find the area of the region that lies in both
r= a(1+cos@) and r = a sin @
@ = theta sign
2007-11-09 03:28:43 · 2 answers · asked by !z@@h. (はりざ ) 4 in Science & Mathematics ➔ Mathematics
1+ cos @ = sin @ to find intercepts
sin@-cos@=1
Define t so sin @-cos@ = A sin (@-t)=A( sin @ cos t-cos @ sin t)
A cos t =1
A sin t = 1 squaring
A = sqrt(2) and tan t= 1 so t = pi/4
so sqrt(2) *cos (@-pi/4) = 1WRONG
cos (@-pi/4) =1/2 sqrt 2
@-pi/4 = -pi/4 and @ -pi/4 = pi/4
so @=0 and @=pi/2 as integration limits
Edit
It should be
sqrt(2)*sin (@-pi/4)=1
so @-pi/4 = pi/4 @=pi/2
and @-pi/4= 3pi/4 so @ = pi
Area between the curves= 1/2a^2Int (pi/2,pi)(1+cos @))^2-(sin^2@) *d@
=1/2a^2 Int( 1+2cos @ +cos2@)d@
=1/2a^2(@ +2sin@ +1/2 sin 2@) (pi/2,pi)
= 1/2 a^2( pi-pi/2-2) = 1/2a^2( 2-pi/2) Taken positive
2007-11-09 06:21:21 · answer #1 · answered by santmann2002 7 · 1⤊ 0⤋
Hint: Integrate from 0 to pi/2, and from pi/2 to pi.
2007-11-09 03:46:31 · answer #2 · answered by sahsjing 7 · 1⤊ 0⤋