# Surface area of revolution about y

1. Apr 15, 2015

### Tarpie

1. The problem statement, all variables and given/known data

Find the surface area obtained by rotating the curve
$y = x^2/4 - ln(x)/2$

$1 \leq x \leq 2$

2. Relevant equations
$2π \int f(x)\ \sqrt{1+(f'x)^2} dx$

3. The attempt at a solution
I can't seem to isolate for x in terms of y. I raised both sides to e and seperated the exponents and rearranged. Giving me $e^x^2/x^2 = e^(4y)$ but I'm stuck here.

Thanks

2. Apr 15, 2015

### SammyS

Staff Emeritus
Are you revolving this about the y-axis or about the x-axis?

3. Apr 15, 2015

### Tarpie

Sorry y-axis. Could've sworn i wrote it

4. Apr 15, 2015

### SammyS

Staff Emeritus
Yes, it's in the title. Just wanted to make sure. -- It is always best to include all information in the Original Post as well.

About the y-axis: should that be x rather than f(x) in the integral.

$\displaystyle\ 2π \int_1^2 x\, \sqrt{1+(f'(x))^2} dx\$

Also, why do you want to find x in terns of y ?

Last edited: Apr 16, 2015
5. Apr 15, 2015

### Staff: Mentor

The LaTeX interpreter choked on your equation, and I can't make sense of it either. What is the left side supposed to be? With LaTeX, if an exponent is more than one character, you need braces. IOW, this -- $e^2$ (that is, e^2) is OK, but $e^2x$ (using e^2x) doesn't render the same as e^{2x} would.

Yes. All the information should be in the post, not just in the title.

6. Apr 15, 2015

### Tarpie

I see now. That x in your equation i'm used to thinking of as f(y), just like when you rotate the curve about the x-axis it's a y; i also see that as an f(x).

So for each radius which is each x value on the curve i thought I had to rewrite it as a function of y integrate over the y interval...just like how y=f(x) i thought x had to be f(y)...didnt realize you can just stick the actual x and y in by themselves, much more convenient.

Thanks a ton

7. Apr 15, 2015

### Tarpie

Does this apply to volume as well or just surface area, because for volumes i've always rearranged for x as a function of y if it asks to rotate a curve about the y axis.

8. Apr 16, 2015

### Tarpie

I would suspect it's just surface area because of that invariant arc length term

9. Apr 16, 2015

### Tarpie

ya...because each point on that arc length is mapped to a certain x value regardless of whether in terms of y or x...so you can just use that x and the x interval instead of flipping it and using the y interval in terms just because it's about y...like rotation of height only sideways............anyways im rambling

thank you

10. Apr 16, 2015

### Staff: Mentor

Not necessarily. This is true if the function happens to be one-to-one on the interval in question (your function is one-to one on the interval).

A big problem is that what you show as your relevant equation isn't relevant. Since the curve is being revolved around the y-axis, the radius is NOT f(x), but is instead x. The relevant formula here would be $2\pi\int_1^2 x \sqrt{1 + (f'(x))^2}dx$. Did you draw a sketch of the curve and another of the surface of revolution? Doing that will help you derive the formula or at least use the right one.