Question

I need to calculate the double integral...

Answer 1

\[\int\limits_{D}^{}\int\limits_{}^{} \frac{ dxdy }{ (x^2+y^2)^2}\]

Answer 2

D: \[x^2+y^2 = 4x, x^2+y^2=8x, y=0, y=x\]

Answer 3

i need to use polar coordinates for this..
and i was about to limit rho to \[2\sqrt(x) \le \rho \le 2\sqrt(2x)\]

Answer 4

but what am i supposed to do with y=0 and y=x?

Answer 5

do you have multiple D's ?
the first two look like circles
but y=0, y=x needs another bound

Answer 6

just one D for both ..it says that D is limited with a circle

Answer 7

Sorry, I don't understand. Can you post the problem?

Answer 8

it's not in english but i'll try to explain it again

Answer 9

Calculate the double integral (the one i posted above) using polar coordinates, where D is limited with circle with x^2+y^2 = 4x, x^2+y^2=8x , y=0, y=x

Answer 10

I think the picture looks like this

Answer 11

Yes, but what region are we talking about?

Answer 12

meaning?

Answer 13

I'm guessing D is this??

Answer 14

because of the y=x?

Answer 15

and y=0. But it's not obvious (to me) what they mean.

Answer 16

but if y=0 then how can it be above 0?

Answer 17

I'm thinking each of those curves (or lines) are boundaries to D
the region I outlined is my *guess* as to what they mean.

Answer 18

but why not the whole area

Answer 19

then we ignore y=x (it's not used)

Answer 20

right but then it can be both left and tight side of the line y=x

Answer 21

the left side ignores (does not need) y=0

Answer 22

i am not getting that, what do you mean it does not need y=0..what does y=0 actually do?

Answer 23

y=0 is the x-axis. I assume it is part of the boundary to region D
if we pick a region that does not touch y=0 then why bother to specify y=0 as part of the boundary?

Answer 24

okay i get it now.. so how do i put limits for rho now?

Answer 25

the small circle has the equation
r= 4 cos t
and the larger has the equation
r= 8 cos t
those will be the lower and upper limits on r as a function of angle t
t will go from 0 to pi/4 (x-axis to y=x)

Answer 26

so
\[\int\limits_{0}^{\pi/4}\int\limits_{4\cos \theta}^{8\cos \theta}\frac{ d \rho d \theta }{ (\rho^4) } \]

Answer 27

isn't it dx dy ==> r dr dt ?

Answer 28

well (x^2+y^2) = rho^2

Answer 29

and 0 to pi/4 should be limits for theta

Answer 30

does your r = rho?

Answer 31

or r as r in circle

Answer 32

yes, all of that is ok. but dx dy is not dr dt
it's r dr dt
so you end up with
\[ \int_0^\frac{\pi}{4} \int_{4 \cos \theta}^{8 \cos \theta} \frac{d\rho \ d\theta}{\rho^3}\]

Answer 33

yes, r means rho and t means theta

Answer 34

oh right jacobian

Answer 35

also how do i know that rho goes from 4 cos t to 8 cos t , i mean how do i know it's cos?

Answer 36

It's one of the few polar graphs I remember.

Answer 37

It's the last example at the bottom of the page
http://colalg.math.csusb.edu/~devel/IT/main/m11_conic/src/s07_polar-graphs.html

Answer 38

so a here is r of the circle?
for the small one r = 2 and for the big one r = 4?

Answer 39

yes. a circle with center at (a, 0) (in (x,y) coords) and radius a has the polar equation
\[ r = 2 a \cos \theta \]

Answer 40

okay, got it, thanks!

Answer 41

How about this one
\[\int\limits_{}^{}\int\limits_{}^{}\sqrt{a^2-x^2-y^2}dxdy\]
D is limited with circle:
\[x^2+y^2 = a^2,y \ge x , y \le x \sqrt3\]

Answer 42

rho goes from 0 to a
the two lines limit the angle y= x is at pi/4
the other is atan(sqr(3)) = pi/3

Answer 43

i understand for rho, but agan for theta, atan(sqr3)?
where did i get tan?

Answer 44

this is the area?