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anonymous
 5 years ago
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(9x^2) and the outer integral from 0 to 3 of 3xy dy dx
anonymous
 5 years ago
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(9x^2) and the outer integral from 0 to 3 of 3xy dy dx

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Owlfred
 5 years ago
Best ResponseYou've already chosen the best response.0Hoot! 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
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0ok I know I have a problem since my answer is negative (and I'm in the first quadrant)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok I have a picture now

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x goes from 1/4y to sqrt(9y^2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wouldn't it also be the same to have x goes from 0 to 3 and y goes from 4x to sqrt(9x^2)??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{3}\int\limits_{y/4}^{\sqrt{9y ^{2}}} 3xy dxdy\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you make the graph you can see that if you set x to go from 03 than y will go from 0 till 4x or 0 till sqrt(9x^2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so I guess that way you have to split the integral to two parts

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok I'm going to try to solve.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am doing it now too, do you have the answer for it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have 33.75(3/32)*ln(3). Does this look right??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok lets do it step by step

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0first you do the x integral to get 3x^2y/2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now you put back the integral values

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01/2{3(9y^2)y3/16y^3)}

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have 1/2{3(9y^2)y3/16y^2)}

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it is y^3 you have 3x^2y and you put 1/4 y for x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now expanding the brackets gives: 1/2(2751/16y^3)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now we integrate for y that is: 1/2(27y51/4*16y^4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we only need to put back 3 as for 0 everything is 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01/2(27*351*3^4/(4*16)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and I dont know what that is :D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0just to confirm: we have \[(1/2)* ((27*3)((51*(3^4))/(4*16))\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks for your help! My webGrader still doesn't like it but I'll ask my professor tomorrow!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am not hundred percent sure but I guess it is ok, if you have the answer you could write a reply tomorrow

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In 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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