# double intergration

• Mar 15th 2010, 01:11 AM
karljones
double intergration
hi i need to find area enclosed in the shape (x^2+y^2-4y=0) - a circle? From pi/3 rads and pi/4 rads. How to do this? Do I need to convert to polar form?
• Mar 15th 2010, 01:41 AM
sa-ri-ga-ma
Quote:

Originally Posted by karljones
hi i need to find area enclosed in the shape (x^2+y^2-4y=0) - a circle? From pi/3 rads and pi/4 rads. How to do this? Do I need to convert to polar form?

(x^2+y^2-4y=0)
Rewrite this as
x^2 + y^2 - 4y + 4 = 4
Or
x^2 + (y-2)^2 = 4.......(1)
Here r = 2.
Now put x = r*cosθ and (y-2) = r*sinθ . Find the value of x or y at the angles π/3 and π/4.
Then find the intg(x*dy) between the above limits to get the area.
• Mar 15th 2010, 02:16 AM
karljones
hey thanks for your reply sa ri ga ma. pls see attached. The area is in green, so i don't think r can be constant?
• Mar 15th 2010, 02:21 AM
mr fantastic
Quote:

Originally Posted by karljones
hi i need to find area enclosed in the shape (x^2+y^2-4y=0) - a circle? From pi/3 rads and pi/4 rads. How to do this? Do I need to convert to polar form?

$\displaystyle x^2 + y^2 - 4y = 0 \Rightarrow r^2 - 4r \sin \theta = 0 \Rightarrow r = 4 \sin \theta$ (r = 0 is trivial).
• Mar 15th 2010, 02:40 AM
karljones
So then i need to evalute the following:

integral (theta = pi/3 and pi / 4) integral (r=0 and r= 4 sin theta) 1.r dr dtheta ???

is this correct? many thanks.
• Mar 15th 2010, 03:20 AM
tom@ballooncalculus
correct