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DylsexicChciken

Revolution of surfaces versus solids

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In revolution of solids:

What variable you integrate with respect to, i.e. the independent variable, matters. But in revolution of surfaces, the variable you integrate with respect to doesn't matter. Why is that?

E.g., you have a segment of a graph bounded by [latex] y=x^2 [/latex] and [latex] y=1[/latex]. You rotate this about the y axis. To find the area of the solid, you integrate:

[latex] \int_0^{1} \pi y\,dy[/latex]

Where you have to use the variable y because the radius is in terms of x, and at different values of y, the value of x stays near different values longer because the graph is curved. i.e., the values of x, the radius, would be different from the above for a graph bounded by [latex] y=x^{10} [/latex] and [latex] y=1[/latex]. And it is precisely for this reason, I believe, that we use the variable y to find x.

 

But when it comes to revolution of surface areas:

 

Example: You revolve the graph of [latex] y=x^2 [/latex] from (0,0) to (1,1) around the y-axis. In this case the radius is the value of x. But this is what I don't get. Shouldn't you use y to find x instead of being able to just use the value of x? Like the revolution of solids, the x value, or the radius, changes from [latex]y=0[/latex] to [latex]y=1[/latex], or the value of x stays near different values longer because the graph is curved.

For reference the formula for the surface area of the graph above [latex] y=x^{2} [/latex] from (0,0) to (1,1) is:

[latex] \int_0^{1} 2 \pi x \sqrt{1 +4x^2} \,dx[/latex], where x in circumference [latex]2\pi x[/latex] is the radius and the square root is derived from the distance formula for the (slant) length(height) of approximating "bands" around the revolved surface.

Consider revolving [latex] y=x^{10}[/latex] around the y-axis and formulating the integral again, with limits of integration as (0,0) to (1,1), same as before. The radius in this case is also x integrated from from (0,0) to (1,1)!

Likewise, the formula for [latex] y=x^{10} [/latex] from (0,0) to (1,1):

[latex] \int_0^{1} 2 \pi x \sqrt{1 +100x^{18}} \,dx[/latex]

So how can it be? The radius is the same for both graphs?

So how can we say that the radius is x in both cases of surface of revolution? This would imply that the radius of an approximating circle takes a value of radius uniformly between the infinitesimal values between 0 and 1 when it does not, or that the radius from 0 to 1 on the two graphs have the same radius at different intervals of the graph.

 

To clarify my problem, consider the two graphs again:

 

Graph of [latex] y=x^2[/latex]

Graph of [latex] y=x^{10}[/latex]

If we rotate each around the y-axis. Then we fit rectangular parallelepipeds of variable lengths(length being parallel to the major axis) but same width and height, with major axis parallel to the x-axis, into the center of the two revolved graphs such that the ends of the rectangle touches the boundaries of the revolved surface area, then the sum of the lengths of the rectangular parallelepiped that were fitted would be different for the two graphs, wouldn't it?

Let [latex]t_i[/latex] denote the length of the ith rectangle fit into [latex] y=x^2[/latex] and [latex]k_i[/latex] denote the length of the ith rectangle fit into [latex] y=x^{10}[/latex], then:

[latex] \sum_{i=1}^{n}t_i \neq \sum_{i=1}^{n}k_i. [/latex]

Then how can we say the radius of the two graphs for the integral for finding their surface areas are both x, integrated from (0,0) to (1,1), when the sum of the radius or diameter are different?

Edited by DylsexicChciken

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What variable you integrate with respect to, i.e. the independent variable, matters. But in revolution of surfaces, the variable you integrate with respect to doesn't matter. Why is that?

 

 

I am not sure if I have read your question correctly but

 

If you do a simple definite intergration, between limits

 

say [math]\int {ydx} [/math]

 

This will get you the area beteen the curve and the x axis.

It can be thought of as a summation of strips parallel to the y axis.

 

similarly [math]\int {xdy} [/math]

 

will get you the area between the curve and the y axis.

