Help with area of surface of revolution $x=\frac{1}{3}(y^2+2)^{\frac{3}{2}}, 4 \le y \le 5$

The question is:

Find the exact area of the surface obtained by rotating the curve about the x-axis.

$$x=\frac{1}{3}(y^2+2)^{\frac{3}{2}}, 4 \le y \le 5$$

I'm really confused by how the solution is presented. The formulas I just studied show two possibilities.

When rotated about the x-axis: $$S= \int 2 \pi y ds$$

When rotated about the y-axis: $$S= \int 2 \pi x ds$$

And of course, ds is: $$ds= \int \sqrt{1+(\frac{dy}{dx})^2} dx$$

Or: $$ds= \int \sqrt{1+(\frac{dx}{dy})^2} dy$$

So putting this altogether I got: $$\frac{dx}{dy}=\frac{1}{2}(y^2+2)^{\frac{1}{2}}2y$$

Which simplifies to: $$\frac{dx}{dy}=y(y^2+2)^{\frac{1}{2}}$$

So then:

$$1+ (\frac{dx}{dy})^2=1+y^2(y^2+2)$$

I assumed that I then setup my integral in the following way (which seems wrong to me): $$2 \pi \int_{4}^{5}\frac{1}{3}(y^2+2)^{\frac{3}{2}}(\sqrt{1+y^2(y^2+2)}dy$$

But everything about that seems wrong. The solution shows them going from my last confident step to this:

$$2 \pi \int_{1}^{2}y(y^2+1)dy$$

I know how to integrate that, but I don't know how they got here. Can someone help me to understand this?

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You were okay up to

$$1+ (\frac{dx}{dy})^2 \ = \ 1+y^2(y^2+2) \ = \ y^4 + 2y^2 + 1$$

$$\Rightarrow \ \sqrt{1+ (\frac{dx}{dy})^2} \ = \ \sqrt{y^4 + 2y^2 + 1} \ = \ \sqrt{(y^2 + 1)^2} \ \ .$$

So

$$S= \int 2 \pi \ y \ \ ds \ = \ \int 2 \pi \ y \ \ (y^2 + 1 ) \ \ dy \ .$$

The $\ " y " \$ in the surface area integral is the "radius arm" extending from the $\ x$-axis to the surface. Since you are integrating along the $\ y$-direction, you don't replace it with a function (least of all with your function $\ x \ = \ f(y) \$ ), but you leave it as $\ " y " \$ .

What I'm not following in your post is why the interval is given as $\ 4 \ \le \ y \ \le \ 5 \$ initially, but the limits of the integration became 1 to 2 ...

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Thanks that made it much more clear. But like you, I don't understand the changing of the limits of integration from 1 to 2. Perhaps a mistake in the solution? – free_mind Oct 31 '13 at 3:20
If that's what the solution showed, it looks like someone lost track of the original question. It sure doesn't look like 1 and 2 connect in any simple way to 4 and 5 using that function... – RecklessReckoner Oct 31 '13 at 3:25
Ah, looks to me like the solver tried to do the $\ u$-substition without actually transforming the integrand(?!). Except they did it half-way and backwards: putting 2 in for $\ y \$ gets you 5 , but 1 doesn't give you 4 ; and, in any case, you want $\ u = y^2 + 1 \$ , which will take 4 and 5 to 17 and 26... – RecklessReckoner Oct 31 '13 at 3:34

The mistake you have made is you are substituting x in place y in the formula. Just keep y in the S

S = Int (2piy(1+ (dx/dy)^2)^.5 dy

dx/dy = y*(SQRT(y^2+2))

(dx/dy)^2 = y^2*(Y^2+2)

1+(dx/dy)^2 = y^2(Y^2+2)+1

= (y^2+1)^2

If you take the square root of it then it is simply (y^2+1)

Now S = Int [(2piy(y^2+1)]dy

It is easy to find the integral for y from 4 to 5

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