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i got the answer to π/8 and 1, but that can't be true?

|dw:1352299213766:dw|

not quite sure i understand. i mean on polar coordinates the set would be:
Dp={(x,z,y)┤|-1≤r≤0 ,π/2≤θ≤3π/2} right?
and then the following (see file)

how did you change the limits?

|dw:1352299738473:dw|

don't know if you can see it but there is a square root around 1-x^2

Unless i have misunderstod something?

it's the left half of a circle, I agree

sorry .. i completely missed sqrt() ..

well kinda my falt i should have written sqrt() .. insted of √(1-x^2

what bounds did you use?

|dw:1352300314417:dw|

yeah, that's what I got

oh no, r^3 though...

is it not r^2?

you have r^2 from the x^2 and r from dA

yeah .. it's r^3

can evne find that in my book that extra r, but seems to be right yea

can't* even*

But is this polar set wrong then? Dp={(x,z,y)┤|-1≤r≤0 ,π/2≤θ≤3π/2}

\[ dx \;dy = r \; dr \; d\theta \]

no wait can see it my self.. it is just 0<=r<=1

right, the angle covers the negative part

I still get pi/8 by the way...

Just think i need to take a few more looks becuase i keep geting a wrong result.

So we can all agree to this: (sorry for have pulling both you through all thís)

That's what I got

Alright thanks alot to both of you.

Welcome.