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Homework Help: Area of a circle by integration

  1. Mar 6, 2013 #1
    My curiosity was piqued by another poster who's trying to find the area of a lune using calculus.

    I wanted to do this now. So I don't highjack his thread, I'm making a new post about it.

    First, the geometric picture of a lune that constitutes my basic plan of attack. First I want to solve a special case of a lune, where the larger circle intersects the smaller circle at the smaller circle's widest point. (However, I can't even get there. I can't find the area of a quarter or semi circle by integration. So as you will see, I'm going to end this post by asking for help with this easier problem).

    Consider the smaller circle, of radius ##r##, as being centered at the origin, and bigger circle, of radius R, offset along the y axis by -b.

    The two equations which express the height ##y## of these circles as a function of x in the first quadrant are:

    ##
    f(x) = \sqrt{r^{2}-x^{2}} \\
    g(x) = \sqrt{R^{2}-x^{2}} - b \\
    ##

    Now the area of the smaller circle above the x axis should be equal to

    ##A_{f} = 2\int_{0}^{r}f(x)dx##

    I should be able to calculate this if I can find an antiderivative for F. This turns out to be hard.

    ##f(x) = \sqrt{r^{2}-x^{2}} \\
    = \sqrt{r^{2}(1-\frac{x^2}{r^2}} \\
    = r\cdot \sqrt{1-(\frac{x}{r})^{2}} \\
    ##


    Then after having a heck of a time trying to find an antiderivative for that, I realized that I need to fish for something "pi-ish".

    Recognizing that ##r \cos \theta = x## in the first quadrant, we have that ##\cos \theta = \frac{x}{r}##

    ##
    f(\theta) = r \cdot \sqrt{1- \cos ^{2}\theta} \\
    = r \cdot \sin \theta \\
    ##

    And in retrospect, I could have just started there in polar coordinates.

    Now, considering the first quandrant in a clockwise sense, we start when

    ##cos \theta = \frac{0}{r}## and stop when ##\cos \theta = \frac{r}{r} = 1 ##. So, in radians,
    ##cos^{-1}(0) = \frac{\pi}{2} \\
    cos^{-1}{1} = 0##

    So we integrate from ##\frac{\pi}{2}## to ##0##.

    ##A = -\int_{0}^{\frac{\pi}{2}} r \sin \theta d\theta##

    So now, it seems this should be easy: i just need an antiderivative for ##r \sin \theta##. I think it's ##-r \cos \theta## by the rule for constants.

    The answer for a quarter circle should be ## -r( -\cos 0 + \cos \frac{pi}{2} ) = r##

    But this answer can't be right, because it doesn't include pi. How am I getting off track here?
     
    Last edited: Mar 6, 2013
  2. jcsd
  3. Mar 6, 2013 #2
    for the antiderivative of f(x) which is a semi circle if i understand you correctly,
    try the substitution of $$x=ru$$ so $$u = \frac {x}{r}$$
    you should be able to handle the rest
     
  4. Mar 6, 2013 #3
    I have ##f(\theta)## and I wish to integrate with respect to ##\theta##. I'm sorry: I'm not saying that your input isn't useful, but I'm not sure how to use it.
     
  5. Mar 6, 2013 #4
    The equation for your first circle is [tex]r={\sqrt{x^2+y^2}}[/tex] where r is the radius and can be treated as a constant. To find the area of a quarter of this circle you would just do [tex]∫^\frac{\pi}{2}_0rdθ[/tex]

    Edit: I'm not sure this is correct. I believe I'm missing something.
     
    Last edited by a moderator: Mar 7, 2013
  6. Mar 6, 2013 #5
    My substitution that I gave was before you did the polar equation stuff so that you would have an easier time finding the area of a semi-circle

    From what i can gather, integrating polar equations is much more complex than regular functions and uses double integrals.

    Sorry if this doesn't help much...
     
  7. Mar 6, 2013 #6
    Yes, it would be a double integral. So [tex]∫^{\frac{\pi}{2}}_0∫^r_0r(dr)(dθ)[/tex]

    That would give [tex]{\frac{1}{2}}r^2∫^{\frac{\pi}{2}}_0dθ[/tex]Which would give [tex]{\frac{1}{4}}{\pi}r^2[/tex]

    Which you can see, is 1/4 the formula we have for area of a circle. [tex]{\pi}r^2[/tex]
     
  8. Mar 6, 2013 #7

    SammyS

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    The problem in that link you gave is exactly the situation you describe above.

    The following post is from the poster of the tread in the above link.

    He/she never posted a reply in that thread but gives advice below.

    How ironic.

     
  9. Mar 6, 2013 #8
    Your problem was in your substitution, you successfully substituted for ##x## but you did not substitute for " ## dx ## ".

