RKOwens4 said:Homework Statement
The title really says it all. Find the area of the region inside r=6sin(theta) but outside r=3.
Homework Equations
The Attempt at a Solution
I first found the x values where the two curves intersect and came up with .52359877 and 2.61799388. I then integrated 6sin(theta) (or, -6cos(theta)) from .52359877 to 2.61799388. The answer I got was 10.3923, but that's incorrect.
In polar coordinates, r represents the distance from the origin and theta represents the angle. Therefore, r=6sin(theta) is a polar equation that describes a circle with a radius of 6 units centered at the origin, and r=3 describes a smaller circle with a radius of 3 units also centered at the origin.
The area inside r=6sin(theta) but outside r=3 is the area of the annulus (or ring) formed by the two circles. This can be calculated by subtracting the area of the smaller circle from the area of the larger circle, which gives an area of 27π square units.
The area of a circle is given by the formula A=πr^2, where r is the radius. To calculate the area of the annulus, we can use the formula A=π(R^2-r^2), where R is the radius of the larger circle and r is the radius of the smaller circle. In this case, R=6 and r=3, so the area is A=π(36-9)=27π square units.
The values of r=6sin(theta) and r=3 represent the radii of the two circles that make up the annulus. These values determine the size and shape of the annulus and are important in calculating the area and other properties of the region.
Yes, this equation can be applied to other shapes or regions that can be described using polar coordinates. However, the specific values of r=6sin(theta) and r=3 will change for different shapes or regions. The concept of finding the area inside one shape but outside another can also be applied to other coordinate systems, such as Cartesian coordinates.