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Simon Bridge

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The curve is y=9-x^2, between 0 < y < 9 and you are rotating about the y axis.

It looks like you have applied a formula instead of understanding it.

That leads to uncertainty about the answer.

i.e. could you find the area dA of the strip on the surface between y and y+dy?

Lastly: you have not asked a question - do you have an issue with what you have done?

This information helps us to direct replies to where you need them.

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If not could you point out the beginning of my mistake.

- #4

Simon Bridge

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dx/dy is incorrect, but not in a way that will affect the result.

You should also double-check your algebra in simplifying the integrand.

$$\sqrt{9-y}\cdot \sqrt{1+\left(\frac{1}{2}\frac{1}{\sqrt{9-y}}\right)^2}=\left[1+\frac{1}{4}\frac{1}{9-y}\right]^{1/2}\sqrt{9-y}=\cdots$$... how do you take the √(9-y) inside the other square root?

- #5

STEMucator

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Use a vertical slice. The region is bounded between ##-3 ≤ x ≤ 3##.

##r_{in} = x - dx##

##r_{out} = x##

##r_{ave} = \frac{2x - dx}{2}##

##r_{out} - r_{in} = dx##

##h = 9 - x^2##

So your volume element would be:

##dV = 2\pi*r_{ave}*(wall \space thickness)*height##

Simplify that and ignore any ##(dx)^2## terms.

##r_{in} = x - dx##

##r_{out} = x##

##r_{ave} = \frac{2x - dx}{2}##

##r_{out} - r_{in} = dx##

##h = 9 - x^2##

So your volume element would be:

##dV = 2\pi*r_{ave}*(wall \space thickness)*height##

Simplify that and ignore any ##(dx)^2## terms.

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- #6

Simon Bridge

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... to findSo your volume element would be:...

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:

What wrong with summing the areas of the annuli between ##y## and ##y+dy## - which is what OP is doing?

- #7

STEMucator

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... to findsurface 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?

I just found vertical slices to be convenient. You could also write:

##x = ± \sqrt{9 - y}##, ##0 ≤ y ≤ 9## and use horizontal slices.

Surface area though I didn't see that my bad. Thought it was a volume of revolution problem.

- #8

Simon Bridge

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... 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.

Well sort of - though it is a solid of revolution problem - the task is not to find the volume of the solid.

There's also a niggle: your formula was for cylinder method - not vertical slices.

These details are worth marks.

But good to see someone else weighing in: if you don't speak up you don't learn stuff.

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- #9

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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]

- #10

SammyS

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Early on in this thread, Simon pointed out that you should include

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]

None of your subsequent posts do that.

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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

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Can anyone spot a/the mistake?

- #13

SammyS

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Use a common denominator inside the radical sqrt(1+(1)/(4(9-y)), then combine the two radical expressions into one radical expression.

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

I think you can integrate that.

- #14

Simon Bridge

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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]

https://www.physicsforums.com/showpost.php?p=3977517&postcount=3

... is is extremely worthwhile learning this method.

... 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.

The only way to get confident is to do it.

... 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 distance of the surface from the y-axis is x(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))))

The total surface area is the sum of all the annuli: $$S=2\pi\int_{S}x\; ds$$

... in this case: ##ds = \sqrt{1+(dx/dy)^2}\; dy##, so you need to differentiate x(y). $$\left(\frac{dx}{dy}\right)^2 = \frac{1}{4}\frac{1}{9-y}$$

$$ S=2\pi\int_0^9 (9-y)^{^{\frac{1}{2}}}\left(1+\frac{1}{4(9-y)}\right)^{\!\!\frac{1}{2}}\;dy$$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 also got a nudge in that direction in post #4.

... you should know several techniques for integrating square-root functions.... how do you take the √(9-y) inside the other square root?

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Thanks everyone for your help with this one. I talked with my professor last night and her first question was why I set it up as [itex] x = \sqrt{9-y} [/itex] She told me to try it again the way it is already set up.

Honestly I did it the first way because it was my intuition and I didn't question it.

Here is the solution I got after she made that suggestion.

