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.
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)
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