Surface area of an astroid

[cloudy387]

Consider the upper half of the astroid described by \(\displaystyle x^{2/3}+y^{2/3}=a^{2/3}\), where \(\displaystyle a>0\) and \(\displaystyle \left | x \right | \leq a\). Find the area of the surface generated when this curve is revolved about the x-axis. Use symmetry. Type an exact answer in terms of \(\displaystyle \pi\). Note that the function describing the curve is not differentiable at 0. However, the surface area integral can be evaluated using methods that are known.

I graphed it in Desmos with some arbitrary interval for \(\displaystyle a\)

Since I'm revolving around the x-axis I'm going to rewrite the equation \(\displaystyle y=\left(a^{2/3}-x^{2/3}\right)^{3/2}\) and its derivative \(\displaystyle y'=\frac{-(a^{2/3}-x^{2/3})^{1/2}}{x^{1/3}}\). Now, in framing the integral expression for finding the surface area \(\displaystyle S\), I arrive at \(\displaystyle S=2\pi\int_{c}^{d}\left(a^{2/3}-x^{2/3}\right)^{3/2}\left(\frac{a^{1/3}}{x^{1/3}}\right)dx\)

And here I'm stuck. I'm not sure how to find my interval \(\displaystyle [c, d]\) with which I can calculate a definite integral. Since the function is not differentiable at 0, my approach would be to set the lower limit of integration as 0 and double the result, thus making my surface area integral \(\displaystyle S=4\pi\int_{0}^{d}...\) , but I don't know what to do about \(\displaystyle a\).

As far as "methods that are known" that the question mentions, I am currently in the middle of a chapter on applications of the integral, and in the preceding sections we explored determining volumes using limits of integration and area expressions as well as calculating arc lengths. And now the two are combined somewhat, in calculating surface areas of "objects of revolution".

 

[MarkFL]

I would say you want to set up an improper integral:

[MATH]S=2\pi\lim_{t\to0}\left(\int_t^a \left(a^{2/3}-x^{2/3}\right)^{3/2}\left(\frac{a^{1/3}}{x^{1/3}}\right)\,dx\right)[/MATH]

 

[Steven G]

It looks like you are using the shell method since you have 2pi in front of the integral.
Given that you are rotating around the x-axis should you be integrating wrt x or y?

 

[Steven G]

I would say you want to set up an improper integral:

[MATH]S=2\pi\lim_{t\to0}\left(\int_t^a \left(a^{2/3}-x^{2/3}\right)^{3/2}\left(\frac{a^{1/3}}{x^{1/3}}\right)\,dx\right)[/MATH]

Shouldn't you be doubling that quantity?

 

[MarkFL]

Sure. It seems the OP will likely double after integrating, since that's the way it was set up.

 

[LCKurtz]

I would try parameterizing the curve with [MATH]x=a\cos^3(t),~y=a\sin^3(t)[/MATH] with [MATH]0\le t\le \frac \pi 2[/MATH], setting up in terms of [MATH]t[/MATH] and doubling the answer.
Disclaimer: I haven't tried it yet but still...

[Edit, added]: Looks like it works out nicely to [MATH]\frac{12\pi a^2}{5}[/MATH] if I didn't make any mistakes.

 

[Steven G]

I would try parameterizing the curve with [MATH]x=a\cos^3(t),~y=a\sin^3(t)[/MATH] with [MATH]0\le t\le \frac \pi 2[/MATH], setting up in terms of [MATH]t[/MATH] and doubling the answer.
Disclaimer: I haven't tried it yet but still...

No disclaimer needed. Students need to think in a logical way and you method, whether it works out nicely or not, is something that should be tried.

 

[cloudy387]

I would say you want to set up an improper integral:

[MATH]S=2\pi\lim_{t\to0}\left(\int_t^a \left(a^{2/3}-x^{2/3}\right)^{3/2}\left(\frac{a^{1/3}}{x^{1/3}}\right)\,dx\right)[/MATH]

Unfortunately, I'm not familiar with improper integrals. When I google the term, I see mention that they cannot be computed using normal Riemann integrals which as far as I can tell is what my text has exclusively dealt with thus far. I would like to explore this approach but I am confused as to \(\displaystyle a\) being the upper limit of integration.

It looks like you are using the shell method since you have 2pi in front of the integral.
Given that you are rotating around the x-axis should you be integrating wrt x or y?

Good point, I hadn't considered that. I know from my experience with the shell method rotating around the x-axis I would be integrating with respect to y and, therefore, working with g(y) instead of f(x). However, thus far with arc length and now areas of surfaces of revolution we have integrated f(x) when rotating about the x-axis and g(y) when rotating about the y-axis. The formula does employ 2\(\displaystyle \pi\), implying shell method, but it also employs squared values within the integral, which reminds me of disk/washer method. If I rewrite for g(y) instead of f(x) I end up at the same roadblock.

EDIT: Now that I think about it, I'm conflating concepts here: shell and disk/washer methods are for areas under the curve whereas here I'm working just with surface area.

I would try parameterizing the curve with [MATH]x=a\cos^3(t),~y=a\sin^3(t)[/MATH] with [MATH]0\le t\le \frac \pi 2[/MATH], setting up in terms of [MATH]t[/MATH] and doubling the answer.
Disclaimer: I haven't tried it yet but still...

[Edit, added]: Looks like it works out nicely to [MATH]\frac{12\pi a^2}{5}[/MATH] if I didn't make any mistakes.

I will have to take a look at this. As with the improper integral method, I haven't encountered parameterized curves before, and I'm confused as to how to get from the original function given to the sine and cosine expressions.

Your answer does indeed check out. MarkFL's improper integral did, as well. According to my book's index, improper integrals are broached in the first few sections of the very next chapter and parameterized curves are covered a few chapters down the line. This may be a problem I should set aside and return to later?