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.

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- February 13th 2007, 06:10 PMrcmangopolar 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. - February 13th 2007, 06:18 PMThePerfectHacker
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) - February 13th 2007, 07:36 PMrcmango
sorry, very new to coordinate systems, is INT for integral, integer, or interior?

thanks. - February 13th 2007, 10:12 PMSoroban
Hello, rcmango!

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

Quote:

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

- February 15th 2007, 01:39 PMrcmango
i moved this to urgent section.