Question

Find the area of the region which is bounded by the polar curves: theta=pi and r=3(theta), interval from 0<=theta=>pi

Answer 1

Help please

Answer 2

OK

Answer 3

have answer choices

Answer 4

In polar coordinates, the integral to find area is:
A = [a, b]∫ (1/2)r^2 dt

Here, we have a = -π/2, b = π/2, and r = 3 + 2sin(t), so:
A = [a, b]∫ (1/2)r^2 dt
= [-π/2, π/2]∫ (1/2)[3 + 2sin(t)]^2 dt
= [-π/2, π/2]∫ (1/2)[9 + 12sin(t) + 4sin^2(t)] dt
= [-π/2, π/2]∫ 9/2 + 6sin(t) + 2sin^2(t) dt

Recall that ∫ sin^2(t) dt = (1/2)[t - sin(t)cos(t)] (you can show this using the power reduction formula for sin^2(x)), so we have:
[-π/2, π/2]∫ 9/2 + 6sin(t) + 2sin^2(t) dt
= [9t/2 - 6cos(t) + {t - sin(t)cos(t)}] | [-π/2, π/2]
= [9(π/2)/2 - 6cos(π/2) + {π/2 - sin(π/2)cos(π/2)}] - [9(-π/2)/2 - 6cos(-π/2) + {-π/2 - sin(-π/2)cos(-π/2)}]
= 11π/2

The inner loop created on 7π/6 ≤ t ≤ 11π/6 (sketch it, plug in numbers, etc. do whatever you can to figure out the limits for t…sometimes it's hard). So the area is:
A = [a, b]∫ (1/2)r^2 dt
= [7π/6, 11π/6]∫ (1/2)[3 + 6sin(t)]^2 dt
= [27t/2 - 18cos(t) - 9sin(t)cos(t)] | [7π/6, 11π/6]
= -(27√3)/2 + 9π
≈ 4.89165

Answer 5

are you giving me the steps to follow?

Answer 6

yes

Answer 7

how did you find the a,b? is that 0,pi in this case

Answer 8

I have another method, similar slightly different approach if you would like it

Answer 9

please

Answer 10

ok so to start draw a picture. this will be your saving grace for everything in calc 2,3

Answer 11

can theta=pi be considered a polar curve?

Answer 12

what part should I draw?

Answer 13

well it is your bound if \[\theta=\pi\] then what does that look like? It is the set of all radius' where theta =pi similar to if I said the line y=4. Do you understand this?

Answer 14

I would like you to draw the entire graph and shade the area we will be integrating over