Find the area of the region that lies inside both curves. r = 9 sin(𝜃), r = 9 cos(𝜃)

Answer 1

If you graph this, you will see that there are two circles. When you set the two polar equations together, the intersection point is:

9sin(θ)=9cos(θ) --> tan(θ) = 1 --> θ =π/4

The region to find the area is the area of 9sin(θ) from 0 to π/4 PLUS the area of 9cos(θ) from π/4 to π/2.

A = (1/2)∫(9sin(θ))²dθ + (1/2)∫(9cos(θ)²dθ

However, the region is symmetrical about π/4, so only need to double the first integral to save us some work.

A = 81∫sin²(θ)dθ from 0 to π/4

= (81/2)∫(1-cos(2θ))dθ

= (81/2)[θ-sin(2θ)/2] evaluated at θ=π/4 and 0

=(81/2)(π/4 -1/2) = (81/2)((π-2)/4) = (81π - 162)/8 ≅ 11.56

Answer 2

Graph this!

You will see 2 circles, one center on each co-ordinate axis (in Cartesian co-ordinates)

The easiest way to answer this MAY BE to convert to Cartesian coordinates.

r=9 cos θ

x=9 cos²θ

r² = 81 cos²θ = x² + y² = 9x

Complete the square: (x-4.5)² + y² = 4.5²

Similarly for the other circle

x² + (y-4.5)² = 4.5²

Now convert the implicit equations to function form and use the the positive sign on the radical on the circle centered on the x axis and the negative sign on the radical for the circle centered on the y axis.

Integrate the difference of the 2 functions from 0 to 4.5. .

There may be a way to do this directly in polar co-ordinates, but I don't see it right off.