find the area of of the region inside r=5sin(T) but outside r=3 on the polar coordinate system?

Question

You rock pascal!!!!!!!!!!!!

Answer 1

First, find the limits of integration -- these will be the points where 5 sin θ = 3, i.e. sin θ = 3/5. For θ between 0 and 180, these points are arcsin (3/5) and π-arcsin (3/5). The limits of integration for r are 3≤r≤5 sin θ. So setting this up as a double integral:

[arcsin (3/5), π-arcsin (3/5)]∫[3, 5 sin θ]∫r dr dθ

Don't forget the r when integrating in polar coordinates! It's very important. Anyway, moving on to the actual integral:

[arcsin (3/5), π-arcsin (3/5)]∫r²/2 | [3, 5 sin θ] dθ
[arcsin (3/5), π-arcsin (3/5)]∫25/2 sin² θ - 9/2 dθ

Recall that sin² θ = (1-cos(2θ))/2, so this is:

[arcsin (3/5), π-arcsin (3/5)]∫25/4 (1-cos (2θ)) - 9/2 dθ

Simplifying a bit:

[arcsin (3/5), π-arcsin (3/5)]∫7/4 - 25/4 cos (2θ) dθ

Integrating:

7/4 θ - 25/8 sin (2θ) | [arcsin (3/5), π-arcsin (3/5)]

Now substitute:

7/4 (π-2 arcsin (3/5)) - 25/8 (sin (2π-2 arcsin (3/5)) - sin (2 arcsin (3/5))

Now simplify the first one:

7/4 (π-2 arcsin (3/5)) - 25/8 (-sin (2 arcsin (3/5)) - sin (2 arcsin (3/5))
7/4 (π-2 arcsin (3/5)) - 25/8 (-2 sin (2 arcsin (3/5)))

Use the double angle formula:

7/4 (π-2 arcsin (3/5)) - 25/8 (-4 sin (arcsin (3/5)) cos (arcsin (3/5)))

Simplifying:

7/4 (π-2 arcsin (3/5)) + 25/2 (3/5 cos (arcsin (3/5)))

Using the fact that for angles between -π/2 and π/2 (read: in the range of the arcsin function) cos θ = √(1-sin² θ):

7/4 (π-2 arcsin (3/5)) + 25/2 (3/5 √(1-sin² (arcsin (3/5))))

Simplifying:

7/4 (π-2 arcsin (3/5)) + 25/2 (3/5 √(1-9/25))
7/4 (π-2 arcsin (3/5)) + 25/2 (3/5 * 4/5)
7/4 (π-2 arcsin (3/5)) + 6

And we are done.