    If you do this you will get that: $$dx = -r \sin (\theta ) d\theta$$

    Which will give the integral: $$\int_0^{\pi / 2} r^2 \sin^2 \theta$$
    which you can solve using a double angle formula to get the result that you wanted.
     
  10. Mar 6, 2013 #9
    Wouldn't it be power reducing formula?
     
    Last edited by a moderator: Mar 7, 2013
  11. Mar 6, 2013 #10
    That's probably what you call it. Notice that the power reducing formulas are really just rewritten versions of double angle formulas - so I tend to call them all a "double angle formula" because I try to memorize as few formulas as possible.
     
  12. Mar 6, 2013 #11
    Aha I see.
     
  13. Mar 7, 2013 #12
    I know what's wrong with my formula for ## f(\theta)##: ##f(\theta) = y##, but it should equal r. I need to find an expression for r in terms of theta.
     
  14. Mar 7, 2013 #13
    R in terms of theta is sqrt(cos^2(theta)+sin^2(theta)), which is just 1, a constant.

    Mod note: I combined three of your posts into one. Please don't flood the thread with one-line posts. Use the edit function is you think of something to add after you've already posted.
     
    Last edited by a moderator: Mar 7, 2013
  15. Mar 7, 2013 #14

    vela

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    You're using f to mean different things. In your original post, f(x) is the height of the circle for a given x. Now you're thinking of r=f(θ) for the polar representation of the circle. The two f's don't represent the same thing.

    I think you're confusing yourself because you're trying to jump to the method you learned about calculating areas using polar coordinates after starting with Cartesian. As hapefish noted, your mistake in the first attempt was not doing the substitution correctly because you didn't deal with the dx in the original integral. Forget about polar coordinates for a second. Just look at it the problem as a regular u-substitution: x=r cos u.

    If you want to calculate the area using polar coordinates, start by writing down the equation for a circle about the origin in polar coordinates.
     
  16. Mar 7, 2013 #15
    There was no dx in anything i wrote. I was simply concerning myself with writing true expressions for y. My ultimate plan was to integrate one of those expressions, but as you can see, I hadn't gotten that far. I have (now) seen the examples online of integrating my original function ##y = f(x) = r \cdot \sqrt{1- (\frac{x}{r})^{2}}##

    by writing

    ##A(x) = \int_{0}^{\pi/2} f(x)dx \\
    = \int_{0}^{\pi/2} r \cdot \sqrt{1- (\frac{x}{r})^{2}} \cdot dx \\
    ##

    And performing u substitution, but none of that explains why my true expressions for y in terms of ##\theta## later don't integrate properly. The answer there appears to be that the FTC is proved in terms of the limit definition, and the limit definition in turn is applicable only to situations where we can write infinite riemann sums: i.e., cartesian coordinate systems. A different version of the FTC will be necessary for polar calculus. And yes, I am mixing up my coordinate systems in an awful way here :).
     
  17. Mar 7, 2013 #16

    vela

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    You wrote
    $$A_f = 2\int_0^r f(x)\,dx.$$ Do you really not see the dx there?
     
  18. Mar 7, 2013 #17
    Look below that, where I am back to algebra on ##f(x)##, no dx in sight.
     
  19. Mar 7, 2013 #18

    vela

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    It's like you're not even bothering to try understand what hapefish said. This is what you did:
    $$A_f = 2\int_0^r f(x)\,dx = 2\int f(r \cos \theta)\,dx = 2 \int f(r \cos \theta)\,d\theta.$$ The f(r cos θ) is what you called f(θ)=r sin θ. That was fine. It's the last step that's wrong where you simply replaced dx by dθ.
     
  20. Mar 7, 2013 #19
  21. Apr 1, 2013 #20
    Since the time I posted this and today, I've done maybe 150 more integrals, and I'm ready to tackle this.


    First the area of a circle centered on the origin is

    ##A = -4 \int_{0}^{r}\ \sqrt{r^{2}-x^{2}} dx \\
    = -4 \int_{0}^{r} r \sqrt{1 - \frac{x^{2}}{r^{2}}} dx \\
    = -4r\int_{0}^{r} \sqrt{1 - \frac{x^{2}}{r^{2}}} dx \\
    ##

