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Frostbite
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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
 2 years ago
 2 years ago
Frostbite Group Title
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
 2 years ago
 2 years ago

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

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

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

Frostbite Group TitleBest ResponseYou've already chosen the best response.0
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)
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.2
how did you change the limits?
 2 years ago

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

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

Frostbite Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

Frostbite Group TitleBest ResponseYou've already chosen the best response.0
Unless i have misunderstod something?
 2 years ago

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

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

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

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
what bounds did you use?
 2 years ago

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

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

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

Frostbite Group TitleBest ResponseYou've already chosen the best response.0
is it not r^2?
 2 years ago

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

experimentX Group TitleBest ResponseYou've already chosen the best response.2
yeah .. it's r^3
 2 years ago

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

Frostbite Group TitleBest ResponseYou've already chosen the best response.0
can't* even*
 2 years ago

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

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

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

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
right, the angle covers the negative part
 2 years ago

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

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

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

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
That's what I got
 2 years ago

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

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
Welcome.
 2 years ago
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