Math Help Forum: polar coordinates

Find the area of the region bounded by r = 3 + 2sin(theta)

heres an example pic that comes with the problem also: http://us.st11.yimg.com/us.st.yimg.c...s_1932_3273205

need help.

That picture makes as much sense as division by zero.
Look at my improved, better, nicer, greater picture below.

You need to find,
INT INT 1*dA
Where the region is the polar figure shown below.

The bounds for theta are from 0 to 2*pi because we condisder the full figure. And the bounds for r are 3+2sin(theta) and 0 because there is no interior figure.
Thus,
INT (from 0 to 2*pi) INT (from 0 to (3+2sin(theta) ) r dr d(theta)

sorry, very new to coordinate systems, is INT for integral, integer, or interior?

thanks.

Hello, rcmango!

By the way, you can stop posting the blank graph paper, okay?

Find the area of the region bounded by: r = 3 + 2sinθ

From the questions you posted, it seems that you know nothing about polar curves.

Are you familiar with the cardioid and its variations so you can make a sketch?

Do you know the area formula for polar coordinates?

. . . . . . . . . .β
. . A . = . ½ ∫ r² dθ
. . . . . . . . . α


This graph is symmetric to the 90° line (the "y axis")
. . so we can integrate from -π/2 to π/2 and multiply by 2


. . A . = . 2 x ½ ∫ (3 + 2sinθ)² dθ

i moved this to urgent section.