#double integral in polar coord
50 messages · Page 1 of 1 (latest)
Basically r = 2cos(theta) goes through all possible radii from -2 to 2 within 0 to pi already
i just understand the question actually, its bcz asked above xy plane
but thank you for the helping
it's because what
This is irrelevant to the integral being computed
could you check it the question sentence
which question?
question in the photo that i sent here
I replied lol
.
if you say so
when i read the question again i got the point
the boundries are x^2 + y ^2 =2x and above xy-plane
play with this
what
You will see that the circle did one rotation within (0,pi)
that is because as I said cosine reaches radii between -2 and 2 within 0 to pi
i already drew the graph
it's more about understanding why 0 to pi is sufficient here
i thought same thing for first
but if you check the photo again the teta limits are 0 and pi
that's the above xy plane
bacc
above the plane implies half circle which would be from 0 to pi btw
So essentially
i know
bacc
because it likes you
also i don't what you are saying
i try to understand
,rcw
so what is your doubt?
for first i didn't get it bcz he scaled the whole of in this circle
but as i read the question again i got the problem
so everything is ok now, thx for your helping
😂
¯_(ツ)_/¯