Parametrics question

**Mathstud28** · June 20th 2008, 08:41 PM

Hello everyone, here is my question. Now my brain may be muddled after thirty two hours without sleep, but

Consider for a second

$I=\int_a^{b}\sqrt{r^2-x^2}$

$I$ is describing the area under a portion of a semi-circle of radius $r$ .

For sake of numerics but not straying completely away from calculus, instead of doing a nasty trig sub, would it be correct/easier to say

$[a,b]$ describes $\frac{1}{n}$ of this semicircle, where $n$ is obviously to be determined case wise.

So if we think about it we can parametrize our circle in terms

$x(\theta)=r\cos(\theta)$

and $y(\theta)=r\sin(\theta)$

So could we not say that

$I=\int_a^{b}\sqrt{r^2-x^2}dx=r^2\int_0^{\frac{\pi}{n}}\cos^2(\theta)d\th eta$ ?

Which would be a much easier calculation. Is this correct in its assumption?

Once again, if this is soo apparently obvious sorry for posting, but if anyone could give me a feasible reason why this is not easier but at the same time using calculus, please do tell.

Mathstud

**mr fantastic** · June 20th 2008, 09:04 PM

Originally Posted by Mathstud28

Hello everyone, here is my question. Now my brain may be muddled after thirty two hours without sleep, but

Consider for a second

$I=\int_a^{b}\sqrt{r^2-x^2}$

$I$ is describing the area under a portion of a semi-circle of radius $r$ .

For sake of numerics but not straying completely away from calculus, instead of doing a nasty trig sub, would it be correct/easier to say

$[a,b]$ describes $\frac{1}{n}$ of this semicircle, where $n$ is obviously to be determined case wise.

So if we think about it we can parametrize our circle in terms

$x(\theta)=r\cos(\theta)$

and $y(\theta)=r\sin(\theta)$

So could we not say that

$I=\int_a^{b}\sqrt{r^2-x^2}dx=r^2\int_0^{\frac{\pi}{n}}\cos^2(\theta)d\th eta$ ?

Which would be a much easier calculation. Is this correct in its assumption?

Once again, if this is soo apparently obvious sorry for posting, but if anyone could give me a feasible reason why this is not easier but at the same time using calculus, please do tell.

Mathstud

The simplest way to check would be to perform both integrations. But how do you propose finding the value of n (assuming that the previous analysis supports your hypothesis)?

My feeling is that it doesn't work, simply because of the region you're effectively doing a double integration over before switching to polars ......

**topsquark** · June 21st 2008, 03:48 AM

Originally Posted by Mathstud28

Hello everyone, here is my question. Now my brain may be muddled after thirty two hours without sleep, but

Consider for a second

$I=\int_a^{b}\sqrt{r^2-x^2}$

$I$ is describing the area under a portion of a semi-circle of radius $r$ .

For sake of numerics but not straying completely away from calculus, instead of doing a nasty trig sub, would it be correct/easier to say

$[a,b]$ describes $\frac{1}{n}$ of this semicircle, where $n$ is obviously to be determined case wise.

So if we think about it we can parametrize our circle in terms

$x(\theta)=r\cos(\theta)$

and $y(\theta)=r\sin(\theta)$

So could we not say that

$I=\int_a^{b}\sqrt{r^2-x^2}dx=r^2\int_0^{\frac{\pi}{n}}\cos^2(\theta)d\th eta$ ?

Which would be a much easier calculation. Is this correct in its assumption?

Once again, if this is soo apparently obvious sorry for posting, but if anyone could give me a feasible reason why this is not easier but at the same time using calculus, please do tell.

Mathstud

I don't see any reason it shouldn't work, except for a little problem in how you put in your substitution: You didn't take the square root. When you do this you get exactly the same expression you would get if you plugged $x = r~sin(\theta)$ into the integral.

