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

- Frostbite

- jamiebookeater

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- Frostbite

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

- experimentX

|dw:1352299213766:dw|

- experimentX

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

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- Frostbite

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

- experimentX

how did you change the limits?

- experimentX

|dw:1352299738473:dw|

- Frostbite

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

- Frostbite

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

Unless i have misunderstod something?

- TuringTest

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

- experimentX

sorry .. i completely missed sqrt() ..

- Frostbite

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

- TuringTest

what bounds did you use?

- experimentX

|dw:1352300314417:dw|

- TuringTest

yeah, that's what I got

- TuringTest

oh no, r^3 though...

- Frostbite

is it not r^2?

- TuringTest

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

- experimentX

yeah .. it's r^3

- Frostbite

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

- Frostbite

can't* even*

- Frostbite

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

- experimentX

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

- Frostbite

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

- TuringTest

right, the angle covers the negative part

- TuringTest

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

- Frostbite

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

- Frostbite

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

##### 1 Attachment

- TuringTest

That's what I got

- Frostbite

Alright thanks alot to both of you.

- TuringTest

Welcome.

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