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Simple integration problem

  1. Feb 26, 2007 #1
    Find the area of the surface swept out when the portion of the astroid [tex]x=a cos^3\theta, y=a sin^3\theta \mbox{ between } \theta=0 \mbox{ and } \theta=\pi \\ [/tex] rotates about the x-axis.
    [tex] ds=\frac{ds}{d\theta}d\theta \mbox{ where } \frac{ds}{d\theta}=\sqrt{(\frac{dx}{d\theta})^2 + (\frac{dy}{d\theta})^2} \\ [/tex]. Now [tex] \frac{dx}{d\theta}=-3a cos^2(\theta)sin(\theta) \\
    \frac{dy}{d\theta}=3a sin^2(\theta)cos(\theta)[/tex]. Therefore [tex] \frac{ds}{d\theta}=\sqrt{9a^2(cos^4\theta sin^2\theta + sin^4\theta cos^2\theta)} \\ [/tex]. I simplify the expression as follows:
    [tex] \frac{ds}{d\theta}=3a \sqrt{ cos^2\theta(\frac{1}{2}(1 + cos2\theta))(\frac{1}{2}(1 - cos2\theta)) + sin^2\theta(\frac{1}{2}(1 - cos2\theta))(\frac{1}{2}(1 + cos2\theta))} \\ [/tex]. This simplifies to
    [tex] ds=3a\sqrt{(\frac{1}{4} - \frac{1}{4}cos^22\theta)(sin^2\theta + cos^2\theta)} d\theta[/tex]. Therefore the integral required is: [tex] \int_0^{\pi}2\pi6a^2\sqrt{\frac{1}{4} - \frac{1}{4}cos^22\theta}sin^3\theta d\theta[/tex]
    (1)Am I on the correct path as regards simplifying the express I got for [tex]ds/d\theta [/tex]?
    (2) Did I simplify the expression correctly? Because the integral = zero with me.
     
  2. jcsd
  3. Feb 26, 2007 #2

    Dick

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    First off, how can the integral be zero? The integrand is non-negative for all of your theta values.
     
  4. Feb 26, 2007 #3
    set up as double integral knowing
    [tex] \int_0^{\pi}\int_0^{r}rd\rd\theta[/tex]

    r^2=x^2 + y^2

    EDIT: I can't get my drdtheta to show, but picture them there anyway
     
  5. Feb 26, 2007 #4

    Dick

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    Why? He's not working in polar coordinates.
     
  6. Feb 26, 2007 #5
    Given the fact that he is working with theta in radians and given x and y explicity, why not? It may make the integration problem easier.

    It also sounds like this is a volume problem. Rotate about the x-axis? Are you sure you are finding surface area and not volume?
     
  7. Feb 26, 2007 #6

    Dick

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    Did you even read the question?
     
  8. Feb 26, 2007 #7
    I must be integrating it wrong: my first step gives: [tex] =3a^2\pi( (1-cos2\theta)^ \frac{1}{2} (-cos\theta+\frac{cos^3\theta}{3}-\int_0^{\pi} (-cos\theta+\frac{cos^3\theta}{3})\frac{2cos2\theta sin2\theta}{(1-cos^2 (2\theta))^\frac{1}{2}d\theta)[/tex]
     
    Last edited: Feb 26, 2007
  9. Feb 26, 2007 #8

    Dick

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    There is a much simpler way to set up the integration. Factor sin^2(theta)*cos^2(theta) outside of the square root. Be careful because the cos will need to come out with an absolute value (why?).
     
    Last edited: Feb 26, 2007
  10. Feb 26, 2007 #9
    I have never seen something factored "outside of the integral", I don't know what you mean. If I saw this type of integral, I then might figure out why the cos will come out with an absolute value.
     
  11. Feb 26, 2007 #10

    Dick

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    Sorry, sorry. I meant 'factored out of the square root'. Typo!
     
  12. Feb 26, 2007 #11
    I am using a book that allows you teach yourself Calculus. Such a type of a book wouldn't throw you in at the deep end so it must be fairly straight forward and simple integral to solve. I still don't see a sin^2(theta)*cos^2(theta) term in the ds/d(theta) expression.
     
  13. Feb 26, 2007 #12

    Dick

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    [tex] \frac{ds}{d\theta}=\sqrt{9a^2(cos^4\theta sin^2\theta + sin^4\theta cos^2\theta)} \\ [/tex]

    sin^2(theta)*cos^2(theta) is a common factor of both terms in the square root. And it IS a simple integral.
     
  14. Feb 26, 2007 #13
    Much thanks for your help.
     
  15. Feb 27, 2007 #14
    This simple integral is giving me zero. [tex]\frac{ds}{d\theta}=\sqrt{cos^2\theta sin^2\theta} \\[/tex] is what is left of the original expression for [tex]ds/d\theta \\ [/tex]. Therefore the required integral is: [tex]6a^2\pi \int_0^{\pi} cos(\theta) sin(\theta) sin^3(\theta) d\theta = 6a^2 \pi\int_0^{\pi} sin^4(\theta) cos(\theta) d\theta \\ [/tex]. Let [tex] u = sin\theta \mbox{ => } \frac{du}{d\theta}= cos(\theta) \mbox{. Therefore } cos(\theta)d\theta = du \\ [/tex] .Therefore, [tex] \int_0^{\pi} u^4 cos(\theta)d\theta = \int_0^{\pi} u^4 du = \frac{u^5}{5} = \frac{sin^5(\theta)}{5}= 0 [/tex], for the given limits, pi and zero.
     
  16. Feb 27, 2007 #15

    Dick

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    Well, you can't say I didn't warn you! cos(theta) can be either plus or minus. So when you bring it out of the square root you have to put an absolute value on it.
     
  17. Feb 27, 2007 #16
    That is something I don't thing I know, what do you mean put an absolute value on it! Have I to find an identity for cos^2(theta)sin^2(theta)?
     
    Last edited: Feb 27, 2007
  18. Feb 27, 2007 #17

    Dick

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    No. You are almost done. It just means that for theta in the range 0 -> pi/2 you have cos(theta) in your integral and from pi/2 -> pi you should have -cos(theta). Since the 'cos(theta)' should always be positive. So split the integral into two ranges. (Actually you will find both ranges give the same contribution).
     
  19. Feb 27, 2007 #18
    It appears to have lost my last reply. There is something I don't understand. What do you mean put an absolute value on the cos(theta)! Do I have to express cos^2(theta)sin^2(theta) as another identity?
     
  20. Feb 27, 2007 #19

    Dick

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    No. No. No. I quoted your last post! If cos(theta) is negative, then what you 'pull out' of the square root needs to be -cos(theta)*sin(theta). Because that is the POSITIVE square root of cos^2*sin^2. Split the integral at pi/2!
     
  21. Feb 27, 2007 #20
    I would never have thought of splitting the integral, I thought of changing the limits to pi/2 to -pi/2 but I had no good reason to do so, other than it might give the correct answer. Thanks very much for the help.
     
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