-Dan

Edit: There is also a problem with your integration limits. The integral finds the area between the curve and the x-axis. Your limits would suggest that you are trying to find the area of a sector (wedge) of the circle. These will obviously not have the same area.

**mathwizard** · June 21st 2008, 01:40 PM

Originally Posted by Mathstud28

Hello everyone, here is my question. Now my brain may be muddled after thirty two hours without sleep, but

Consider for a second

$I=\int_a^{b}\sqrt{r^2-x^2}$

$I$ is describing the area under a portion of a semi-circle of radius $r$ .

For sake of numerics but not straying completely away from calculus, instead of doing a nasty trig sub, would it be correct/easier to say

$[a,b]$ describes $\frac{1}{n}$ of this semicircle, where $n$ is obviously to be determined case wise.

So if we think about it we can parametrize our circle in terms

$x(\theta)=r\cos(\theta)$

and $y(\theta)=r\sin(\theta)$

So could we not say that

$I=\int_a^{b}\sqrt{r^2-x^2}dx=r^2\int_0^{\frac{\pi}{n}}\cos^2(\theta)d\th eta$ ?

Which would be a much easier calculation. Is this correct in its assumption?

Once again, if this is soo apparently obvious sorry for posting, but if anyone could give me a feasible reason why this is not easier but at the same time using calculus, please do tell.

Mathstud

I don't think I understood what you are trying to say. Are you trying to determine n in terms of a and b... or what?

btw, there are some errors in your post; the integrand after the parametrization should be just cosine theta without the square; Also, you have $x = b = r cos \theta$ or $\theta = arccos(b/r)$ . Similarily for the lower limit. I don't see how you've got pi/n and 0 for your limits of the new integral.

Please elaborate..

PS: It's really annoying to keep typing [math*] and then [/math*] and then type those silly words again... can we suggest this forum to invent a new ... ( (huh???

) , what do you call this?)

**bobak** · June 21st 2008, 01:55 PM

You may hate me for this, but have you consider using geometry for this ?

Supposing the case that a is negative and b is positive, then you know that the area of a sector is $1/2 r^2 \theta$ so in this case $\theta = \pi - \arccos a/r - \arccos b/r$

Then add the areas of the triangles.

Bobak

**topsquark** · June 21st 2008, 04:48 PM

It's really annoying to keep typing [math*] and then [/math*] and then type those silly words again.

When you open a new post window you will see a toolbar just above the text box. On the second line start from the right and go over three tools. You will see a $\Sigma$ . Highlight the Latex code you want inside the tags and click the $\Sigma$ . This will insert both tags for you.

-Dan

**topsquark** · June 21st 2008, 04:51 PM

Originally Posted by bobak

You may hate me for this, but have you consider using geometry for this ?

Supposing the case that a is negative and b is positive, then you know that the area of a sector is $1/2 r^2 \theta$ so in this case $\theta = \pi - \arccos a/r - \arccos b/r$

Then add the areas of the triangles.

Bobak

Am I missing something here? The area represented by the integral looks something like the picture below. It is NOT a sector of the circle!

-Dan

**mathwizard** · June 21st 2008, 06:20 PM

Originally Posted by topsquark

When you open a new post window you will see a toolbar just above the text box. On the second line start from the right and go over three tools. You will see a $\Sigma$ . Highlight the Latex code you want inside the tags and click the $\Sigma$ . This will insert both tags for you.

-Dan

Cool. I didn't know that; it would save me a lot of typing! Thanks.

**mr fantastic** · June 21st 2008, 06:41 PM

Originally Posted by topsquark

Am I missing something here? The area represented by the integral looks something like the picture below. It is NOT a sector of the circle!

-Dan

My point exactly ...... (see post #2)

**Mathstud28** · June 21st 2008, 09:39 PM

I have been thinking about this, and you guys are correct. I do not know what I was thinking when I made this.

