Area of the region

[Leons]

Refer to Figure 1, Find the area of the region lying between the inner and outer loops of limacon ?=1−2sin?.



can someone check my answer whether it is correct or not?

 

[Harry_the_cat]

Just by looking at the diagram, your final answer seems incorrect. I assume it is the pink area you are trying to find. The area of the circle with radius 1 is pi. The area you have found is quite a lot less than pi. Shouldn't it be more? Why have you selected the integration limits that you have?

 

[Dr.Peterson]

Refer to Figure 1, Find the area of the region lying between the inner and outer loops of limacon ?=1−2sin?.
View attachment 32937


can someone check my answer whether it is correct or not?


View attachment 32938

Think carefully about your integral, and you'll see that you've calculated the area of a different region of the picture than you were asked for.

How did you think you were finding the area between the loops? How can you supplement your work to do that?

 

[Leons]

Think carefully about your integral, and you'll see that you've calculated the area of a different region of the picture than you were asked for.

How did you think you were finding the area between the loops? How can you supplement your work to do that?

area of outer loop - area of inner loop= area between the loops?

 

[Dr.Peterson]

area of outer loop - area of inner loop= area between the loops?

Yes.

Which part did you calculate? How can you find the other?

(Which is not to say that the details of your work are without error as far as you went ...)

 

[Steven G]

There is only one value that you can subtract from -3sqrt(3) and get -3sqrt(3). That value is 0.
What I am saying is that -3sqrt(3) - 0 = -3sqrt(3) and no other number than 0 will work. So -3sqrt(3) - -3sqrt(3) is NOT -3sqrt(3). You know this. Just take things slower.
5 - 3 = 2, so 5 - 1 is not 2. That is 5pi - pi is not 2pi!

 

[nasi112]

Refer to Figure 1, Find the area of the region lying between the inner and outer loops of limacon ?=1−2sin?.
View attachment 32937


can someone check my answer whether it is correct or not?


View attachment 32938

You have calculated the area within the inner loop.

\(\displaystyle A_{inner} = \int_{\pi/6}^{5\pi/6} \frac{1}{2}(1 - 2\sin\theta)^2 \ d\theta\)

Now, you need to calculate the area within the outer loop.

\(\displaystyle A_{outer} = \int_{5\pi/6}^{\pi/6 + 2\pi} \frac{1}{2}(1 - 2\sin\theta)^2 \ d\theta\)

After that you can find the required area as:

\(\displaystyle A = A_{outer} - A_{inner}\)