# find the area of region enclosed by astroid

by vande060
Tags: astroid, enclosed, region
 P: 187 1. The problem statement, all variables and given/known data find the region enclosed by astroid x= acosϑ 3 y=asinϑ 3 the astroid looks like this picture in my book, with the shaded region stretching for x (-a,a) and for y(-a,a) 2. Relevant equations i think the formula is ..... from (-a to a) ∫ y * x' dϑ 3. The attempt at a solution a = ∫ acosϑ 3 * 3asinϑ 2 cosϑ a = 3a2∫ cosϑ 4 * sinϑ 2 a = 3a2∫ cosϑ 4 * (1-cosϑ 2 ) a = 3a2∫ cosϑ 4 - cosϑ6 is this one the right track, if so I can evaluate the rest of that integral myself
 HW Helper P: 6,164 Your formula would be: $$A=\int_{-a}^a y dx = \int_{\pi}^0 y \frac {dx} {d\theta} d\theta$$ Please note that this will be only the area above the x-axis. To get the total area, you will need to double that. Your derivative of x should be: $$\frac {dx} {d\theta}=\frac {d} {d\theta} (a\cos^3\theta)=3a\cos^2\theta\cdot -\sin\theta$$ Substituting gives: $$A=\int_{\pi}^0 (a\sin^3\theta) \cdot (3a\cos^2\theta\cdot -\sin\theta) d\theta =3a^2\int_0^{\pi} \sin^4\theta \cos^2\theta d\theta$$ I think you can take it from there. So you were on the right track, but made a mistake in substitution. Also, you should take note of the integral boundaries. They can not just be left out.[/edit]
 P: 187 quick question, how did you decide the bound from pi to 0 and then you flipped them, it has something to do with the negative sign right ?
HW Helper
P: 6,164

## find the area of region enclosed by astroid

 Quote by vande060 quick question, how did you decide the bound from pi to 0 and then you flipped them, it has something to do with the negative sign right ?
In the first equation we shift from x to theta.
A value of x=-a corresponds to a value theta=pi, and x=+a corresponds to theta=0.
So the integral is from pi to 0.

In the third equation, there is a minus sign in the equation.
Flipping the boundaries means flipping the sign, or in this case removing the minus sign.
P: 187
 Quote by I like Serena In the first equation we shift from x to theta. A value of x=-a corresponds to a value theta=pi, and x=+a corresponds to theta=0. So the integral is from pi to 0. In the third equation, there is a minus sign in the equation. Flipping the boundaries means flipping the sign, or in this case removing the minus sign.
okay i was scratching my head about this one but i think you get it by just substituting x for a and solving for theta right?

-a = acos3ϑ

-1 = cos3ϑ

ϑ = pi

a = acos3ϑ

1 = cos3ϑ

ϑ = 0

did i arrive at that correctly?
HW Helper
P: 6,164
 Quote by vande060 okay i was scratching my head about this one but i think you get it by just substituting x for a and solving for theta right? -a = acos3ϑ -1 = cos3ϑ ϑ = pi a = acos3ϑ 1 = cos3ϑ ϑ = 0 did i arrive at that correctly?
Yes. That is entirely correct!

Actually, what I did is look at the graph you posted.
With theta being the angle with the positive x-axis, a value of pi corresponds to x=-a.
Substituting pi in the expression for x shows that this is correct.

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