Find the area of region enclosed by astroid

[vande060]

Homework Statement

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)

Homework Equations

i think the formula is ... from (-a to a) ∫ y * x' dϑ

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

 

[I like Serena]

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<br /> =3a^2\int_0^{\pi} \sin^4\theta \cos^2\theta d\theta<br />

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]

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]

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]

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?

 

[I like Serena]

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.