Question

Convert to polar coordinates and evaluate the double integral. Below:

Answer 1

so far:
4-x^2=y^2
y^2+x^2=4
r^2 = 4, r = 2
x = rcostheta, x=2costheta

Answer 2

4xydxdy = 4(2)(2costheta)(2sintheta) dr d(theta)= 32costhetasintheta dr d(theta) ?

Answer 3

you will want to leave r as a variable, and not plug in 2 before integrating. r is not constant

Answer 4

the way i do these is draw it (i think it's crucial)

Answer 5

the bounds of theta are from pi/4 (45 degrees) to pi/2 (90 degrees)

and the bounds of r are from 0 to 2

let me know if you have questions thus far while i write the rest

Answer 6

nope

Answer 7

x = rcostheta
y = rsintheta

dydx = dA = r dr dtheta

Answer 8

\[\int\limits_{0}^{\sqrt2}\int\limits_{0}^{\sqrt{4 - x^2}}4xydydx = \int\limits_{\pi/4}^{\pi/2}\int\limits_{0}^{2}4rcos\theta rsin\theta r dr d\theta= 4\int\limits_{\pi/4}^{\pi/2}\int\limits_{0}^{2}r^3\cos\theta \sin\theta dr d\theta\]

Answer 9

I think I got the answer
is it 1?

Answer 10

haven't done it yet :P ill let u know soon

Answer 11

ok lol. and thanks

Answer 12

i did it really fast and i don't think it's one

Answer 13

i get\[8 ( 1 - \frac{ 1 }{ \sqrt2 })\]i could be wrong though

Answer 14

before you do the outside integral do you get 4sinthetacostheta?

Answer 15

1/4r^4sinthetacostheta from 0 to 2, (2^4)/4 = 4

Answer 16

i get 16 costheta sin theta

Answer 17

keeping the 4 from the begininng

Answer 18

oh duh

Answer 19

ok thanks

Answer 20

I have a similar problem with triple integrals that I'm going to post if you want to help out there too

Answer 21

i'll check it out :)