This time you are summing strips parallel to the x axis.

 

So you are calculating a different area.

 

Is this what you mean by the variable of integration matters?

 

When you are integrating (summing) disks to find a volume you can often arrange the strips to cover the required volume in either direction, so long as you pick the defining disk correctly.

 

Have I pickup up your query correctly?

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Have I pickup up your query correctly?

 

 

I understand that you use values of y to find values of x when a solid is formed from revolution around the y axis. This is because for each values of y, the resulting values of x from the equation [latex] y=x^2 [/latex] is unique. In other words, the radius of a discrete cross section, perpendicular to the y-axis, for [latex] y=x^2 [/latex] is unique from the cross section radius found in revolution of [latex] y=x^{10} [/latex], for any boundaries, which I chose (0,0) and (1,1) for simplicity.

 

However, for me this intuition is thrown out the window when it comes to surface areas. Finding the surface area, you need the circumference of an approximating section of a cone and the (slanted) height, which is given by the formula [latex] 2\pi xh [/latex](the integration formula is similar), where x is the radius and h is the height. This is what is referred to as a "band", like a number of discrete rubber bands wrapped around the revolved graph, and the approximation of surface area gets better as there are more and thinner rubber bands. So the problem, if you have followed through above, is that x, directly, is the radius instead of having to use y to find x. The formula for the integral of revolution of surface area assumes that the sum of radius of cross sections or rubber bands, perpendicular to y-axis, of [latex] y=x^2 [/latex] and [latex] y=x^{10} [/latex] are equal and no longer unique for two different graphs.

Edited by DylsexicChciken

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OK do you understand where this formula comes from?

 

5e856592fb6840b4f5fc2fe08ed1284e-1.png,

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I understand that you use values ofto y to find values of x when a solid is formed from revolution around the y axis. This is because for each values of y, the resulting values of x from the equation [latex] y=x^2 [/latex] is unique. In other words, the radius of a discrete cross section, perpendicular to the y-axis, for [latex] y=x^2 [/latex] is unique from the cross section radius found in revolution of [latex] y=x^{10} [/latex], for any boundaries, which I chose (0,0) and (1,1) for simplicity.

 

However, for me this intuition is thrown out the window when it comes to surface areas. Finding the surface area, you need the circumference of an approximating section of a cone and the (slanted) height, which is given by the formula [latex] 2\pi xh [/latex](the integration formula is similar), where x is the radius and h is the height. So the problem, if you have followed through above, is that x, directly, is the radius instead of having to use y to find x. The formula for the integral of revolution of surface area assumes that the sum of radius of cross sections, perpendicular to y-axis, of [latex] y=x^2 [/latex] and [latex] y=x^{10} [/latex] are equal and no longer unique.

 

If you're using x as the radius, then you're revolving around the y-axis. Which means you're integrating over y. So to evaluate the integral you need x in terms of y, and x(y) will depend on y(x). So I'm not really sure what the problem is.

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OK do you understand where this formula comes from?

 

5e856592fb6840b4f5fc2fe08ed1284e-1.png,

[latex] \int 2 \pi x \sqrt{1+\big(\dfrac{dy}{dx}\big)^2} \,dx [/latex]
My textbook says, for rotation about y-axis, the formula above is used.
Edited by DylsexicChciken

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Yes that's correct, you are revolving the arc ab length Sab around the x axis to generate the area.

 

It is simplest to understand if you start with a straight line parallel to the x axis (dy/dx=0) and generate a cylinder.

 

You know the surface area is circumference times length of line ie

 

[math]{A_{ab}} = 2\pi y\left( {{x_b} - {x_a}} \right)[/math]

 

So you know the formula works.

 

So all you are doing is having a fancy line or staking up hoops or circles to form the area.

Edited by studiot

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If you're using x as the radius, then you're revolving around the y-axis. Which means you're integrating over y. So to evaluate the integral you need x in terms of y, and x(y) will depend on y(x). So I'm not really sure what the problem is.