    Now the trig substitution that I wasn't able to properly do last month

    ## \frac{x}{r} = \sin \theta \\
    x = r \sin \theta \\
    dx = r \cos \theta d\theta \\
    \theta = \arcsin(x/r) \\

    ##

    so

    ##
    A = -4r\int_{0}^{r} \sqrt{1 - \frac{x^{2}}{r^{2}}} dx \\
    = -4r \int_{0}^{\arcsin(r/r)} \sqrt{1 - \sin^{2}\theta} r \cos \theta d\theta \\
    = -4r^{2} \int_{0}^{\pi/2} \sqrt{\cos^{2}\theta} \cos \theta d\theta \\
    = -4r^{2} \int_{0}^{\pi/2} \cos^{2} \theta d\theta \\
    = -4r^{2} \int_{0}^{\pi/2} \frac{1}{2}(1+\cos2\theta) d\theta \\
    = -4r^{2} \int_{0}^{\pi/2} 1/2 d\theta - 2r^{2} \int_{0}^{\pi/2} \cos2\theta d\theta \\
    = -4r^{2}(0 - \frac{\pi}{4}) - 2r^{2} (\sin(2*0) - \sin2\frac{pi}{2}) \\
    = \pi r^{2} - 2r^{2} (0 - 0) \\
    = \pi r^{2} \\
    ##

    This was way out of my league last month!

    Now, I need an integral for the area of part of a larger circle of radius R. The problem is that the easy trick of evaluating the integral for a whole quadrant doesn't work. For this integral, I am going to change my coordinate system so that the big circle is centered at the origin once again.

    This diagram gives the idea

    lune.png

    I want to get the area with green hatches, but the integral I'm about to form will be a slice of pie that includes the shaded region that plus a triangle of size 1/2(r*h) below it. So I will simply subtract out that triangle.

    Because I am only considering a lune for which the smaller circle is eclipsed at its widest point, for any given R, there is only one distance, h, at which this kind of eclipse can happen, namely, when ## h = \sqrt{R^{2} - r^{2}} ##.

    I need to set my limits of integration so that I stop integrating when
    ##\sin \theta = h \\
    \theta = \arcsin(h) ##

    But arcsin is only defined on [-1,1]. I have no guarantee that ## \sqrt{R^{2} - r^{2}}## will be so bounded: in fact, I know that any circle with a radius larger than r should be able to eclipse a circle of radius r at that circle's widest point, that is, R can be arbitrarily large. Thus ## \sqrt{R^{2} - r^{2}}## can be arbitrarily larger than 1, and I'm stuck...

    EDIT: oh, wait, it's

    ##R \sin \theta = h \\
    \theta = \arcsin(h/R) ##

    So I'll chew on that for a bit.

    Edit 2:

    So ## 0 < \sqrt{R^{2} - r^{2}} / R <*1 ##, and that looks good again.

    Now the limits of integration are ##0, arcsin(\sqrt{R^{2} - r^{2}} / R)## and I can use my formula from above, remembering to subtract out the triangle :)
     
    Last edited: Apr 1, 2013
  22. Apr 1, 2013 #21

    SammyS

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    Do you need any help or comments.

    Don't forget, you appear to be finding just half the area of the complete lune.
     
  23. Apr 1, 2013 #22
    Yes, thanks for pointing it out. The only help I need is someone else to check the work and say whether it looks right or not, since I think I have posted the answer :). I also wanted to record this posterity, in case anyone else wanted to find the answer to this. It's not really easy to google right now.
     
  24. Apr 1, 2013 #23

    SammyS

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    What you have done looks right.

    Where have you posted the answer?

    Which answer ? -- The area of the lune, or the area shaded in green in your figure?
     
  25. Apr 2, 2013 #24
    Above I posted what I believe is the correct derivation for the area of a circle, which was the question I posted here. And as for the original question I posed, I suppose I haven't quite put it together, but it's basically there. But I believe the answer is this:

    ##\left(\text{Area of the smaller semicircle}\right) - 2\left(\text{Area of the green region}\right)##

    ## \text{Lune}(r,R) = \left(\frac{1}{2}\pi r^{2}\right) -
    R^{2} \left( \int_{\arcsin(h)}^{\pi/2} \theta d\theta + \int_{\arcsin(h)}^{\pi/2} \cos2\theta d\theta\right) - \left(rh\right)##

    where

    ## h = \frac{\sqrt{R^{2} - r^{2}}}{R} ##.

    This assumes that R > r, but finding the other lune created by the overlap of two circles is easy and requires no more hard work. Also, using materially the same argument to find the limits of integration on the small circle would yield an expression for the area of any lune in terms of r, R, h, where h is their offset.
     
    Last edited: Apr 2, 2013
  26. Apr 2, 2013 #25

    SammyS

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    There should not be a leading negative sign.
    ##\displaystyle A = 4 \int_{0}^{r}\ \sqrt{r^{2}-x^{2}} dx ##

    You are using you limits of integration in reverse order.
    ##\displaystyle \int_{0}^{\pi/2} 1/2 d\theta##
    ##\displaystyle
    = \left(\ \frac{\theta}{2}\ \right|_{\ 0}^{\ \pi/2}##


    ##= \frac{ \pi}{4}-0##​

    The two mistakes do cancel .[/QUOTE]
     
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