Area Polar Graph Homework: Find Region's Area

[_N3WTON_]

Homework Statement

Find the area of the region.
Interior of: r = 2 - sin(b)

Homework Equations

A = 1/2 ∫ (r)^2 dr

The Attempt at a Solution

I really don't have any idea how to approach this problem. I don't understand how to determine my limits of integration. The only part of the problem I have accomplished is finding that b = arcsin(2). Also, I have found that the graph of this equation makes a convex limacon.

 

[Mark44]

Homework Statement

Find the area of the region.
Interior of: r = 2 - sin(b)

Homework Equations

A = 1/2 ∫ (r)^2 dr

The Attempt at a Solution

I really don't have any idea how to approach this problem. I don't understand how to determine my limits of integration. The only part of the problem I have accomplished is finding that b = arcsin(2). Also, I have found that the graph of this equation makes a convex limacon.

You should be able to figure out the integration limits from the graph. You could integrate from b = 0 to b = $2\pi$. Alternatively, you could integrate from b = $\pi/2$ to b = b = $3\pi/2$ and double that result.

 

[_N3WTON_]

You should be able to figure out the integration limits from the graph. You could integrate from b = 0 to b = $2\pi$. Alternatively, you could integrate from b = $\pi/2$ to b = b = $3\pi/2$ and double that result.

ok so would my equation look something like ∫ [(1-sin(b))^2 db] with the limits of integration going from 0 to 2pi?

 

[vanhees71]

Looks good, provide $b$ is the polar angle. Why don't you use LaTeX for typesetting formulas? It's no big deal to type it and it makes everything much easier to read and understand!

 

[_N3WTON_]

Looks good, provide $b$ is the polar angle. Why don't you use LaTeX for typesetting formulas? It's no big deal to type it and it makes everything much easier to read and understand!

I don't know how to use latex, nor do I know where to find it...sorry

 

[_N3WTON_]

Ok, I am confused about this problem now. The website for my calculus textbook offers a 24/7 tutoring service. When I asked the tutor about this problem he said I could not integrate from 0 to 2pi; however, he was also not sure how to do the problem. So can I safely assume that he was wrong about being unable to integrate from 0 to 2pi?

 

[vanhees71]

Well, could you post the full question? I don't know, why one shouldn't integrate over the full range of the angle, because $r(b)=2-\sin b>0$ for all $b \in [0,2 \pi]$.

 

[LCKurtz]

Yes, like this:

##\frac 1 2\int_{0}^{2\pi} (2-\sin\theta)^2d\theta##

which displays like this: $\frac 1 2\int_{0}^{2\pi} (2-\sin\theta)^2d\theta$

 

[Mark44]

When I asked the tutor about this problem he said I could not integrate from 0 to 2pi; however, he was also not sure how to do the problem.

It doesn't give me a lot of confidence when someone tells me I can't do a problem a certain way, but then doesn't know how to do the problem.

So can I safely assume that he was wrong about being unable to integrate from 0 to 2pi?

If you look at the graph, it's pretty obvious that you can integrate using those limits, regardless of what the tutor is saying.

 

[_N3WTON_]

Well, could you post the full question? I don't know, why one shouldn't integrate over the full range of the angle, because $r(b)=2-\sin b>0$ for all $b \in [0,2 \pi]$.

I did post the full question.

 

[_N3WTON_]

It doesn't give me a lot of confidence when someone tells me I can't do a problem a certain way, but then doesn't know how to do the problem.

If you look at the graph, it's pretty obvious that you can integrate using those limits, regardless of what the tutor is saying.

Thank you...my first instinct was to integrate from 0 to 2pi; however, the tutor swayed me lol...