Question

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

Answer 1

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!

Answer 2

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

Answer 3

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

Answer 4

ok I have a picture now

Answer 5

y goes from 0 to 3

Answer 6

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

Answer 7

what do you think?

Answer 8

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)??

Answer 9

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

Answer 10

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)

Answer 11

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

Answer 12

ok- I'm going to try to solve.

Answer 13

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

Answer 14

no I don't :(

Answer 15

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

Answer 16

I got different :-)

Answer 17

ok lets do it step by step

Answer 18

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

Answer 19

I have that!

Answer 20

now you put back the integral values

Answer 21

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

Answer 22

agreed?

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

agreed?

Answer 27

yep!

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

and I dont know what that is :D

Answer 32

8.2265625

Answer 33

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

Answer 34

yep!

Answer 35

charming number

Answer 36

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

Answer 37

I am not hundred percent sure but I guess it is ok, if you have the answer you could write a reply tomorrow

Answer 38

thanks again!

Answer 39

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.