Use a double integral to find the area of the smaller region bounded by the spiral rtheta=1, the circles r=1 and r=3, and the polar axis. I just need help with setting up the integral.
To find the area of the smaller region bounded by the given curves, we can set up a double integral in polar coordinates.
First, let's understand the region we are working with. We have a spiral curve r(θ) = 1, two circles centered at the origin with radii r = 1 and r = 3, and the polar axis (θ = 0).
To set up the integral, we need to consider the boundaries for both r and θ. Let's start with θ.
Since the spiral curve r(θ) = 1 is not defined for all θ, we need to find the values of θ where the spiral intersects the two circles. To do this, we set r(θ) = 1 equal to the equations of the circles: r = 1, and r = 3.
For r = 1, we have:
1 = 1
This equation is satisfied for all values of θ, so there are no intersections and we don't need to include this value in our integral.
For r = 3, we have:
1 = 3
This equation has no solution within our given region, so we can disregard it as well.
Now let's consider the boundaries for r. The inner circle has r = 1, and the outer circle has r = 3. This means our integral for the region will go from r = 1 to r = 3.
For the boundaries of θ, we need to find the angle where the spiral curve intersects the polar axis. The spiral intersects the polar axis at θ = 0 and θ = 2π.
Therefore, the double integral to find the area of the smaller region is:
∬(over the region) dA = ∫(θ=0 to θ=2π) ∫(r=1 to r=3) r dr dθ
In this double integral, r represents the radial coordinate and θ represents the angular coordinate. The "dA" represents the differential area element, which is equal to r dr dθ in polar coordinates.
Remember to evaluate this double integral to find the actual area of the region.