Calculating the Surface Area of a Rotating Astroid in Simple Steps

[John O' Meara]

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

 

[Dick]

First off, how can the integral be zero? The integrand is non-negative for all of your theta values.

 

[Plastic Photon]

set up as double integral knowing
\int_0^{\pi}\int_0^{r}rd\rd\theta

r^2=x^2 + y^2

EDIT: I can't get my drdtheta to show, but picture them there anyway

 

[Dick]

set up as double integral knowing
\int_0^{\pi}\int_0^{r}rd\rd\theta

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.

 

[Plastic Photon]

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?

 

[Dick]

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?

 

[John O' Meara]

I must be integrating it wrong: my first step gives: =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)

 

[Dick]

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?).

 

[John O' Meara]

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.

 

[Dick]

Sorry, sorry. I meant 'factored out of the square root'. Typo!

 

[John O' Meara]

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.

 

[Dick]

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

sin^2(theta)*cos^2(theta) is a common factor of both terms in the square root. And it IS a simple integral.

 

[John O' Meara]

Much thanks for your help.

 

[John O' Meara]

This simple integral is giving me zero. \frac{ds}{d\theta}=\sqrt{cos^2\theta sin^2\theta} \\ is what is left of the original expression for ds/d\theta \\. Therefore the required integral is: 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 \\. Let u = sin\theta \mbox{ =&gt; } \frac{du}{d\theta}= cos(\theta) \mbox{. Therefore } cos(\theta)d\theta = du \\ .Therefore, \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, for the given limits, pi and zero.

 

[Dick]

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.

 

[John O' Meara]

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)?

 

[Dick]

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).

 

[John O' Meara]

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?

 

[Dick]

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!

 

[John O' Meara]

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.