Question

Help with integration with polar coordinates?

The main problem I'm having is finding the bounds in polar coordinates. The problem is:

Integration of 3sqrt(x^2+y^2)dydx from 0 to sqrt(4-x^2) and from -2 to 2. If someone would tell me how to convert the bounds to polar coordinates, that would be great because I know the integrand converted is 3sqrt(r^2)

Answer 1

do you know the answer? final answer?

Answer 2

answer choices:
a) 18
b) 15pi/2
c) 8pi
d) 4pi
e) none of the above

but I don't know the final answer.

Answer 3

hmmmmm let me check my working first. are you 2nd year?

Answer 4

3rd year

Answer 5

that was weird!

Answer 6

my sister have this in her 2nd year, so idk if the problem in 3rd year is different.

Answer 7

I got 16pi...

Answer 8

no no, it's 8pi, answer is c

Answer 9

thank you, but can you tell me how to convert the bounds or how you got the answer?

Answer 10

okay,
theta bound:

y bounds from y=0 to y=sqrt(4-x^2) <-> x^2+y^2=4=2^2 which is the equation of circle center (0,0) radius 2
so if you draw this on paper, you'll se y bounds from y=0 to upper hafl of the circle. then you can see theta run from 0 to pi
0<=theta<=pi

Answer 11

r bound:

r runs from center (0,0) to circle x^2+y^2=4
so 0<=r<=2

Answer 12

so: omega={ (r,theta): 0 <= r <=2, 0 <= theta <= pi }
then do the integration.