Surface area of revolution about y

[Tarpie]

Homework Statement

[/B]
Find the surface area obtained by rotating the curve
[itex]y = x^2/4 - ln(x)/2 [/itex]

[itex]1 \leq x \leq 2[/itex]

Homework Equations

[itex]2π \int f(x)\ \sqrt{1+(f'x)^2} dx [/itex]

The Attempt at a Solution

I can't seem to isolate for x in terms of y. I raised both sides to e and separated the exponents and rearranged. Giving me [itex] e^x^2/x^2 = e^(4y) [/itex] but I'm stuck here.

Thanks

 

[SammyS]

Homework Statement

[/B]
Find the surface area obtained by rotating the curve
[itex]y = x^2/4 - ln(x)/2 [/itex]

[itex]1 \leq x \leq 2[/itex]

Homework Equations

[itex]2π \int f(x)\ \sqrt{1+(f'x)^2} dx [/itex]

The Attempt at a Solution

I can't seem to isolate for x in terms of y. I raised both sides to e and separated the exponents and rearranged. Giving me [itex] e^x^2/x^2 = e^(4y) [/itex] but I'm stuck here.

Thanks

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

 

[Tarpie]

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

 

[SammyS]

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

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 ?

 

[Mark44]

I can't seem to isolate for x in terms of y. I raised both sides to e and separated the exponents and rearranged. Giving me e^x^2/x^2 = e^(4y) but I'm stuck here.

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.

It is always best to include all sure information in the Original Post as well.

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

 

[Tarpie]

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

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

 

[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.

 

[Tarpie]

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

 

[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 I am rambling

thank you

 

[Mark44]

ya...because each point on that arc length is mapped to a certain x value regardless of whether in terms of y or x

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.

...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 I am rambling