 

The problem is the value of the radius. Surface areas can be calculated with either y variable or x variable. Solid revolutions can only be done with variable whose axis is perpendicular to the radius. I understand the integral for solid, but not surface area.

 

e.g. from the example above for [latex]y=x^2[/latex], we can use as radius [latex]x[/latex] or [latex] \sqrt{y} [/latex]. If we use [latex]x[/latex] as the radius, that's when it confuses me. I understand why you would use [latex] \sqrt{y} [/latex] as the radius. This is because for each values of y, there is a unique value of x. But when you use x, you're assuming the radius is no longer unique, and hence the same radius x can be used for a graph of [latex]y=x^{10}[/latex], for example in the bounds of (0,0) and (1,1) of both graphs. But clearly the shape of the two graphs are different, then how can we claim they have the same radius x, where x takes values in [0,1]? We can use x as the radius because surface area integral works for both variables as can be verified.

 

If it's still difficult to understand the question, I can scan a page from my book.

 

Yes that's correct, you are revolving the arc ab length Sab around the x axis to generate the area.

 

It is simplest to understand if you start with a straight line parallel to the x axis (dy/dx=0) and generate a cylinder.

 

You know the surface area is circumference times length of line ie

 

[math]{A_{ab}} = 2\pi y\left( {{x_b} - {x_a}} \right)[/math]

 

 

So you know the formula works.

 

So all you are doing is having a fancy line or staking up hoops or circles to form the area.

 

I understand that the formula works, but I can't get an intuitive grasp of why it works when you integrate with a variable whose axis is parallel to a cross-sectional radius of the surface. This implies that the radius is the same in the bounds [0,1] for two different graphs. This is because we assume x takes values uniformly in [0,1], right?

 

As you can see, the x taken as the radius implies the two graphs have the same cross-sectional radius, e.g. each discrete rubber band has an average radius that eventually adds up to the same number for two different graphs:

 

[latex] \int_0^{1} 2 \pi x \sqrt{1 +4x^2} \,dx [/latex]

 

[latex] \int_0^{1} 2 \pi x \sqrt{1 +100x^{18}} \,dx [/latex]

 

But the formula is correct!

Edited by DylsexicChciken

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When you change the variable of integration from dx to dy you also change the limits of integration.

 

Since y = f(x) you can calculate the new limits, given their x values.

Edited by studiot

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When you change the variable of integration from dx to dy you also change the limits of integration.

 

Since y = f(x) you can calculate the new limits, given their x values.

 

So limits of integration with bounds (0,0) to (1,1) for [latex] y=x^n [/latex] is a special case where the limits of integration are equal for both variables since [latex] 1^n=1 [/latex]. I guess that's why I was confused. Wow, I've outdid myself this time.

 

But what about solid of revolution? We can't use both variables for solid of revolution?

Edited by DylsexicChciken

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The problem is the value of the radius. Surface areas can be calculated with either y variable or x variable. Solid revolutions can only be done with variable whose axis is perpendicular to the radius. I understand the integral for solid, but not surface area.

 

e.g. from the example above, we can use as radius [latex]x[/latex] or [latex] \sqrt{y} [/latex]

 

I still don't understand what the problem is. Let's work out the surface obtained by revolving your two example functions around the y-axis, y=x2 and y=x10, over the interval y=[0,2]. So we have:

 

[math]A_1 = \int_0^2 2 \pi x \sqrt{1+\left ( \frac{dx}{dy} \right )^2} dy = 2 \pi \int_0^2 y^{1/2} (1+y^{-1}/4)^{1/2} dy \approx 13.61[/math]

 

[math]A_2 = \int_0^2 2 \pi x \sqrt{1+\left ( \frac{dx}{dy} \right )^2} dy = 2 \pi \int_0^2 y^{1/10} (1+y^{-9/5}/100)^{1/2} dy \approx 14.01[/math]

 

What here do you take issue with?

