Quick question about finding area for polar coordinates

[Bigworldjust]

Homework Statement

Find the area of the shaded region.
r=sqrt(θ)

Homework Equations

A = integral from a to b 1/2r^2dθ

The Attempt at a Solution

I know how to solve the question, I just don't know what to use for a and b. I tried 0 and 2pi but I am getting the wrong answer. Here is a picture:

http://www.webassign.net/mapleplots/a/8/0c017c152b2f6bb8c6df656e3a1bc2.gif

Thank you for the help. I just need to know what parameters to use for a and b :).

 

[Curious3141]

Homework Statement

Find the area of the shaded region.
r=sqrt(θ)

Homework Equations

A = integral from a to b 1/2r^2dθ

The Attempt at a Solution

I know how to solve the question, I just don't know what to use for a and b. I tried 0 and 2pi but I am getting the wrong answer. Here is a picture:

http://www.webassign.net/mapleplots/a/8/0c017c152b2f6bb8c6df656e3a1bc2.gif

Thank you for the help. I just need to know what parameters to use for a and b :).

Which quadrant does the shaded area reside in?

Which values of \theta define that quadrant?

 

[SammyS]

If you are to find the area of the shaded region, the values of θ bounding the region look quite obvious.

You will also need to know r as a function of θ. It appears that r is directly proportional to θ. The constant of proportionality should be given in your problem.

 

[andrien]

can you tell the answer because it does not seem very hard

 

[Curious3141]

If you are to find the area of the shaded region, the values of θ bounding the region look quite obvious.

You will also need to know r as a function of θ. It appears that r is directly proportional to θ. The constant of proportionality should be given in your problem.

Actually, his original problem statement gives r = \sqrt{\theta}, which makes the integration even easier.

 

[SammyS]

Actually, his original problem statement gives r = \sqrt{\theta}, which makes the integration even easier.

OOPS!

I missed that. r=sqrt(θ) probably would have been more recognizable with a radical or fractional exponent.

Yes, that makes things even easier!

I think we've given all the clues we should until we hear back from OP .

 

[Bigworldjust]

Haha, wow sorry about that guys. It was late last night and I was trying to study so I was over thinking everything. Got the answer now, thanks :P.

 

[DryRun]

Yeah, i think the answer should be: \frac{1}{2}\left[(\ln (2\pi)-\ln (\frac{3\pi}{2})\right]

 

[Bigworldjust]

Yeah, i think the answer should be: \frac{1}{2}\left[(\ln (2\pi)-\ln (\frac{3\pi}{2})\right]

Yeah, that was it, lol. Thanks.

 

[ehild]

Yeah, i think the answer should be: \frac{1}{2}\left[(\ln (2\pi)-\ln (\frac{3\pi}{2})\right]

Are you sure?

ehild

 

[Bigworldjust]

Are you sure?

ehild

Regardless, I got the answer, lol. It was form 3pi/2 to 2pi. Sorry about asking such a dumb question guys, hah.

 

[DryRun]

Are you sure?

ehild

Well, the integration gives:
A=\frac{1}{2}\left[(\ln \theta)\right] ^{\theta=2\pi}_{\theta=\frac{3\pi}{2}}=\frac{1}{2}\left[(\ln (2\pi)-\ln (\frac{3\pi}{2})\right]
So, to answer your question, i am sure. Thanks for asking.

 

[skiller]

Well, the integration gives:
A=\frac{1}{2}\left[(\ln \theta)\right] ^{\theta=2\pi}_{\theta=\frac{3\pi}{2}}=\frac{1}{2}\left[(\ln (2\pi)-\ln (\frac{3\pi}{2})\right]
So, to answer your question, i am sure. Thanks for asking.

Where does the \ln \theta come from?

 

[DryRun]

Where does the \ln \theta come from?

A=\frac{1}{2}\int \frac{1}{\theta}\,.d\theta

 

[skiller]

A=\frac{1}{2}\int \frac{1}{\theta}\,.d\theta

Are you sure? Please explain. I fail to see where you get \frac{1}{\theta} from.

 

[DryRun]

Are you sure? Please explain. I fail to see where you get \frac{1}{\theta} from.

Show your work and i'll tell you where you went wrong.

 

[skiller]

Show your work and i'll tell you where you went wrong.

