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Frostbite
 4 years ago
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
Frostbite
 4 years ago
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
 4 years ago
Best ResponseYou've already chosen the best response.0i got the answer to π/8 and 1, but that can't be true?

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2dw:1352299213766:dw

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2if expression is like above then just evaluate it with usual process ... doesn't seem that difficult.

Frostbite
 4 years ago
Best ResponseYou've already chosen the best response.0not 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
 4 years ago
Best ResponseYou've already chosen the best response.2how did you change the limits?

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2dw:1352299738473:dw

Frostbite
 4 years ago
Best ResponseYou've already chosen the best response.0don't know if you can see it but there is a square root around 1x^2

Frostbite
 4 years ago
Best ResponseYou've already chosen the best response.0well 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
 4 years ago
Best ResponseYou've already chosen the best response.0Unless i have misunderstod something?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1it's the left half of a circle, I agree

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2sorry .. i completely missed sqrt() ..

Frostbite
 4 years ago
Best ResponseYou've already chosen the best response.0well kinda my falt i should have written sqrt() .. insted of √(1x^2

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1what bounds did you use?

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2dw:1352300314417:dw

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1yeah, that's what I got

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1oh no, r^3 though...

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1you have r^2 from the x^2 and r from dA

Frostbite
 4 years ago
Best ResponseYou've already chosen the best response.0can evne find that in my book that extra r, but seems to be right yea

Frostbite
 4 years ago
Best ResponseYou've already chosen the best response.0But is this polar set wrong then? Dp={(x,z,y)┤1≤r≤0 ,π/2≤θ≤3π/2}

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2\[ dx \;dy = r \; dr \; d\theta \]

Frostbite
 4 years ago
Best ResponseYou've already chosen the best response.0no wait can see it my self.. it is just 0<=r<=1

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1right, the angle covers the negative part

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1I still get pi/8 by the way...

Frostbite
 4 years ago
Best ResponseYou've already chosen the best response.0Just think i need to take a few more looks becuase i keep geting a wrong result.

Frostbite
 4 years ago
Best ResponseYou've already chosen the best response.0So we can all agree to this: (sorry for have pulling both you through all thís)

Frostbite
 4 years ago
Best ResponseYou've already chosen the best response.0Alright thanks alot to both of you.
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