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Find the area of region enclosed by astroid

by vande060
Tags: astroid, enclosed, region
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vande060
#1
Apr15-11, 05:20 PM
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
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I like Serena
#2
Apr16-11, 10:43 AM
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P: 6,187
Your formula would be:

[tex]A=\int_{-a}^a y dx = \int_{\pi}^0 y \frac {dx} {d\theta} d\theta[/tex]

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:

[tex]\frac {dx} {d\theta}=\frac {d} {d\theta} (a\cos^3\theta)=3a\cos^2\theta\cdot -\sin\theta[/tex]

Substituting gives:

[tex]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
[/tex]

I think you can take it from there.
So you were on the right track, but made a mistake in substitution.

[edit]Also, you should take note of the integral boundaries. They can not just be left out.[/edit]
vande060
#3
Apr18-11, 10:12 AM
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
?

I like Serena
#4
Apr18-11, 10:19 AM
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Find the area of region enclosed by astroid

Quote Quote by vande060 View Post
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.
vande060
#5
Apr18-11, 10:27 AM
P: 187
Quote Quote by I like Serena View Post
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?
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#6
Apr18-11, 10:36 AM
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P: 6,187
Quote Quote by vande060 View Post
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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