Question

Evaluate the iterated Integral:
(pic added)

Answer 1

Now it'll be nice for someone to back me up on this since I'm new to these, but I'll give it a shot. So since dx is the inner differential, I'll pretend that any y's are just constants and I will integrate with respect to x.

\[\int\limits_{0}^{1}\int\limits_{0}^{1}(1-\frac{ x ^{2}+y ^{2} }{ 2 })dxdy\]
\[\int\limits_{0}^{1}\int\limits_{0}^{1}(1-\frac{ x ^{2} }{ 2 }+\frac{ y ^{2} }{ 2 })\]
Now again, pretending y's are just constants I'll get:
\[x-\frac{ x ^{3} }{ 6 }+\frac{ xy ^{2} }{ 2 }\]
Now integral from 0 to 1. In this case, 0 does not get us anywhere, so I'll just plug in 1:
\[1-\frac{ 1 }{ 6 }+\frac{ y ^{2} }{ 2 }=\frac{ 5 }{ 6 }+\frac{ y ^{2} }{ 2 }\]
Now that I'm left with only y, I can integrate with respect to y and finish the integral.
\[\int\limits_{0}^{1}(\frac{ 5 }{ 6 }+\frac{ y ^{2} }{ 2 })dy\]
think you could finish from there? :P

Answer 2

@Psymon, just one small error:
\[\int\limits_{0}^{1}\int\limits_{0}^{1}\left(1-\frac{ x ^{2}+y ^{2} }{ 2 }\right)dxdy=\int\limits_{0}^{1}\int\limits_{0}^{1}\left(1-\frac{x^2}{2}\color{red}{-\frac{y^2}{2}}\right)dxdy\]

Answer 3

@SithsandGiggles Why is it - y^2/2 ?

Answer 4

@Andysebb Yes, he is correct about the negative, he caught my mistake. This is the reason:

\[-\frac{ x ^{2}+y ^{2} }{ 2 } = -(\frac{ x ^{2} }{ 2 }+\frac{ y ^{2} }{ 2 }) = -\frac{ x ^{2} }{ 2 } - \frac{ y ^{2} }{ 2 }\]

Answer 5

@SithsandGiggles Thank you for the error correction, my bad :P

Answer 6

@Psymon ah I see it now. Thanks

Answer 7

I'm working out the problem right now following through your work. If I have any questions i'll post here. Thanks for your help!

Answer 8

Yeah, np ^_^ I'll have to do this stuff soon anyway, so might as well get the practice in xD