Isn't the area equal to
\frac{1}{2}\int_\frac{3\pi}{2}^{2\pi}r^2\, d\theta

 

[DryRun]

Isn't the area equal to
\frac{1}{2}\int_\frac{3\pi}{2}^{2\pi}r^2\, d\theta

You have the integrand wrong. Check the first post for the correct integrand.

 

[SammyS]

Isn't the area equal to
\frac{1}{2}\int_\frac{3\pi}{2}^{2\pi}r^2\, d\theta

Yes, and \displaystyle r=\sqrt{\theta} so your guess is as good as mine concerning where \displaystyle \frac{1}{\theta} comes from !

 

[skiller]

Yes, and \displaystyle r=\sqrt{\theta} so your guess is as good as mine concerning where \displaystyle \frac{1}{\theta} comes from !

I think we can safely assume that 1/2r^2 has been misinterpreted as 1/(2r^2).

 

[DryRun]

Homework Statement

Find the area of the shaded region.
r=sqrt(θ)

Homework Equations

A = integral from a to b 1/2r^2dθ

The Attempt at a Solution

I know how to solve the question, I just don't know what to use for a and b. I tried 0 and 2pi but I am getting the wrong answer. Here is a picture:

http://www.webassign.net/mapleplots/a/8/0c017c152b2f6bb8c6df656e3a1bc2.gif

Thank you for the help. I just need to know what parameters to use for a and b :).

The original integral posted by the OP is:
A=\int^b_a \frac{1}{2r^2}\,.d\theta
Since the OP has confirmed that he got the answer through this interpretation...

 

[Bigworldjust]

The original integral posted by the OP is:
A=\int^b_a \frac{1}{2r^2}\,.d\theta
Since the OP has confirmed that he got the answer through this interpretation...

Oh no that is wrong, lol. Sorry about that I didn't type it out correctly, because I didn't know how to format it on this website. What the others said as the formula were correct. Sorry about the misinterpretation.

 

[ehild]

The area of the shaded region is \int_{3\pi/2}^{2\pi}{\frac{1}{2}r^2d\theta}
It is not the same as the integral of \frac{1}{2r^2} over the shaded region.

In the second case, the integral is 0.5\int_{\frac{3 \pi} {2}}^{2\pi} \int_0^{\sqrt{\theta}}\frac{1}{r}drd\theta

ehild

 

[skiller]

Show your work and i'll tell you where you went wrong.

I don't think you should be saying this sort of thing!

You have the integrand wrong.

No I haven't.

Check the first post for the correct integrand.

I don't need to; it's a commonly known integrand and even if it wasn't it's easily derived. On top of that, having double-checked the first post, it was given correctly anyway.

Yes, and \displaystyle r=\sqrt{\theta} so your guess is as good as mine concerning where \displaystyle \frac{1}{\theta} comes from !

Indeed!

The original integral posted by the OP is:
A=\int^b_a \frac{1}{2r^2}\,.d\theta

No it isn't. That's your incorrect interpretation of it.

Since the OP has confirmed that he got the answer through this interpretation...

I think the OP may have accidentally incorrectly said he came to the same answer as yours, but it was pretty obvious that they knew the integrand but was simply unsure of which upper and lower limits to take.

Sorry about that I didn't type it out correctly.

I think you typed it ok enough. As I said earlier, 1/2r^2 might be misinterpreted as 1/(2r^2) by some, but, if anything, the former is correct. (It reminds my of the "priority by juxtaposition" argument in the old "48÷2(9+3)" debate. (But don't get me started on that old chestnut!)

The area of the shaded region is \int_{3\pi/2}^{2\pi}{\frac{1}{2}r^2d\theta}
It is not the same as the integral of \frac{1}{2r^2} over the shaded region.

ehild

I think most (if not all) posting in this thread know this to be the case.

My guesses are one of the following (for "he", read "he or she"):

a) sharks is winding us up by picking up on an ambiguously written integrand (although I believe it is correctly written) and following through with that to the erroneous solution, in which case he really shouldn't be "confirming" a result that he knows is definitely wrong - especially in a Homework thread;

b) sharks believes he is being genuine, but only by the fact that he has assumed that the given integrand is correct (although he has incorrectly misinterpreted it), in which case he really shouldn't be "confirming" a result that he does not know necessarily to be correct - especially in a Homework thread;

c) sharks genuinely got it wrong by actually believing his integrand was correct whether he'd read it in the OP or not, in which case he really shouldn't be "confirming" results that he definitely cannot know to be true, as they are not - especially in a Homework thread.