This trick would be useful for two cases

Let $I=\int_a^{b}\sqrt{r^2-x^2}dx$

And then $I_1=\int_{-r}^{r}\sqrt{r^2-x^2}dx$

Obviously the discussed " $n$ " here would be one therefore

$I_1=\int_{-r}^{r}\sqrt{r^2-x^2}dx=r^2\int_0^{\pi}\cos^2(\theta)d\theta=\frac{ r^2}{2}\bigg[\theta+\frac{1}{2}\sin(2\theta)\bigg]\bigg|_{0}^{\pi}=\frac{1}{2}\pi{r^2}$

Which I suppose would be an easy way to find this integral without falling back on gruesome geometry and seeing this a semi-circle

The other would be

$I_2=\int_0^{r}\sqrt{r^2-x^2}dx=\int_{-r}^{0}\sqrt{r^2-x^2}dx$

Here it is obvious that $n=2$

So we would have

$I_2=r^2\int_0^{\frac{\pi}{2}}\cos^2(\theta)d\theta =\frac{r^2}{2}\bigg[\theta+\frac{1}{2}\sin(2\theta)\bigg]\bigg|_{0}^{\frac{\pi}{2}}=\frac{1}{4}\pi{r^2}$

Which would be a calculus way of finding the area of a quarter-circle.

The main failing of my attempt would be to find n, except for special cases you need to already know the area

. Not much use integrating then, yeah?

The only other use I could see would be that given a question worded as such

" Find the area of one-(whatevereth) of a circle"

You could use this since

If one were asked to calculate the area of $\frac{1}{52}$ of a circle then

$I_{52}=r^2\int_0^{\frac{\pi}{52}}\cos^2(\theta)d\t heta$

Otherwise, this is useless, thanks for everyone's reply.

Mathstud

**Mathstud28** · June 21st 2008, 10:07 PM

Originally Posted by mathwizard

btw, there are some errors in your post; the integrand after the parametrization should be just cosine theta without the square;

What are you talking about?

Since the orientation of the parameterization is counterclockwise

$A=\int_a^{b}x(\theta)y'(\theta)d\theta$

So it would be $\cos^2(\theta)$

Also, you have

or

Where did you get this b stuff from?

I don't see how you've got pi/n and 0 for your limits of the new integral.

As for the limits of integration, how is it not that? The zero should be obvious, the $\frac{\pi}{n}$ would come from the amout of $\theta$ needed to complete $\frac{1}{n}$ th of the circle?

So...am I missing something? I appreciate that you care enough to point out what you believe are errors, but you should probably double check. But don't fret! I have been guilty of this many-a-time.

**mathwizard** · June 21st 2008, 10:15 PM

Originally Posted by Mathstud28

What are you talking about?

Since the orientation of the parameterization is counterclockwise

$A=\int_a^{b}x(\theta)y'(\theta)d\theta$

So it would be $\cos^2(\theta)$

Where did you get this b stuff from?

As for the limits of integration, how is it not that? The zero should be obvious, the $\frac{\pi}{n}$ would come from the amout of $\theta$ needed to complete $\frac{1}{n}$ th of the circle?

So...am I missing something? I appreciate that you care enough to point out what you believe are errors, but you should probably double check. But don't fret! I have been guilty of this many-a-time.

Ah, I'm sorry; that post was nonsense...

My brain was fried at that time and I didn't understand what you meant in your post (English is not my native, btw);

Umm, it's 00:20 am right here... I'll come back to it tomorrow...

**bobak** · June 21st 2008, 11:00 PM

Originally Posted by topsquark

Am I missing something here? The area represented by the integral looks something like the picture below. It is NOT a sector of the circle!

-Dan

True but you can spilt it up into a sector and two triangles, refer to my attachment. To get the area of a sector you need to know the angle suspending it. to figure out the angles between the triangles and the x-axis you use just inverse trig ( $\arccos{ \frac{a}{r}}$ etc... ) the add the area of the triangles.

I am fairly confident this problem can be solved using geometry alone.

