Question

Evaluate the integral ∫∫R(x2−2y2)dA, where R is the first quadrant region between the circles of radius 3 and radius 6

Answer 1

\[\int\limits_{}^{}\int\limits_{R}^{} x^2-2y^2 dA\]

Answer 2

yes

Answer 3

it didn't copy and paste well

Answer 4

thats ok

Answer 5

this is for calc 3 isnt it

Answer 6

yes

Answer 7

im in that class too

Answer 8

my very last section of the class

Answer 9

nice we are doing stokes thrm

Answer 10

ok so for these i like to draw them first

Answer 11

does it say if they are centered at the origin?

Answer 12

no it doesn'

Answer 13

lol then a guess well assume they are. lol

Answer 14

i would use polar

Answer 15

do you know polar?

Answer 16

yes that's what this is on

Answer 17

what does
x=????
y=????

Answer 18

well with radius 3 and 6 is all it says

Answer 19

and in the first quadrant

Answer 20

no in polar

Answer 21

how do i convert to polar?

Answer 22

what is x = to and y = to

Answer 23

ok so you gotta memorize this

Answer 24

seriously i've had to teach myself this entire calc 3 cause my teacher is really unhelpful

Answer 25

x=rcos(theta)
y=rsin(theta)

Answer 26

stick that in your head

Answer 27

you have to use it for all the polar ones if that start in x and y

Answer 28

alright

Answer 29

oh and this one too:
dA = r (dr) (dtheta)

Answer 30

then we just sub in those into our integral

Answer 31

is r my radius?

Answer 32

and to find theta we use arccos(x/r)?

Answer 33

\[\int\limits_{?}^{?}\int\limits_{?}^{?}(rcos \theta)^2 - 2 (rsin \theta)^2 r dr d \theta\]

Answer 34

so then we get that

Answer 35

the next step is to find our limits of integration

Answer 36

lets recall the drawing