I really don't know if it's a or b or c.

 

[DryRun]

I don't think you should be saying this sort of thing!

No I haven't.

I don't need to; it's a commonly known integrand and even if it wasn't it's easily derived. On top of that, having double-checked the first post, it was given correctly anyway.

Indeed!

No it isn't. That's your incorrect interpretation of it.

I think the OP may have accidentally incorrectly said he came to the same answer as yours, but it was pretty obvious that they knew the integrand but was simply unsure of which upper and lower limits to take.

I think you typed it ok enough. As I said earlier, 1/2r^2 might be misinterpreted as 1/(2r^2) by some, but, if anything, the former is correct. (It reminds my of the "priority by juxtaposition" argument in the old "48÷2(9+3)" debate. (But don't get me started on that old chestnut!)

I think most (if not all) posting in this thread know this to be the case.

My guesses are one of the following (for "he", read "he or she"):

a) sharks is winding us up by picking up on an ambiguously written integrand (although I believe it is correctly written) and following through with that to the erroneous solution, in which case he really shouldn't be "confirming" a result that he knows is definitely wrong - especially in a Homework thread;

b) sharks believes he is being genuine, but only by the fact that he has assumed that the given integrand is correct (although he has incorrectly misinterpreted it), in which case he really shouldn't be "confirming" a result that he does not know necessarily to be correct - especially in a Homework thread;

c) sharks genuinely got it wrong by actually believing his integrand was correct whether he'd read it in the OP or not, in which case he really shouldn't be "confirming" results that he definitely cannot know to be true, as they are not - especially in a Homework thread.

I really don't know if it's a or b or c.

Correct answer: d) You really have too much free time and obviously enjoy flaming (read the forum rules about this).

As far as I'm concerned, the OP didn't write the integral correctly (as he admitted) and anyone could have made the honest mistake of misinterpreting the integrand. If i wasn't sure, i would not have replied, and even if i did, i would have made it clear that i wasn't sure of my suggestion. This is a prime example of why I've grown tired of suggesting to use LaTeX for more clarity, especially when expressions are involved.

Show your work and i'll tell you where you went wrong.

I only said that for your own sake, as i thought you didn't know, and as per the forum rules, other members are not allowed to do your homework (by giving full answers), but only help and guide through hints/suggestions, and homework helpers are always reminding members about showing their work/efforts. There's nothing more to it.

 

[skiller]

As far as I'm concerned, the OP didn't write the integral correctly (as he admitted) and anyone could have made the honest mistake of misinterpreting the integrand. If i wasn't sure, i would not have replied, and even if i did, i would have made it clear that i wasn't sure of my suggestion. This is a prime example of why I've grown tired of suggesting to use LaTeX for more clarity, especially when expressions are involved.

I agree about using LaTeX for clarity, and I could tell from the fact that you used LaTeX so fluently and from the other posts I've seen from you on this site that you obviously are far from a dunce when it comes to maths. But at the time I saw your posts, you did come across as someone who genuinely didn't get the answer right.

I only said that for your own sake, as i thought you didn't know, and as per the forum rules, other members are not allowed to do your homework (by giving full answers), but only help and guide through hints/suggestions, and homework helpers are always reminding members about showing their work/efforts. There's nothing more to it.

Ditto, I thought you didn't know. People who may know you from this site may well have known what you were doing, but I didn't. Newcomers to that thread would I dare say think the same.

It's one thing to give hints (that's why my initial replies to you were very short one-liners - I was trying not to give the full answer to you!), but to give an incorrect answer on the basis that you think the provided information was written incorrectly is quite another. (And I still stand by what 1/2r^2 means - even if you think it's ambiguous, you should say so.)

I was only trying to correct you as I thought it looked bad to have an incorrect answer from you posted in someone's Homework question. There was no "flaming" intention on my part. Sorry if I gave that impression.

 

[DryRun]

I agree about using LaTeX for clarity, and I could tell from the fact that you used LaTeX so fluently and from the other posts I've seen from you on this site that you obviously are far from a dunce when it comes to maths. But at the time I saw your posts, you did come across as someone who genuinely didn't get the answer right.

