what is the polar area between r=5sinx and r=5cosx
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.
To find the polar area between the curves r=5sin(x) and r=5cos(x), you need to determine the region enclosed by the two curves.
First, let's plot the curves on the polar coordinate system. The equation r=5sin(x) describes a curve that resembles a "cardioid" shape, while the equation r=5cos(x) represents a circle.
To find the points of intersection between the two curves, we can set them equal to each other and solve for x:
5sin(x) = 5cos(x)
Divide both sides by 5 to simplify the equation:
sin(x) = cos(x)
Now, using the identity sin(x) = cos(x - π/4), we can rewrite the equation as:
cos(x - π/4) = cos(x)
For two cosine functions to be equal, their arguments must be equal:
x - π/4 = x
Solving for x, we find:
π/4 = 0
This result is not possible, which means the two curves do not intersect. Thus, we cannot determine an enclosed region between the two curves, and the polar area is equal to zero.