Area of Region Common to r=4sinθ & r=4cosθ

[tandoorichicken]

Find the area of the region common to [itex]r=4\sin\theta[/itex] and [itex]r=4\cos\theta[/itex]

How can I find the limits of integration? I know one of the limits is pi/4 by setting r1 = r2 but I can't find the other one because it is different for each equation. The graph shows it intersecting at (0,0), but that's a different theta for each equation.

 

[HallsofIvy]

The easy way to do this is to use the symmetry. Integrate [itex]r= 4 sin(\theta)[/itex] for [itex] \theta[/itex]= 0 to [itex]\frac{\pi}{4}[/itex], then double it.

 

[M98Ranger]

To find the limits of integration for this problem, we can use the fact that the two equations intersect at (0,0). This means that at this point, both r values are equal to 4. So, we can set r1 = r2 and solve for theta to find the common limit of integration.

For r = 4sinθ, we have 4sinθ = 4, which gives us θ = π/6.

Similarly, for r = 4cosθ, we have 4cosθ = 4, which gives us θ = 0.

Therefore, the limits of integration for the region common to both equations are θ = 0 and θ = π/6.

To find the area of this region, we can use the formula for the area of a polar region, which is A = 1/2∫r^2dθ.

Substituting in our limits of integration and the given equations, we get:

A = 1/2∫(4sinθ)^2dθ - 1/2∫(4cosθ)^2dθ

= 1/2∫16sin^2θdθ - 1/2∫16cos^2θdθ

= 1/2(8θ - 1/2sin2θ)∣∣∣π/6 0 - 1/2(8θ + 1/2sin2θ)∣∣∣0 π/6

= 1/2(4π/3 - √3) - 1/2(0 + 0)

= 2π/3 - √3

Therefore, the area of the region common to r = 4sinθ and r = 4cosθ is 2π/3 - √3 units squared.