# Simple integration problem

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 $$\frac{dx}{d\theta}=-3a cos^2(\theta)sin(\theta) \\ \frac{dy}{d\theta}=3a sin^2(\theta)cos(\theta)$$. 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.

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

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
Homework Helper
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.

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
Homework Helper
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?

I must be integrating it wrong: my first step gives: $$=3a^2\pi( (1-cos2\theta)^ \frac{1}{2} (-cos\theta+\frac{cos^3\theta}{3}-\int_0^{\pi} (-cos\theta+\frac{cos^3\theta}{3})\frac{2cos2\theta sin2\theta}{(1-cos^2 (2\theta))^\frac{1}{2}d\theta)$$

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

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

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

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 $$u = sin\theta \mbox{ => } \frac{du}{d\theta}= cos(\theta) \mbox{. Therefore } cos(\theta)d\theta = du \\$$ .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
Homework Helper
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.

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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Dick
Homework Helper
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).

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