Double integral in polar coordinates stuck :(

[bangshank]

I've been stuck on this problem for hours and the longer I stare at it the less I am getting away from finding a solution..


I can do parrt (i) and (ii) however I am stuck on part (iii).

I have found the integration limits to be theta from 0-pi and r from 0-2rsin(theta)

I am unsure how to integrate y^2sin(x), I change it to polar form and get r^2sin^2(theta)sin(rcos(theta)) and then get stuck

Any help would be much appreciated

 

[Deleted member 4993]

I've been stuck on this problem for hours and the longer I stare at it the less I am getting away from finding a solution..

View attachment 22892
I can do parrt (i) and (ii) however I am stuck on part (iii).

I have found the integration limits to be theta from 0-pi and r from 0-2rsin(theta)

I am unsure how to integrate y^2sin(x), I change it to polar form and get r^2sin^2(theta)sin(rcos(theta)) and then get stuck

Any help would be much appreciated

While solving (i)

what did you substitute for x & y to convert to polar system?

what polar expression did you get for the region D?

 

[bangshank]

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

r^2=x^2+y^2

The polar expression I got for the region D was r^3 rdrd(theta)

 

[Deleted member 4993]

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

r^2=x^2+y^2

The polar expression I got for the region D was r^3 rdrd(theta)

How did you get that? With he substitution you had proposed:

D is a closed curve defined by:

x² + y² = 2y

r² = r * sin(Θ)

r = sin(Θ)

Now what?

 

[HallsofIvy]

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

r^2=x^2+y^2

The polar expression I got for the region D was r^3 rdrd(theta)

That's the integrand, NOT an expression for the region.