Simple integration problem

  • #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.
 

Answers and Replies

  • #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.
 
  • #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
 
  • #4
Dick
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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
Why? He's not working in polar coordinates.
 
  • #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?
 
  • #6
Dick
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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?
Did you even read the question?
 
  • #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]
 
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  • #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?).
 
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  • #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.
 
  • #10
Dick
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Sorry, sorry. I meant 'factored out of the square root'. Typo!
 
  • #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.
 
  • #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.
 
  • #13
Much thanks for your help.
 
  • #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.
 
  • #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.
 
  • #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)?
 
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  • #17
Dick
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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)?
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).
 
  • #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?
 
  • #19
Dick
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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?
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!
 
  • #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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