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

- anonymous

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- 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

ok- I know I have a problem since my answer is negative (and I'm in the first quadrant)

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- anonymous

ok I have a picture now

- anonymous

y goes from 0 to 3

- anonymous

x goes from 1/4y to sqrt(9-y^2)

- anonymous

what do you think?

- 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

\[\int\limits_{0}^{3}\int\limits_{y/4}^{\sqrt{9-y ^{2}}} 3xy dxdy\]

- 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

so I guess that way you have to split the integral to two parts

- anonymous

ok- I'm going to try to solve.

- anonymous

I am doing it now too, do you have the answer for it?

- anonymous

no I don't :(

- anonymous

I have 33.75-(3/32)*ln(3). Does this look right??

- anonymous

I got different :-)

- anonymous

ok lets do it step by step

- anonymous

first you do the x integral
to get 3x^2y/2

- anonymous

I have that!

- anonymous

now you put back the integral values

- anonymous

1/2{3(9-y^2)y-3/16y^3)}

- anonymous

agreed?

- anonymous

I have 1/2{3(9-y^2)y-3/16y^2)}

- anonymous

it is y^3
you have 3x^2y and you put 1/4 y for x

- anonymous

now expanding the brackets gives: 1/2(27-51/16y^3)

- anonymous

agreed?

- anonymous

yep!

- anonymous

now we integrate for y
that is: 1/2(27y-51/4*16y^4

- anonymous

we only need to put back 3 as for 0 everything is 0

- anonymous

1/2(27*3-51*3^4/(4*16)

- anonymous

and I dont know what that is :D

- anonymous

8.2265625

- anonymous

just to confirm: we have \[(1/2)* ((27*3)-((51*(3^4))/(4*16))\]

- anonymous

yep!

- anonymous

charming number

- anonymous

Thanks for your help! My webGrader still doesn't like it but I'll ask my professor tomorrow!

- 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

thanks again!

- 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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