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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
 one year ago
 one year 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
 one year ago
 one year ago

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

experimentXBest ResponseYou've already chosen the best response.2
dw:1352299213766:dw
 one year ago

experimentXBest 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.
 one year ago

FrostbiteBest 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)
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
how did you change the limits?
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
dw:1352299738473:dw
 one year ago

FrostbiteBest 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
 one year ago

FrostbiteBest 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
 one year ago

FrostbiteBest ResponseYou've already chosen the best response.0
Unless i have misunderstod something?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
it's the left half of a circle, I agree
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
sorry .. i completely missed sqrt() ..
 one year ago

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

TuringTestBest ResponseYou've already chosen the best response.1
what bounds did you use?
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
dw:1352300314417:dw
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
yeah, that's what I got
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
oh no, r^3 though...
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
you have r^2 from the x^2 and r from dA
 one year ago

FrostbiteBest ResponseYou've already chosen the best response.0
can evne find that in my book that extra r, but seems to be right yea
 one year ago

FrostbiteBest 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}
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
\[ dx \;dy = r \; dr \; d\theta \]
 one year ago

FrostbiteBest ResponseYou've already chosen the best response.0
no wait can see it my self.. it is just 0<=r<=1
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
right, the angle covers the negative part
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
I still get pi/8 by the way...
 one year ago

FrostbiteBest 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.
 one year ago

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

FrostbiteBest ResponseYou've already chosen the best response.0
Alright thanks alot to both of you.
 one year ago
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