Question

(double integral)R 1/(1+x^2+y^2)^2; R is the first quadrant. Solve the integral

Answer 1

\[\int\int{1\over1+x^2+y^2}dA\]in the first quadrant?

Answer 2

if so, convert to polar coordinates

Answer 3

1/(1+x^2+y^2)^2

Answer 4

right sorry\[\int\int{1\over(1+x^2+y^2)^2}dA\]let's convert these to polar coordinates and see what happens...

Answer 5

how do you do that?

Answer 6

for polar coordinates\[x=r\cos\theta\]\[y=r\sin\theta\]\[dA=rdrd\theta\]

Answer 7

isn't there a given region?

Answer 8

oh k

Answer 9

there is but its long wait i am gonna try to write it

Answer 10

like integrate over the first Q of x^2+y^2=1 or something?

Answer 11

well we need the region to find the bounds on r I think

Answer 12

limit b to infinity, (integral)alfa to beta (integral) a to b g(r,theta) r dr dtheta

Answer 13

this is what it says in the book

Answer 14

that is just a general formula for what I am showing you
nevermind, just do the polar conversion I mentioned earlier

Answer 15

oh okay lol

Answer 16

how do i go about solving it

Answer 17

do the substitution I mentioned earlier and you will see the integration is easy
I just don't know what the bounds on r are supposed to be

Answer 18

okay

Answer 19

i don't even know how to set it up

Answer 20

yeah, it looks like the bounds on r are from 0 to 1

so just plug in the subs for conversion to polar coordinates I gave above and simplify\[x=r\cos\theta\]\[y=r\sin\theta\]\[dA=rdrd\theta\]plug that into your integral\[\int\int{1\over(1+x^2+y^2)^2}dA\]

Answer 21

(double integral) 1/(1+(r cos theta)^2 + (r sin theta)^2)^2 r dr dtheta ?

Answer 22

yes, now simplify that using trig identities
hint: factor out r^2 where possible, that should reveal a trig identity

Answer 23

costheta + sintehta is cos2theta right?

Answer 24

I don't think so, but that's not the identity I had in mind
let's look at just what is in the parentheses in the denom...

Answer 25

\[1+(r\cos\theta)^2+(r\sin\theta)^2\]simplify this using the hint I gave you

Answer 26

hmm....

Answer 27

I'll let you think about it for a minute while I help someone else :)
hint: MAJOR trig identity here after factoring out r^2

Answer 28

1+r^2? lol sorry i suck at math

Answer 29

that is right!
and how did you make it to multivariable calc if you suck at math?

Answer 30

i mean i don't suck but i like blank out sometimes

Answer 31

anyway, our integral is now\[\int\int{1\over(1+r^2)^2}rdrd\theta\]

Answer 32

we are in Q1, so what are the bounds on theta?

Answer 33

0 to 2pi?

Answer 34

that would be a full circle (we would go through all 4 quadrants that way)
we want just Q1
so think again

Answer 35

hmm. 0 to pi

Answer 36

oh wait its 0 to pi/2 right?

Answer 37

yes :)

Answer 38

now the only thing I am unsure of is the limits of r, since no region is given
I think it may be from 0 to 1 but am not sure...
there is no equation for the region R in your problem?

Answer 39

no not in this problem but in problem before its 0 to infinity and 1 to infinity. but in this problem its not given

Answer 40

let me check my notes...

Answer 41

my notes are not helping me
the phrasing is strange for me, but since Q1 is an infinite region making the bounds on r 0 to infinity is plausible
never seen it before though, but it fits with your notes it seems

Answer 42

ye i don't know why i gives limits for r in this problem. its there in all the other problem

Answer 43

it does \(not\) give the limits for r, right?

Answer 44

if not, then the integral wil then be\[\int_{0}^{\frac\pi2}\int_{0}^{\infty}{1\over(1+r^2)^2}rdrd\theta\]if they are given, use them instead
gotta go, good luck!

Answer 45

thanks