Bobak

**mathwizard** · June 22nd 2008, 03:10 PM

Originally Posted by Mathstud28

Hello everyone, here is my question. Now my brain may be muddled after thirty two hours without sleep, but

Consider for a second

$I=\int_a^{b}\sqrt{r^2-x^2}$

$I$ is describing the area under a portion of a semi-circle of radius $r$ .

For sake of numerics but not straying completely away from calculus, instead of doing a nasty trig sub, would it be correct/easier to say

$[a,b]$ describes $\frac{1}{n}$ of this semicircle, where $n$ is obviously to be determined case wise.

So if we think about it we can parametrize our circle in terms

$x(\theta)=r\cos(\theta)$

and $y(\theta)=r\sin(\theta)$

So could we not say that

$I=\int_a^{b}\sqrt{r^2-x^2}dx=r^2\int_0^{\frac{\pi}{n}}\cos^2(\theta)d\th eta$ ?

Which would be a much easier calculation. Is this correct in its assumption?

Once again, if this is soo apparently obvious sorry for posting, but if anyone could give me a feasible reason why this is not easier but at the same time using calculus, please do tell.

Mathstud

I apologize if am getting annoying, but I'm afraid that I still do not completely understand what you are trying to use 1/n of the area of the semi-circle and stuff.

What you are using is basically polar coordinates integration given by

$ \int \int r dr d\theta = \int 1/2 r^2 d\theta $
where the limits of the inner integral is from 0 to r and the limits of the outer integral is from 0 to pi for your area of the semicircle.

For your original integral

$I=\int_a^{b}\sqrt{r^2-x^2}dx$

if you are using the substitution $x = r cos \theta$ and $dx = -r sin \theta d\theta$ ,

then your integral would become

$\int -r^2 sin^2 \theta d\theta$

running from $\theta = arccos(a/r)$ to $\theta = arccos(b/r)..$
[tex]I=\int_a^{b}\sqrt{r^2-x^2}dx=r^2\int_0^{\frac{\pi}{n}}\cos^2(\theta)d\th eta

In your original post you said that
$r^2 \int_0^{\frac{\pi}{n}}\cos^2(\theta)d\theta$

I suspect you are actually using

$\int \sqrt{r^2 - y^2} dy$ with the substitution $y = r sin \theta$

am I correct?

In that case your variable is $\theta$ and (you claimed) the upper limit of the integral is $\pi /n$ in your post...

however, from the limits of the original integral you have x = a for lower and x = b for upper limit. Now, the problem is that you are actually integrating with respect to y if my reasoning on your derivation of

$r^2 \int cos^2 \theta d\theta $

is correct; if the limits of the integral of $\sqrt{r^2 - y^2} dy$ is c to d, then we have

$ y = d = r sin \theta \implies \theta = arcsin(d/r) $
&

$y = c = r sin \theta \implies \theta = arcsin(c/r).$

Do you think $arcsin(d/r) = \pi /n$ & $arcsin(c/r) = 0$ ?

**Mathstud28** · June 22nd 2008, 03:52 PM

Originally Posted by mathwizard

I apologize if am getting annoying, but I'm afraid that I still do not completely understand what you are trying to use 1/n of the area of the semi-circle and stuff.

What you are using is basically polar coordinates integration given by

$ \int \int r dr d\theta = \int 1/2 r^2 d\theta $
where the limits of the inner integral is from 0 to r and the limits of the outer integral is from 0 to pi for your area of the semicircle.

For your original integral

$I=\int_a^{b}\sqrt{r^2-x^2}dx$

if you are using the substitution $x = r cos \theta$ and $dx = -r sin \theta d\theta$ ,

then your integral would become

$\int -r^2 sin^2 \theta d\theta$

running from $\theta = arccos(a/r)$ to $\theta = arccos(b/r)..$
$I=\int_a^{b}\sqrt{r^2-x^2}dx=r^2\int_0^{\frac{\pi}{n}}\cos^2(\theta)d\th eta In your original post you said that <html> <head> <style type=$ img.top {vertical-align:15%;} $r^2 \int_0^{\frac{\pi}{n}}\cos^2(\theta)d\theta$

I suspect you are actually using

$\int \sqrt{r^2 - y^2} dy$ with the substitution $y = r sin \theta$

am I correct?

