Surface Area rotated about an axis which is not the x or y axis

[Lily Rose]

Hi. I understand how to solve surface Area using integration when it is to be revolved about the x or y axis. But when the axis is not x or y I have a difficult time solving it. Please help me. Here is the equation

sqrt(x+1) rotated at x=-1 and y=5.
the bounds are 1 to 5.
since y=sqrt(x+1)
x = y^2 -1

I solved for x=-1 and I got ∫ (from 1 to 5) 2pi ((y^2 -1) +1) √(1+4y) dy

and for y=5 ∫ (from 1 to 5) 2pi ((√x+1)+5) √(1+ [itex]\frac{1}{4x+4}[/itex])


pls help me. I really do understand on how to rotate it about the x or y axis. I tried searching online for some tutorials but I can't find any.

 

[raving_lunatic]

Are the axes you're rotating about x = -1 and y = 5? As in, these are two separate parts of one question?

I suggest a change of variables here. You can redefine the axes (shifting everything to the left, right, up or down) and then use the formulae you know to find the surface area of revolution when the axis of revolution is the x or y axis. For example, if you make the substitution x' = x - a, then the line x = a is now the x' = 0 axis, or the y' axis.

 

[LCKurtz]

Hi. I understand how to solve surface Area using integration when it is to be revolved about the x or y axis. But when the axis is not x or y I have a difficult time solving it. Please help me. Here is the equation

sqrt(x+1) rotated at x=-1 and y=5.
the bounds are 1 to 5.
since y=sqrt(x+1)
x = y^2 -1

I solved for x=-1 and I got ∫ (from 1 to 5) 2pi ((y^2 -1) +1) √(1+4y) dy

I assume you have a typo under that radical and you mean$$
2\pi\int_1^5(y^2-1+1)\sqrt {1+4y^\color{red}2}~dy$$Although that isn't the way I would have set it up, at least the integrand is now correct, so you are doing something right. But your limits aren't. When ##x## goes from ##1## to ##5##, what are the corresponding ##y## values? That's what you want for your limits.

and for y=5 ∫ (from 1 to 5) 2pi ((√x+1)+5) √(1+ [itex]\frac{1}{4x+4}[/itex])


pls help me. I really do understand on how to rotate it about the x or y axis. I tried searching online for some tutorials but I can't find any.

In your second one$$
2\pi \int_1^5 (\color{red}{\sqrt{x+1}+5})\sqrt{1+\frac 1 {4x+4}}~dx$$you have the radius of rotation wrong. You want ##y_{upper} - y_{lower} = 5 -\sqrt{x+1}##.

Actually, you seem to understand pretty well what you are doing. Personally I would have done the first integral in terms of ##x## and the second one in terms of ##y## but once you fix the mistakes, you are OK.

 

[BiGyElLoWhAt]

How I look at it is like this:

If you rotate your function around an axis, what is your radius? Well look at it carefully, what are you really doing?

Let's say our function is [itex]y=\sqrt{x}[/itex] and we want to find the surface area. Well what's the radius if we rotate about the x-axis? Well at the x-axis [itex]y=0[/itex] and thus the distance from our function to our axis we're rotating about: [itex]r=\sqrt{x}-0[/itex]

Now let's say we want to rotate our function about [itex]y=-3[/itex]
So what's r? Well it's the distance from our function to our axis.
Or: [itex]|f(x)-a_{rotation}|[/itex], which in our case is [itex]|\sqrt{x}-(-3)|=|\sqrt{x}+3|[/itex]

I would recommend drawing it out.

Then you just plug it into your [itex]∫_{a}^b2\pi r\sqrt{1+f'{x}^2}dx[/itex]
Mod note: The above should be
##\int_a^b 2\pi r\sqrt{1+(f'(x))^2}dx ##

 

[BiGyElLoWhAt]

oops. in the eq. it's f'(x), not f'x

 

[Lily Rose]

I assume you have a typo under that radical and you mean$$
2\pi\int_1^5(y^2-1+1)\sqrt {1+4y^\color{red}2}~dy$$Although that isn't the way I would have set it up, at least the integrand is now correct, so you are doing something right. But your limits aren't. When ##x## goes from ##1## to ##5##, what are the corresponding ##y## values? That's what you want for your limits.

I knew I was wrong there. So I solved for the values of y and I got √2 and √6 for my bounds in my second equation. Am I correct? OH! And thanks for noticing the 4y^2. I didn't notice that I did not square the Y

In your second one$$
2\pi \int_1^5 (\color{red}{\sqrt{x+1}+5})\sqrt{1+\frac 1 {4x+4}}~dx$$you have the radius of rotation wrong. You want ##y_{upper} - y_{lower} = 5 -\sqrt{x+1}##.

as for this one, yeah, I get what your trying to say. I changed my answer.

Actually, you seem to understand pretty well what you are doing. Personally I would have done the first integral in terms of ##x## and the second one in terms of ##y## but once you fix the mistakes, you are OK.

THIS LAST ONE ACTUALLY MADE ME SMILE. Thanks. It means a lot to hear someone say I'm actually right! I took your advice and here are my answers

For equation 1: 2pi ∫(from 1 to 5) ([itex]\sqrt{x+1}[/itex]-(-1)) [itex]\sqrt{1+\frac{1}{4x+4}}[/itex] dx

For equation 2: 2pi ∫(from √2 to √6) (5-[itex]\sqrt{y^2 -1}[/itex]) [itex]\sqrt{1+4y^2}[/itex] dySO! Am I right?

 

[Lily Rose]

Also guys...I have a problem. I solved for the both of them and i got different answers. Aren't they supposed to have the same answers?

 

[LCKurtz]

Also guys...I have a problem. I solved for the both of them and i got different answers. Aren't they supposed to have the same answers?

You should have two different answers. One for revolving about x=-1 and another for revolving about y = 5. The dx and dy versions should agree for each axis.

 

[LCKurtz]

THIS LAST ONE ACTUALLY MADE ME SMILE. Thanks. It means a lot to hear someone say I'm actually right! I took your advice and here are my answers

For equation 1: ##2\pi \int_1^5(\sqrt{x+1}-(-1)) \sqrt{1+\frac{1}{4x+4}} dx##

Correct.

For equation 2: ##2\pi\int_{\sqrt 2}^{\sqrt 6} (\color{red}{5-\sqrt{y^2 -1}})\sqrt{1+4y^2}##dySO! Am I right?

Almost. ##y_{upper} - y_{lower} = 5-y##.

 

[Lily Rose]

##y_{upper} - y_{lower} = 5-y##.

you lost me there. May i ask why is it y instead of [itex]\sqrt{y^2 - 1}[/itex]

 

[LCKurtz]

you lost me there. May i ask why is it y instead of [itex]\sqrt{y^2 - 1}[/itex]

You can express a point on the curve ##y=\sqrt{x+1}## in terms of ##x## as ##(x,\sqrt{x+1})## or in terms of ##y## as ##(y^2-1,y)##. You want the ##y## value (the second coordinate) in terms of ##y## for the dy integral.

 

[Lily Rose]

You can express a point on the curve ##y=\sqrt{x+1}## in terms of ##x## as ##(x,\sqrt{x+1})## or in terms of ##y## as ##(y^2-1,y)##. You want the ##y## value (the second coordinate) in terms of ##y## for the dy integral.

STUPID ME. To be honest I looked at my notes and I was looking for where I got [itex]\sqrt{y^{2}-1}[/itex] and I can't find it anywhere. Maybe I got it from my original integral which was in terms of x then I just removed the inside of the radical. It's things like that which I need to watch out for. Thanks so so so much!