Avon High School SOLUTIONS

AP Calculus BC
Period _____ Score ______ / 10

Skill Builder 9.9: Finding the Area of a Region Bounded by Two Polar
Curves
Use your calculators to both graph the polar equations and find the area of the indicated region.
1.) Common interior of r = 4(1 + sin q ) and 2.) Outside r = 1 + sin q and inside r = 2
r = 4(1 - sin q )
A = A1 + A2 A1 = A2 Þ A = 2 A1 r1 = 1 + sin q r2 = 2
"# $% !
circle radius 2
symmetry
2p
r1 = 4 (1 + sin q ) r2 = 4 (1 - sin q ) 1
A = p ( 2 ) - ò r12 dq
2

! 2 0
p area of circle #$%$
&
r1 = 0 Þ sin q = -1 Þ q = - inside r1
2
p = 7.8539"
r2 = 0 Þ sin q = 1 Þ q =
2
r1 = 4 Þ sin q = 0 Þ q = 0
r2 = 4 Þ sin q = 0 Þ q = 0
0 p /2
1 1
A1 = ò r12 dq + ò r dq = 5.6991!
2
2
2 -p / 2 2 0

A = 2 A1 = 11.3982!
3.) Inside the lemniscate of r 2 = 8cos 2q and 4.) The sonar signal from a submarine is modeled
outside r = 2 by r = a cos 2 q . Find the area of the region
between the two curves for a = 5 and
a = 10 .
A = A1 + A2 A1 = A2 A = 2 A1
"# $%
symmetry

1 p p
4 = 8cos ( 2q ) Þ cos ( 2q ) = Þ 2q = Þ q = A = A1 + A2 A1 = A2 A = 2 A1
"# $%
2 3 6 symmetry
p 6
1
A1 = ò ( r 2 - 22 ) dq = 1.3697 ! r1 = 5cos q r2 = 10 cos 2 q
2

2 -p 6 p 2
1
A = 2 A1 = 2.7394!
A1 = ò
2 -p 2
( r22 - r12 ) dq = 44.1786!

A = 2 A1 = 88.3572!

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 Find the arc length of the polar equation on the given interval. You may use your calculator to evaluate the
integrals for #’s 7 and 8 only.
p 6.) r = eq 0 £ q £ 2p
5.) r = sec q 0 £ q £
4
p /4 2 2p 2
æ dr ö æ dr ö
L= ò
0
r2 + ç ÷ dq
è dq ø
L= ò r +ç 2
÷ dq
è dq ø
0
p /4 2p 2p 2p

ò ( secq ) + ( secq tan q ) dq e 2q + ( eq ) dq =

0
= ò
0
ò
0
2e2q dq = 2 ò eq dq
0
p4
= ò sec 2 q + sec 2 q tan 2 q dq = 2 ( e 2p - e 0 ) = 2 ( e2p - 1)
0
p /4
= ò sec 2 q (1 + tan 2 q )dq
0
p /4 p /4
= ò sec 2 q ( sec 2 q )dq = ò sec 4 q dq
0 0
p4
p
dq = [ tan q ]0 = tan
p 4
= ò sec - tan 0 = 1
2

0
4
7.) r = 2 + sin q 0 £ q £ 2p e
8.) r = p £ q £ 2p
2p
æ dr ö
2 q
L= ò r2 + ç ÷ dq
è dq ø
2π
⎛ dr ⎞
2

L= ∫ r + ⎜ ⎟ dθ
0 2
2p
π
⎝ dθ ⎠
ò r + ( cos q ) dq = 13.3648!
2
= 2
2π

( )
2
0
= ∫
π
r 2 + −eθ −2 dθ = 1.9350…

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 Find the surface area when the polar equation on the given interval is rotated about the given line.
Surface area in polar is very similar to finding surface area in both rectangular (which we covered in August)
and surface area in parametric (which we covered in December).
Surface Area Formula w/ Rectangular Equations b
S = 2p ò f ( x) 1 + [ f ¢( x) ] dx
2

a
Surface Area Formula w/ Parametric Equations b 2 2
æ dx ö æ dy ö
S = 2p ò f (t ) ç ÷ + ç ÷ dt
a è dt ø è dt ø
Surface Area Formula w/ Polar Equations b 2
æ dr ö
S = 2p ò r sin q ( r ) + ç
2
÷ dq
a è dq ø
Note: All three of the above formulae assume that the plane region is being rotated about a HORIZONTAL
axis, hence the reason the “radius” expression that precedes the square root is a “y” value.
p p
9.) r = 2sin q 0 £ q £ q=
2 2
p /2
ó é dr ö ù
2

S = 2p ô ê r cos q r 2 + æç ÷ ú dq
ô ê è dq ø ú
õ0 ë û
p /2
ó é dr ö ù
2

S = 2p ô ê 2sin q × cos q r 2 + æç ÷ ú dq
ô ê è dq ø ú
õ0 ë û
p /2
S = 2p ó é ( 2sin q ) + ( 2 cos q ) ù dq
2 2
ô ê 2sin q × cos q
õ0 ë ûú
p /2
ô éê 2sin q × cos q 4 ( sin 2 q + cos 2 q ) ùú dq
S = 2p ó
õ0 ë û
p /2
S = 8p ò
0
[sin q × cos q ] dq
u = sin q du = cos q dq
p 2
é1 2ù éæ p ö 2 2ù
= 8p ê ( sin q ) ú = 4p êç sin ÷ - ( sin 0 ) ú
ë2 û0 êëè 2ø úû
= 4p (1 - 0 ) = 4p

p
10.) r = e-q 0 £q £ q =0
2
p /2
ó é dr ö ù
2

S = 2p ô ê r sin q r 2 + çæ ÷ ú dq
ô ê è dq ø ú
õ ë û
0
p /2

S = 2p ó ée -q × sin q e-2q + e-2q ù dq

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