Question

Calculating double integral

Answer 1

\[\int\limits_{}^{}\int\limits_{}^{}xydxdy\]

Answer 2

d;
y>= 0, x^2 + y^2 <=1, x^2+y^2-2x<=0

Answer 3

is this the area?

Answer 4

also rho should go from 0 to 1
but where should theta go from? pi/2 to pi?

Answer 5

it looks like we have to divide that into two separate regions

Answer 6

which two?

Answer 7

from theta =0 up to some angle (tbd), rho goes out to 1
but then it uses the other circle as its limit

Answer 8

compare to

Answer 9

The area of of integration is best represented over 3 axis.

Answer 10

these are circles why would we need 3 axis?

Answer 11

and how do i know where to put the lines?

Answer 12

what's the question ?

Answer 13

where do i draw the lines on the picture, how do i know where to draw them?

Answer 14

you can imagine rho going from 0 to 1 at angle 0
and we can do that until we reach point A in this picture

Answer 15

we need to find the angle of that line, because after that angle
the limits on rho change to the red circle

Answer 16

okay, but how do i know my line goes exactly to A ?

Answer 17

where the two circles overlap?

Answer 18

once we pass point A, rho does not go to 1. (Look at the region you shaded in)

can you find the (x,y) coords of A ?

Answer 19

yes it's probably 1/2?

Answer 20

yes, x is 1/2

Answer 21

but why did we move the line?

Answer 22

Do you know how to integrate xy with respect to x?

Answer 23

The "right part" of the region of integration
has rho vary from 0 to 1
and theta from 0 to ?? (at point A)
we can set up that integral and find the area

the "left part" of the region has rho vary from 0 to 2 cos t
t varies from ?? (at point A) to pi/2 when r is tangent to the circle.
that is a different integral that we work out.

Answer 24

**Do you know how to integrate xy with respect to x?***
I would not use rectangular coords for this problem. It's messy enough in polar coords!

Answer 25

find (x,y) of point A, and then find the angle (using inverse tan (y/x) )

Answer 26

You know y would be a constant.. So integrate x dx

Answer 27

@DawnR

Answer 28

i'm here

Answer 29

Use the reverse power rule and you have your answer with a rho or polar coord.

Answer 30

Well the first half.

Answer 31

i am confused now because you are both taking different things, can someone draw me the picture or is the left side the left one from 0 to A ?

Answer 32

i found that for A, x=1/2, what equation do i use to calc y?

Answer 33

The integral of xy with respect to x is y times I when I is the integral of x dx

Answer 34

can you explain why am i doing the integral of xy?

Answer 35

You want the integral of this integral yes?

Answer 36

is the yellow side the left one and the green one the right one?

Answer 37

From the inside out you have the integral of xy with respect to x over the integral with respect to y

Answer 38

if i have integral of xy , where y is a constant that would be x^2/2

Answer 39

so does y = x^2/2?

Answer 40

no, y is a constant remember so then your inside integral becomes yx^2 / 2

Answer 41

okay, and what do i do with that?

Answer 42

Take the second integral with respect to y

Answer 43

okay so that is xy^2/2

Answer 44

You want the integral of that with respect to y since we want the double

Answer 45

can you write that as an equation, because i am not sure i follow?

Answer 46

\[\int\limits \int\limits xy dx dy \rightarrow \int\limits xy dx = I \rightarrow I= y \int\limits x dx = y (\frac{ x^2 }{ 2 })\]

Answer 47

we probably do want to use rectangular coords with this one.

Answer 48

\[\int\limits I dI = \int\limits (\frac{ x^2 }{ 2 })y dy = \left( \frac{ x^2 }{ 2 }\right) \int\limits y dy = \frac{ x^2y^2 }{ 4} + c+ d\]

Answer 49

what is d?

Answer 50

The constant for the second integral

Answer 51

Oh and c technically becomes cx as is typical when taking anti-derivatives.

Answer 52

staying in x y coords