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



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

experimentX
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dw:1352299213766:dw

experimentX
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if expression is like above then just evaluate it with usual process ... doesn't seem that difficult.

Frostbite
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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)

experimentX
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how did you change the limits?

experimentX
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dw:1352299738473:dw

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

Frostbite
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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

Frostbite
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Unless i have misunderstod something?

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

experimentX
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sorry .. i completely missed sqrt() ..

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

TuringTest
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what bounds did you use?

experimentX
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dw:1352300314417:dw

TuringTest
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yeah, that's what I got

TuringTest
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oh no, r^3 though...

Frostbite
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is it not r^2?

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

experimentX
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yeah .. it's r^3

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

Frostbite
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can't* even*

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

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

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

TuringTest
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right, the angle covers the negative part

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

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

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

TuringTest
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That's what I got

Frostbite
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Alright thanks alot to both of you.

TuringTest
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Welcome.