# Simple integration problem

• John O' Meara
In summary, the area swept out by 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 is 3a^2\pi.

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

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

Plastic Photon said:
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?

Plastic Photon said:
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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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.

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.

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

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.

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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John O' Meara said:
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?

John O' Meara said:
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!

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.