Edited by elfmotat

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So limits of integration with bounds (0,0) to (1,1) for [latex] y=x^n [/latex] is a special case where the limits of integration are equal for both variables since [latex] 1^n=1 [/latex]. I guess that's why I was confused. Wow, I've outdid myself this time.

 

But what about solid of revolution? We can't use both variables for solid of revolution?

 

You will also need to watch this when you come to multiple integrals, the limits become functions, not just simple numbers, and this catches many (including me when I first saw it)

Edited by studiot

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I still don't understand what the problem is. Let's work out the surface obtained by revolving your two example functions around the y-axis, y=x2 and y=x10, over the interval y=[0,2]. So we have:

 

[math]A_1 = \int_0^2 2 \pi x \sqrt{1+\left ( \frac{dx}{dy} \right )^2} dy = 2 \pi \int_0^2 y^{1/2} (1+y^{-1}/4)^{1/2} dy \approx 13.61[/math]

 

[math]A_2 = \int_0^2 2 \pi x \sqrt{1+\left ( \frac{dx}{dy} \right )^2} dy = 2 \pi \int_0^2 y^{1/10} (1+y^{-9/5}/100)^{1/2} dy \approx 14.01[/math]

 

What here do you take issue with?

 

The problem was that the limits and radius are the same for two different graphs. That didn't make sense considering they are two different graphs. It turns I used a graph that was a special case where the limits stay the same when converting between variables.

 

But the question arises, why can't we use the same logic that studiot pointed out above this post and integrate using both variables for solids of revolution?

Edited by DylsexicChciken

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f8254f3fd1f9d8cb294c48c227238a4c-1.png
My textbook says, for rotation about y-axis, the formula above is used.

 

 

Are you sure your textbook says about the y axis?

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Are you sure your textbook says about the y axis?

 

Never mind. I just did a random example using both variables for solids of revolution. It turns out you can use either variable. The example is:

 

[latex] y= \sqrt{x}[/latex] revolved around the x-axis.

 

[latex] y^2= x[/latex]

[latex] dx= 2y\,dy[/latex]

 

[latex] \int_0^1 \pi x \, dx = \dfrac{\pi}{2} [/latex]

[latex] \int_0^1 \pi y^2 \big( 2y \, dy \big) = \dfrac{\pi}{2} [/latex]

 

So it actually turns out you can use either variable for solids of revolution and surfaces of revolution. This is knowledge that rare amount of teachers actually explain. This is the case because I recall being told you can only use one variable for solids of revolution.

Edited by DylsexicChciken

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[latex] \int 2 \pi x \sqrt{1+\big(\dfrac{dy}{dx}\big)^2} \,dx [/latex]
My textbook says, for rotation about y-axis, the formula above is used.

 

 

This is incorrect. It looks like a mish-mash of the formula for revolving around the x and y axes. What you should have is:

 

[math]A_x = \int 2 \pi y \sqrt{1+ \left ( \frac{dy}{dx} \right )^2} \, dx[/math]

 

[math]A_y = \int 2 \pi x \sqrt{1+ \left ( \frac{dx}{dy} \right )^2} \, dy[/math]

 

where Ax and Ay are the surface areas from revolving around x and y respectively.

 

 

 

The problem was that the limits and radius are the same for two different graphs. That didn't make sense considering they are two different graphs. It turns I used a graph that was a special case where the limits stay the same when converting between variables.

 

But the question arises, why can't we use the same logic that studiot pointed out above this post and integrate using both variables for solids of revolution?

 

How is the radius the same? [math]y^{1/2} \neq y^{1/10}[/math]

Edited by elfmotat

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Never mind. I just did a random example using both variables for solids of revolution. It turns out you can use either variable. The example is

 

I was coming to that but I was waiting for you to answer my last comment because it is important.

 

I just noticed that you said rotates about the y axis.

 

For the formula I have given and you have used the rotation is about the x axis.

 

For the formula elfmotat has given rotation is about the y axis, as he correctly states.

