Area inside r=6sin(theta) but outside r=3

In summary, you need to find the area of a region inside of a circle, but outside of a circle that has a radius of 3. The x-values are found where the two curves intersect and the area is calculated using the formula: 1/2 Integral from B to A of (f2(theta)^2 - f1(theta)^2). However, because r=3 is inside the circle, the area cannot be calculated. You need to find the area in the first quadrant and double it to get the entire area. Finally, because the outer curve is r=6 sin(theta) and the inner curve is r=3, you can use these functions to find b
  • #1
RKOwens4
33
0

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.
 
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  • #2
Anyone?
 
  • #3
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.

What's with all the decimals? Leave things in terms of pi. I don't believe those are x values. This is a polar coordinate problem, in terms of r and theta. You might start by stating the correct formula for the area between two polar curves expressed in terms of polar coordinates.
 
  • #4
Okay, I found the formula for the area between two polar curves, which is:

1/2 Integral from B to A of (f2(theta)^2 - f1(theta)^2)

But what do I do about the r=3? And how do I find out what B and A are?
 
  • #5
The two polar curves are symmetric about the y-axis, so you can find the area in the first quadrant and double it to get the entire area.

For b, you can use pi/2. For a, find the point of intersection in quadrant 1 of the two curves - use the exact value, not a decimal approximation.

For the outer curve, r = 6 sin(theta). For the inner curve, r = 3. These are the functions you're calling f1 and f2 (not necessarily in order).

This problem is very similar to the other problem you posted.
 

FAQ: Area inside r=6sin(theta) but outside r=3

What is the meaning of r=6sin(theta) and r=3 in this context?

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.

What is the area inside r=6sin(theta) but outside r=3?

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.

How is this area calculated?

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.

What is the significance of the value of r=6sin(theta) and r=3 in this context?

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.

Can this equation be applied to other shapes or regions?

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.

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