Make a sketch to find that it is actually the intersection of two circles.
r=5sin(x) is tangent to the x-axis at the origin, and sitting in quadrants 1 & 2.
r=5cos(x) is tangent to the y-axis at the origin, and sitting in quadrants 4 & 1.
As you can see, it is symmetric about y=x, or θ=π/4.
Integrating between 0 to r of r=5sin(x) from 0 to π4, and between π/4 to &pi/2 of r=5cos(x) will give the required area.
∫&int dA
π/4 5sin(θ)
= &int ∫ rdrdθ
0 0
= ∫ [r²/2] dθ
= ∫ 25sin²(θ)/2 dθ
= (25/4) ∫ (1-cos(2θ))dθ
= (25/4) [θ -(1/2)sin(2θ))] 0 to π/4
= (25/4)(π/4 - (1/2)]
= (25/16)(π - 2)
By symmetry, the other half from π/4 to π/2 has the same area.
So total area enclosed
= (25/8)(π-2)
= 3.57 approx.
what is the polar area between r=5sinx and r=5cosx
1 answer