Edited by studiot

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This is incorrect. It looks like a mish-mash of the formula for revolving around the x and y axes. What you should have is:

 

[math]A_x = \int 2 \pi y \sqrt{1+ \left ( \frac{dy}{dx} \right )^2} \, dx[/math]

 

[math]A_y = \int 2 \pi x \sqrt{1+ \left ( \frac{dx}{dy} \right )^2} \, dy[/math]

 

where Ax and Ay are the surface areas from revolving around x and y respectively.

 

 

 

How is the radius the same? [math]y^{1/2} \neq y^{1/10}[/math]

 

It's not the same radius. But it is when you integrate with respect to x for boundaries (0,0) and (1,1) and use x in place of the radius. Using y value is clearer, but I know intuitively why we use y value.

 

[latex] y= x^2 [/latex]

 

[latex] \sqrt{y}= x [/latex]

 

So therefore, we can use x as the radius.

 

 

 

Are you sure your textbook says about the y axis?

 

The formula:

[latex] \int 2 \pi x \sqrt{1+ \left ( \frac{dy}{dx} \right )^2} \, dx [/latex]

 

is derived for [latex] y= x^2 [/latex] specifically, and in general also applies to [latex] y= x^n [/latex].

 

The original formula is:

 

[latex] S = \int_a^b 2 \pi y \sqrt{1+ \left ( \frac{dy}{dx} \right )^2} \, dx [/latex]

Edited by DylsexicChciken

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It's not the same radius. But it is when you integrate with respect to x for boundaries (0,0) and (1,1) and use x in place of the radius. Using y value is clearer, but I know intuitively why we use y value.

 

The integrals still aren't the same for those boundaries. For y=x2 I get ~5.3, and for y=x10 I get ~7.5.

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For volumes the corresponding formulae are

 

[math]V = \int\limits_a^b {\pi {y^2}} dx[/math]

for rotation about the x axis from x = a to x = b

 

[math]V = \int\limits_c^d {\pi {x^2}} dy[/math]

 

for rotation about the y axis from y = c to y = d

Edited by studiot

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The integrals still aren't the same for those boundaries. For y=x2 I get ~5.3, and for y=x10 I get ~7.5.

 

Yea, the whole integral isn't the same, but the radius is. That's what confused me.

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Yea, the whole integral isn't the same, but the radius is. That's what confused me.

 

But it's not. The radius is only that same at the boundaries. Between (0,0) and (1,1) the radius is never the same. So the integral is not the same. Nothing is the same.

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Note the formulae up to now have been about rotation about different axes.

 

Your book may have been looking at volumes of revolution in different ways.

 

Here is the same volume generated by rotation about the y axis in both cases.

 

The area bounded by the curves x2=4y and y=2x is rotated about the y axis.

Find the volume generated.

 

solving the two equations simultaneously yields x = 0 and 8 ; y = 0 and 16

 

So the method on the left is as per my simple formula and the shaded area is rotated about the y axis

[math]V = \int\limits_0^{16} {\pi \left( {x_1^2 - x_2^2} \right)} dy[/math]

and for the second method we integrate with respect to x using vertical strips of area (y2-y1)dx which rotate about the y axis on a circle of of radius x .

 

[math]dV = 2\pi x\left( {{y_2} - {y_1}} \right)dx[/math]

 

post-74263-0-06771600-1412722621_thumb.jpg

 

 

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But it's not. The radius is only that same at the boundaries. Between (0,0) and (1,1) the radius is never the same. So the integral is not the same. Nothing is the same.

 

The actual radius being not the same was not directly apparent in the expression. If you have [latex]x=y[/latex] and [latex]x= \sqrt{y}[/latex]:

 

[latex] \int_0^1 x dx [/latex]

[latex] \int_0^1 x dx [/latex]

 

They look like the same integral, but are actually different when the x take different y values even when the boundaries are the same. It was just that this fact is not directly apparent to me because I was looking at the integral only and trying to find a difference. It didn't help that I used examples with the same boundaries.

Edited by DylsexicChciken

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