# Surface Area of an Astroid

• Dec 7th 2008, 06:20 PM
Surface Area of an Astroid
Hi all,
I have a presentation tomorrow on a homework problem and I am stuck on the following problem:

Find the surface area of the astroid x^(2/3) + y^(2/3) = 1

Since it is symmetric I leave the bottom part out and revolve the upper part around the y-axis and multiply the result by 2. Limits of integration is from 0 to 1.
So far I think I am on the right track, but when I try to integrate this it gets really ugly.

I transformed the above equation into x=f(y) notation (since it is rotated around y-axis) and got y= (1-x^(2/3))^(3/2)
When plugged into the calculator I get a somewhat reasonable answer of 2Pi for the surface area of the whole asteroid.
Does anyone have any suggestions on the integration (I tried several different things but always get stuck)?

Thanks
• Dec 7th 2008, 06:44 PM
Skalkaz

what is the main problem in the ugly integral? The 3th root -- I think so.

Therefore substitute x = z^3. I hope you know the method of substitution in integral.

if you carry out it will appear (among other things) a term:
$(1-z^2)^{3/2}$

That suggest stong a new substitution: z = cos(w) or z = sin(w).

I hope you can do it in more detail and you can continue.
• Dec 7th 2008, 07:30 PM
Well, I figured that I have to do substitution somewhere down the line.

The Surface area formula is:

$\int 2\pi f(y) \sqrt{1+(dy/dx)^2} dy\$

For the derivative of f(y) ( $x=(1-y^{2/3})^{3/2}$ I got $\frac{1-y^{2/3}}{y^{2/3}}$ (already squared). Then the square root simplifies to $\frac{1}{y^{2/3}}$.

So I'm left with $\int 2\pi (1-y^{2/3})^{3/2}\frac{1}{y^{2/3}}$ and that's where I don't know which way to go. I am not sure about trig substitution because the exponent of y is 2/3 and not 2.
Do I have to simplify more or do a substitution now?
• Dec 8th 2008, 04:38 AM
Skalkaz
Be careful! Your area formula relate to spatial form what you'll get if you rotate a curve on the x axis. The formula means its surface. But now we have a simply plane curve, haven't we?
• Dec 8th 2008, 07:03 AM
So you are saying I have the wrong formula? If so, please enlighten me. This is how I was taught to do it.
You are right, we have a simple plain curve. The formula above is from my math book used with curves in the plain.
• Dec 8th 2008, 12:45 PM
Ok, I think I am one step closer.
I evaluated the square root and I know that this is correct since this is the length of the curve which is solved in my solutions manual.

So I got $2\pi \int \frac{(1-y^{2/3})^{3/2}}{y^{2/3}}dy\$

Skalkaz suggested a trig substitution. I tried this and ended up with an integral of $\int cos^3(\theta)cot(\theta)d\theta\$

Tried a couple of different things to solve this integral but nothing worked. Any more suggestions?
• Dec 8th 2008, 05:27 PM
Skalkaz
Drop that formula if only because it contains pi what is the number of circle, you know roteting on x-axis.

Man get the area of a curve calculating the simply integral. look up:
Integral - Wikipedia, the free encyclopedia
• Dec 8th 2008, 06:42 PM
I think we are talking about two different things. I am not looking for the area under the curve. I want to know what the area of the surface is when the curve is rotated about the y-axis. The formula that I use is the correct one.
I got the problem finally solved. You got me on the right track with the substitution, however, there is no trigonometric substiution needed, simple u-substitution works just fine and is easier to evaluate.
After simplifying the integral we have

$2\pi \int \frac{(1-y^{2/3})^{3/2}}{y^{2/3}}dy\$

For the substitution, let $u=1-y^{2/3}$
then $du=\frac{-2}{3} y^{-1/3}dy$
The integral needs to be multiplied by -3/2 in order to get rid of the negative coefficient of du. This gives

$\frac{-3}{2}*2\pi\int u^{3/2} du\ = -3\pi*\frac{2}{5}u^{5/2}$
If evaluated from 0 to 1 you get $\frac{6\pi}{5}$
Because that is only the upper half of the asteroid, we need to multiply the result by two to get the final result of $\frac{12\pi}{5}$
• Dec 8th 2008, 09:19 PM
Skalkaz
I feel a contradiction what you said

Quote:

Skalkaz Be careful! Your area formula relate to spatial form what you'll get if you rotate a curve on the x axis. The formula means its surface. But now we have a simply plane curve, haven't we?
Quote:

adiman84: You are right, we have a simple plain curve. The formula above is from my math book used with curves in the plain.
Quote:

adiman84: I am not looking for the area under the curve. I want to know what the area of the surface is when the curve is rotated about the y-axis.
never mind! Congratulation for solving!