how to find the area of polar coordinates by not using calculus

Question

Find the EXACT area contained by the graphs r=8cos theta and r=6sin theta

Answer 1

The area bounded by the graphs of r = 8cosθ [circle with radius 4, center at (4,0)] and r = 6sinθ [circle with radius 3, center at (0, 3)] is the sum of two circular segments A₁ and A₂, one from each circle.

 

To find the areas of these two circular segments, the central angles associated with those segments are needed. These can be evaluated once the angle where the graphs intersect is found.

 

The angle at which the graphs intersect is given by

8cosθ = 6sinθ ⇒ tanθ = 4/3 ⇒ θ = 53.1º or 0.927 rad.

 

Since both triangular portions are isosceles triangles, the central angles for these sectors are

For r = 8cosθ: θ₁ = 73.8º or 1.288 rad

For r = 6sinθ: θ₂ = 106.2º or 1.853 rad

 

So, the areas of the circular segments are

A₁ = ½(R₁)²(θ₁ - sinθ₁) = ½(4)²[1.288 - sin(1.288)] = 2.62

A₂ = ½(R₂)²(θ₂ - sinθ₂) = ½(3)²[1.853 - sin(1.853)] = 4.02

 

The area bounded by the graphs is 6.64.

Answer 2

This is a fairly interesting problem with several elements.

The first is to note that 8 cos(θ) - 6 sin(θ) is equal to 10 cos(θ + arctan(3/4) ) . 

This is a standard transformation that can be checked by using the cosine of a sum formula.

The arctan(3/4) is just a phase shift that has no effect on the answer because the graph of the polar

function r = 10 cos( θ + any_phase_shift) is a circle with radius of 5. 

This can be proved by looking at (x -5)² + y² with x =  10 r cos(θ) = 10 cos²(θ)  y = 10 r sin(θ) = 10 sin(θ) cos(θ)

This does work out to be 5² .

All this establishes the the graph is a circle with radius = 5.   So the area =  pi 5² =  25 pi