Area Of A Polar Region

Hello, I am wondering if there is something I am missing with this question. I have solved it in a very long-winded way (method in spoiler). Can anybody else point out an easier way to do it, or is mine the quickest?

The finite region R is bounded by an arc of the curve with equation

r=\frac{a}{1+cos\theta}

and the half-lines with equations

\theta = \theta_1

and

\theta=\theta_2

. Find the area of R, given that a>0.

Spoiler

Any help appreciated :smile:

Reply 1

RichE

15

write

cost + 1 = 2 cos^2(t/2)

and note

sec^4u = sec^2u (1+tan^2u)

Reply 2

Kolya

OP

17

RichE

write

cost + 1 = 2 cos^2(t/2)

and note

sec^4u = sec^2u (1+tan^2u)

Thank you! :smile:

Reply 3

danniella

4

Hello!

Alternatively:
Rewrite 1+cos x as (2 cos²x)² using double angle identities.
That means you're now integrating
a²/(4 cos⁴x/2 ) wrt x.
which
=(a²/4) (sec⁴x/2)
=(a²/4 ) (sec²x + sec²x/2 tan²x/2)

which integrates to
(a²/4) (tan x/2 + 1/3 tan³x/2)

Put in the limits of 0 and pi/2
and you end up with (a²/4) (4/3) = a²/3 as required.

love danniella

Reply 4

Christophicus

18

danniella

Hello!

Alternatively:
Rewrite 1+cos x as (2 cos²x)² using double angle identities.
That means you're now integrating
a²/(4 cos⁴x/2 ) wrt x.
which
=(a²/4) (sec⁴x/2)
=(a²/4 ) (sec²x + sec²x/2 tan²x/2)

which integrates to
(a²/4) (tan x/2 + 1/3 tan³x/2)

Put in the limits of 0 and pi/2
and you end up with (a²/4) (4/3) = a²/3 as required.

love danniella