[itex] y = 9 - x^2 [/itex]

Taking the derivative:

[itex] f(x)' = -2x [/itex]

Squaring the derivative gives: [itex] 4x^2 [/itex] (for future use)

Set up the Surface Area equation as:

[itex] 2pi \int_0^3 x \sqrt{1 + f(x)'^2}\ dx [/itex]

Take integral with u-substitution:

Let [itex] u = 1 + 4x^2 [/itex] then [itex] du = 8x [/itex]

which gives: [itex] ∏/4 \int_0^3 u^{(1/2)} \ du [/itex]

Integrate: [itex] ∏/4 [{(2/3)} u ^{(3/2)}] = ∏/4 [{(2/3)}(1+4x^2)^{(3/2)}] [/itex] evaluated from [0,3]

Sub in the values: [itex] ∏/4[{(2/3)}(37)^{(3/2)}] = 117.84 [/itex]

This is my first attempt at using the Latex code so please forgive me as parts of it still do not look right. Regardless this is the correct answer. ( At least, according to my professor )

Honestly I did it the first way because it was my intuition and I didn't question it.

Here is the solution I got after she made that suggestion.

[itex] y = 9 - x^2 [/itex]

Taking the derivative:

[itex] f(x)' = -2x [/itex]

Squaring the derivative gives: [itex] 4x^2 [/itex] (for future use)

Set up the Surface Area equation as:

[itex] 2pi \int_0^3 x \sqrt{1 + f(x)'^2}\ dx [/itex]

Take integral with u-substitution:

Let [itex] u = 1 + 4x^2 [/itex] then [itex] du = 8x [/itex]

which gives: [itex] ∏/4 \int_0^3 u^{(1/2)} \ du [/itex]

Integrate: [itex] ∏/4 [{(2/3)} u ^{(3/2)}] = ∏/4 [{(2/3)}(1+4x^2)^{(3/2)}] [/itex] evaluated from [0,3]

Sub in the values: [itex] ∏/4[{(2/3)}(37)^{(3/2)}] = 117.84 [/itex]

This is my first attempt at using the Latex code so please forgive me as parts of it still do not look right. Regardless this is the correct answer. ( At least, according to my professor )

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- #16

Simon Bridge

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So \Pi gets you ##\Pi## while \pi gets you ##\pi##.

For the math - you still sound like you don't really understand what is going on...

It is much better to understand than to memorize equations.

The surface of a solid of revolution has a rotation axis (in this case the y axis) and it has a radius that varies with position along the rotation axis.

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.

The equation of the surface is: y(r)=9-r^2

The annulus between r and r+dr has area ##dS=2\pi r \;dl## where ##dl## is the length along the surface between r and r+dr. So $$dl=\sqrt{dr^2 + (y(r+dr)-y(r))^2}\\ \qquad = \sqrt{1+\left(\frac{y(r+dr)-y(r)}{dr}\right)^2}\;dr$$... notice that the remaining squared term inside the radical is the definition of dy/dr ?

Then you get the formula you used above.

In terms of memorized formula - the surface area of a solid of rotaton is: $$S=2\pi\int r\sqrt{dz^2+dr^2}=2\pi\int_z r\sqrt{1+\left(\frac{dr}{dz}\right)^2}\; dz=2\pi\int_r r\sqrt{1+\left(\frac{dz}{dr}\right)^2}\;dr$$ ... where the z-axis is chosen to correspond to the rotation axis, and ##r^2=x^2+y^2## is the distance from z to the surface.

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When you say:

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.

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.

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.

- #18

Simon Bridge

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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.

The rings conform to the surface of the solid - it's like when you divide the area under a graph into trapeziums rather than bars.

On your graph y=f(x), the two points (x,f(x)) and (x+dx, f(x+dx)) have a "diagonal" line segment between them. The length of this line segment is the "width" of the annulus.

Some people would just say that dy=f(x+dx)-f(x) right off.

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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.

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!

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.

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.

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!

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.

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- #20

Simon Bridge

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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!

If you are summing along the x axis, the limits have to change though.

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.

That is perfectly fine and you should have met negative slopes before.

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I want to ask though. When you integrate along the x-axis are the rings still in the same position as integrating along the y-axis? 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?

- #22

Simon Bridge

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When you integrate along the x-axis are the rings still in the same position as integrating along the y-axis?

It doesn't matter.

It is good discipline to think of them as being in different positions and having different widths.

You can see this if you replace the dx and dy terms with deltas instead.

The line between ##\big(x,y(x)\big)## and ##\big(x+\Delta x, y(x+\Delta x)\big)## is not always going to be the same length as the one between ##\big(x(y),y \big)## and ##\big(x(y+\Delta y),y+\Delta y \big)##

The infinitesimal version comes when you take the limit that the delta-thing gets very small.

... 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...

When you get used to the process of setting up the infinitesimals that way, you'll find you no longer need to memorize the formulas and you get more confident with the results.

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