anonymous
  • anonymous
I am working on a double integral of 3xy where R is the region in the first quadrant bounded by x^2+y^2=9, y=4x, y=0. I drew a picture and I have the inner integral from 4x to sqrt(9-x^2) and the outer integral from 0 to 3-- of 3xy dy dx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Owlfred
  • Owlfred
Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!
anonymous
  • anonymous
I need to get a pen and paper for this, good practice for me too :-) next week exam about this. 2min and I am back
anonymous
  • anonymous
ok- I know I have a problem since my answer is negative (and I'm in the first quadrant)

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anonymous
  • anonymous
ok I have a picture now
anonymous
  • anonymous
y goes from 0 to 3
anonymous
  • anonymous
x goes from 1/4y to sqrt(9-y^2)
anonymous
  • anonymous
what do you think?
anonymous
  • anonymous
wouldn't it also be the same to have x goes from 0 to 3 and y goes from 4x to sqrt(9-x^2)??
anonymous
  • anonymous
\[\int\limits_{0}^{3}\int\limits_{y/4}^{\sqrt{9-y ^{2}}} 3xy dxdy\]
anonymous
  • anonymous
if you make the graph you can see that if you set x to go from 0-3 than y will go from 0 till 4x or 0 till sqrt(9-x^2)
anonymous
  • anonymous
so I guess that way you have to split the integral to two parts
anonymous
  • anonymous
ok- I'm going to try to solve.
anonymous
  • anonymous
I am doing it now too, do you have the answer for it?
anonymous
  • anonymous
no I don't :(
anonymous
  • anonymous
I have 33.75-(3/32)*ln(3). Does this look right??
anonymous
  • anonymous
I got different :-)
anonymous
  • anonymous
ok lets do it step by step
anonymous
  • anonymous
first you do the x integral to get 3x^2y/2
anonymous
  • anonymous
I have that!
anonymous
  • anonymous
now you put back the integral values
anonymous
  • anonymous
1/2{3(9-y^2)y-3/16y^3)}
anonymous
  • anonymous
agreed?
anonymous
  • anonymous
I have 1/2{3(9-y^2)y-3/16y^2)}
anonymous
  • anonymous
it is y^3 you have 3x^2y and you put 1/4 y for x
anonymous
  • anonymous
now expanding the brackets gives: 1/2(27-51/16y^3)
anonymous
  • anonymous
agreed?
anonymous
  • anonymous
yep!
anonymous
  • anonymous
now we integrate for y that is: 1/2(27y-51/4*16y^4
anonymous
  • anonymous
we only need to put back 3 as for 0 everything is 0
anonymous
  • anonymous
1/2(27*3-51*3^4/(4*16)
anonymous
  • anonymous
and I dont know what that is :D
anonymous
  • anonymous
8.2265625
anonymous
  • anonymous
just to confirm: we have \[(1/2)* ((27*3)-((51*(3^4))/(4*16))\]
anonymous
  • anonymous
yep!
anonymous
  • anonymous
charming number
anonymous
  • anonymous
Thanks for your help! My webGrader still doesn't like it but I'll ask my professor tomorrow!
anonymous
  • anonymous
I am not hundred percent sure but I guess it is ok, if you have the answer you could write a reply tomorrow
anonymous
  • anonymous
thanks again!
anonymous
  • anonymous
In order to do this in x and y, you would have to break the integral up because the y goes from 0 to 4x over interval 0 to 3/4; and y goes from 0 to3 over interval 3/4 to 3. This interval is good for polar coordinates.

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