In that case your variable is $\theta$ and (you claimed) the upper limit of the integral is $\pi /n$ in your post...

however, from the limits of the original integral you have x = a for lower and x = b for upper limit. Now, the problem is that you are actually integrating with respect to y if my reasoning on your derivation of

$r^2 \int cos^2 \theta d\theta $

is correct; if the limits of the integral of $\sqrt{r^2 - y^2} dy$ is c to d, then we have

$ y = d = r sin \theta \implies \theta = arcsin(d/r) $
&

$y = c = r sin \theta \implies \theta = arcsin(c/r).$

Do you think $arcsin(d/r) = \pi /n$ & $arcsin(c/r) = 0$ ?" alt="I=\int_a^{b}\sqrt{r^2-x^2}dx=r^2\int_0^{\frac{\pi}{n}}\cos^2(\theta)d\th eta

In your original post you said that
$r^2 \int_0^{\frac{\pi}{n}}\cos^2(\theta)d\theta$

I suspect you are actually using

$\int \sqrt{r^2 - y^2} dy$ with the substitution $y = r sin \theta$

am I correct?

In that case your variable is $\theta$ and (you claimed) the upper limit of the integral is $\pi /n$ in your post...

however, from the limits of the original integral you have x = a for lower and x = b for upper limit. Now, the problem is that you are actually integrating with respect to y if my reasoning on your derivation of

$r^2 \int cos^2 \theta d\theta $

is correct; if the limits of the integral of $\sqrt{r^2 - y^2} dy$ is c to d, then we have

$ y = d = r sin \theta \implies \theta = arcsin(d/r) $
&

$y = c = r sin \theta \implies \theta = arcsin(c/r).$

Do you think $arcsin(d/r) = \pi /n$ & $arcsin(c/r) = 0$ ?" />

Ok, here is what I am doing. Imagine this

I have an equation

$x^2+y^2=r^2$

This describes a circle with radius $r$

If we parametrize this by

$x=r\cos(\theta)$

and

$y=r\sin(\theta)$

We see that on $[0,2\pi]$ one entire trace of the circe is drawn.

On $[0,\pi]$ one half of the circle is drawn

On $\left[0,\frac{\pi}{2}\right]$ a quarter of the circle is drawn

Now lets concentrate on the semi-circle part

We can see that for $\frac{1}{n}$ th of the semi-circle it takes

$\left[0,\frac{\pi}{n}\right]$ to complete it.

Now we note that the orientation of this parametrization is counterclockwise.

We then note that are of a counterclockwise parametrized curve is given by

$A=\int_a^{b}x(t)y'(t)dt$

So seeing that
$y'(\theta)=r\cos(\theta)$

We have that

$A_{n}=\int_0^{\frac{\pi}{n}}(r\cos(\theta))(r\sin( \theta))d\theta=r^2\int_0^{\frac{\pi}{n}}\cos^2(\t heta)d\theta$

Where $A_n$ denotes the area of $\frac{1}{n}$ th of the circle.

I hope this is clear now.

I have never seen this, so if you do have a grievance please let me know.

And as was stated eariler, the obvious downfall is that $\frac{1}{n}$ th of the circle is measured by Area so we need to know the area to find [/tex]n[/tex]. So unless a specific case such as a quarter or semi circle is encountered, or if someone stated a question in a way such as "Compute the area of one-___th ofa circle" it is pretty much useless.

Mathstud.

P.S. Don't ever worry about annoying someone, for most of the time it is the best way to learn.

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