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Can some fast calculate the double integral using polar coordinates and cartesian coordinates? D={(x,z,y)|-1≤x≤0,-√(1-x^2 )≤y≤√(1-x^2 )} f(x,y,z)=x^2

Mathematics
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i got the answer to π/8 and 1, but that can't be true?
|dw:1352299213766:dw|
if expression is like above then just evaluate it with usual process ... doesn't seem that difficult.

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Other answers:

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)
1 Attachment
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
well the reason why i set my limit for theta to that is becuase the set should look like something like this: |dw:1352300003766:dw|
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)
1 Attachment
That's what I got
Alright thanks alot to both of you.
Welcome.

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