Ditto, I thought you didn't know. People who may know you from this site may well have known what you were doing, but I didn't. Newcomers to that thread would I dare say think the same.

It's one thing to give hints (that's why my initial replies to you were very short one-liners - I was trying not to give the full answer to you!), but to give an incorrect answer on the basis that you think the provided information was written incorrectly is quite another. (And I still stand by what 1/2r^2 means - even if you think it's ambiguous, you should say so.)

I was only trying to correct you as I thought it looked bad to have an incorrect answer from you posted in someone's Homework question. There was no "flaming" intention on my part. Sorry if I gave that impression.

I help whenever i can. It's my way of giving back to the community for the help that I've received, which I'm very grateful for.

Too many times, I've seen people posting in a hurry, with no LaTeX formatting and poor writings of any expression involved. On top of that, they're usually impatient (i assume from having tried several times and failed at reaching an agreeable solution) so they're liable to omit certain little details, like parentheses. So, even though the order of operations does dictate that the expression should be interpreted as \frac{1}{2}r^2, i (wrongly) assumed that the OP had skipped the parentheses on purpose. For those who aren't comfortable with LaTeX, parentheses can be crucial in delivering the right question, especially since most people will concentrate on the actual problem, rather than verify if the problem itself has been typed correctly. Example: For a newbie/lazy/careless member posting this: 1/2r^2 could have been 1/(2r^2) or 1/2(r^2). If the OP had used the latter, then the problem would have been really obvious at first sight.

Unless everyone starts using LaTeX (or correct use of parentheses) to post their homework, it can be fully expected that this will not be the last time that a "misinterpretation" of an expression in the original problem is at the heart of a wrong suggestion by a homework helper.

 

[skiller]

I help whenever i can. It's my way of giving back to the community for the help that I've received, which I'm very grateful for.

Me too; we're obviously on the same page!

Too many times, I've seen people posting in a hurry, with no LaTeX formatting and poor writings of any expression involved.

That's to be expected, but you can usually interpret what they mean with a quick question or two. Not everyone knows LaTeX.

On top of that, they're usually impatient (i assume from having tried several times and failed at reaching an agreeable solution) so they're liable to omit certain little details, like parentheses. So, even though the order of operations does dictate that the expression should be interpreted as \frac{1}{2}r^2, i (wrongly) assumed that the OP had skipped the parentheses on purpose.

That was, with respect, your mistake to assume they'd got it wrong, when in fact, they'd got it right.

For a newbie/lazy/careless member posting this: 1/2r^2 could have been 1/(2r^2) or 1/2(r^2). If the OP had used the latter, then the problem would have been really obvious at first sight.

To be fair, it was obvious to probably most people.

Unless everyone starts using LaTeX (or correct use of parentheses) to post their homework, it can be fully expected that this will not be the last time that a "misinterpretation" of an expression in the original problem is at the heart of a wrong suggestion by a homework helper.

You need to think about the person posing the question and use your intelligence to suggest what the question is, if it isn't obvious. It isn't that difficult usually.

 

[Bigworldjust]

Lol, honestly I am new to this forum, so I wasn't even familiar with LaTeX :P. Next time, I'll be sure to use it tho.

 

[DryRun]

Lol, honestly I am new to this forum, so I wasn't even familiar with LaTeX :P. Next time, I'll be sure to use it tho.

Heh. At least something positive came out of it, and it'll get oay off my back.

 

[skiller]

Lol, honestly I am new to this forum, so I wasn't even familiar with LaTeX :P. Next time, I'll be sure to use it tho.

Hi BWJ,
Even though LaTeX is great for writing out maths, it certainly wasn't necessary for your question.

Did you get your answer to be...

Spoiler

\frac{7}{16}\pi^2

 

[skiller]

Heh. At least something positive came out of it, and it'll get oay off my back.

Off my back? Have I offended you?

 

[Bigworldjust]

Hi BWJ,
Even though LaTeX is great for writing out maths, it certainly wasn't necessary for your question.

Did you get your answer to be...

Spoiler

\frac{7}{16}\pi^2

Yeah that's the answer I got :D. Thanks for the help everyone, didn't mean to spark such a debate, hah.

 

[skiller]

Yeah that's the answer I got :D. Thanks for the help everyone, didn't mean to spark such a debate, hah.

No probs. Mathematicians always argue. But it's always me who is right.