x as a constant
In either case you get e ^(x+y)
Now, what if i want to integrate ∫∫∫ ze^(x+y) dV to find the mass of the rectangular box determined by 0≤x≤1, 0≤y≤1 and 0≤z≤5.
i tried to solve the integral in the following order..
∫from 0 to 1 ∫from 0 to 1 ∫from 0 to 5 (ze^(x+y))dzdydx
i use the method Calculus 26 described and i got 58.38 but it's wrong!
can you tell mewhat i did wrong, is my set up right?
Thank you