- #1
... to find surface area?So your volume element would be:...
I just found vertical slices to be convenient. You could also write:... to find surface area?
The vertical slice between x and x+dx is a parabola thickness dx
The area of the parabola would be ##\int y(x,z) dz## with apropriate limits - wouldn't it be easier to slice the resulting parabaloid horizontally or use shells?
Anyway: surface area...
What wrong with summing the areas of the annuli between ##y## and ##y+dy## - which is what OP is doing?
... well that's what the title says right? ;)Surface area though I didn't see that my bad. Thought it was a volume of revolution problem.
Early on in this thread, Simon pointed out that you should include words along with your mathematics to explain what it is that you are doing and/or attempting to do.Okay, I made another attempt at this problem but once I start to take the integral things become very ugly.
This is where I am now. http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427e51gddpbmno
If someone could tell me if I'm on the right track and if so which way is the best to integrate and if not then I think I need help setting up the problem. I will be active for the next few hours so we can facilitate the discussion.
[Edit] Sorry for using the WolframAlpha tool but I don't know an easier/faster way to post math formulas [/Edit]
Use a common denominator inside the radical sqrt(1+(1)/(4(9-y)), then combine the two radical expressions into one radical expression.I do apologize. I guess I'm not confident in putting math into words.
I'm trying to follow formulas for similar problems.
I rewrote the problem in terms of x which gave me x=sqrt(9-y)
took the derivative which gave (-1/2)(9-y)^(-1/2)
therefore S=∫2∏x ds
with ds=sqrt(1+(1/4(9-y)))
The next step I have is S= 2pi * integral of ((9-y)^(1/2)*(sqrt(1+(1)/(4(9-y))) from there the integral turns really ugly for me so I'm assuming I've made a mistake somewhere
You can use LaTeX here:[Edit] Sorry for using the WolframAlpha tool but I don't know an easier/faster way to post math formulas [/Edit]
... that's OK, we've all been there. Just give it your best shot and we'll give you pointers.I do apologize. I guess I'm not confident in putting math into words.
... it is better to use maths ;)I'm trying to follow formulas for similar problems.
$$x=\sqrt{9-y}$$I rewrote the problem in terms of x which gave me x=sqrt(9-y)
Reasoning: the area of the annulus (the surface ring) between y and y+dy is dS=2\pi x ds where ds is the length, along the surface, between x(y) and x(y+dy).took the derivative which gave (-1/2)(9-y)^(-1/2)
therefore S=∫2∏x ds
with ds=sqrt(1+(1/(4(9-y))))
$$ S=2\pi\int_0^9 (9-y)^{^{\frac{1}{2}}}\left(1+\frac{1}{4(9-y)}\right)^{\!\!\frac{1}{2}}\;dy$$The next step I have is S= 2pi * integral of ((9-y)^(1/2)*(sqrt(1+(1)/(4(9-y)))
... Ugly integrals are normal in real life, though less common in school problems. You've done tricky integrations before, the way forward is to look for a way to simplify it. SammyS suggests:... from there the integral turns really ugly for me so I'm assuming I've made a mistake somewhere
... a "radical expression" is anything with a square-root symbol.Use a common denominator inside the radical sqrt(1+(1)/(4(9-y)), then combine the two radical expressions into one radical expression.
... you should know several techniques for integrating square-root functions.... how do you take the √(9-y) inside the other square root?
When you say the annulus between y and y+dy I am imagining the small change in the y is accounting for the very small height of the rectangle or rings. This height multiplied by length (2[itex]\pi[/itex]r) accounts for the surface area of one ring? I'm trying to understand the concept so please bare with me if I'm off.Your method starts by working out the area of the annulus between y and y+dy, which, I agree, is more intuitive. The prof's method is to find the area of the annulus between r and r+dr ... (in this case, r=x).
It is the advantage of using polar coordinates.
You are doing fine so far.I really appreciate your help with this, Simon.
When you say:
When you say the annulus between y and y+dy I am imagining the small change in the y is accounting for the very small height of the rectangle or rings. This height multiplied by length (2[itex]\pi[/itex]r) accounts for the surface area of one ring? I'm trying to understand the concept so please bare with me if I'm off.
Technically each narrow ring is a truncated cone.Also when you explain that the prof's method is to find the annulus between r and r+dr I am not able to visualize this. Are the rings still horizontal? The part that says "between r and r+dr" is what I'm not able to see.
Well done.I can definitely see the bigger picture now.
The
$$dl=\sqrt{dr^2 + (y(r+dr)-y(r))^2}\\$$
is the length of the line between (x, f(x)) and (x+dx, f(x+dx) from the Pythagorean theorem.
height and width are not good words for this are they - that's why I was careful to call it the "length along the surface" instead. But that's the idea.Then using that as the "height" or "width" of the truncated cone and multiplying it by 2##\pi##r then summing up the cones from [0,9] gives you the surface area!
That is correct - because, for that equation, y decreases with r.In the equation above, wouldn't $$(y(r+dr)-y(r))$$ give a negative answer? I know it doesn't matter because it will be squared I am just trying to make sure I understand every part of the equation.
It doesn't matter.When you integrate along the x-axis are the rings still in the same position as integrating along the y-axis?
... the rings are neither vertical nor horizontal, they are "diagonal".When I think about integrating along x my mind automatically turns the rings vertical but then the revolution doesn't work so I'm assuming they stay the same?
... which is why I teach it that way: well done :)...what I learned here has spilled over into other integration problems...