Solids of Revolution - Negative Volume

[Dethrone]

I encountered a problem where the answer I got was negative.

Calculate the volume bounded by $y=x^2-5x+6$, $y=0$, about y-axis.

An easy question that is best done with the cylindrical shell method:

$$V=2\pi \int_{2}^{3} x(x^2-5x+6)\,dx$$
$$V=\frac{-5\pi}{6}$$

I think I know why it's negative. The "height" I picked for each shell I represented using $f(x)$ which only returns negative numbers because it is below the x-axis. What is the most conventional way to fix this? Should I include an absolute value in my integration, should I add an negative to my integral...?

EDIT: The height should be $-f(x)$, let me know if that is right.

 

[Dethrone]

$$V=\int_2^3 \pi (r(x))^2 dx$$

where $r(x)=x^2-5x+6$

.

Correct me if I'm wrong, but I do not think that is correct. We are rotating about the y-axis, not the x-axis.

 

[I like Serena]

I encountered a problem where the answer I got was negative.

Calculate the volume bounded by $y=x^2-5x+6$, $y=0$, about y-axis.

An easy question that is best done with the cylindrical shell method:

$$V=2\pi \int_{2}^{3} x(x^2-5x+6)\,dx$$
$$V=\frac{-5\pi}{6}$$

I think I know why it's negative. The "height" I picked for each shell I represented using $f(x)$ which only returns negative numbers because it is below the x-axis. What is the most conventional way to fix this? Should I include an absolute value in my integration, should I add an negative to my integral...?

EDIT: The height should be $-f(x)$, let me know if that is right.

Hey Rido12! ;)

In the cylindrical shell method you are integrating hollow pipes around the y-axis.
The height of each pipe is the upper graph minus the lower graph.

Since you have chosen to integrate between $x=2$ and $x=3$, the upper graph is $y=0$ and the lower graph is $y=f(x)$.
So the height is indeed:
$$h(x) = 0 - f(x) = -f(x)$$

 

[Dethrone]

Thank you! :D

When I was working on that question a couple of days ago, I was basically memorizing formulas. I've realized now that it's best to imagine what you're rotating, be it a shell, a washer, or a disc, and rederive the formula you want.

 

[MarkFL]

I like to begin by computing the volume of an element, in this case a shell:

\(\displaystyle dV=2\pi rh\,dx\)

where:

\(\displaystyle r=x,\,h=0-\left(x^2-5x+6\right)=-x^2+5x-6\)

Hence:

\(\displaystyle dV=2\pi\left(-x^3+5x^2-6x\right)\)

Next, sum up all the elements to get the total volume:

\(\displaystyle V=2\pi\int_2^3 -x^3+5x^2-6x\,dx=\frac{5\pi}{6}\)

 

[I like Serena]

I like to begin by computing the volume of an element

I like to begin by making a drawing. ;)

 

[I like Serena]

Hmm, I've read that someone recently introduced a Desmos plug-in. (Cool)

Lemme try...
[desmos=0,3,-1,6]x^2-5x+6\le y \le 0; 0 \le y \le x^2-5x+6[/desmos]

Hmm, I wonder if we can create something that looks like a body of revolution. (Wondering)

 

[MarkFL]

I like to begin by making a drawing. ;)

Yes, that's actually my first step too, but I was too lazy to put together a plot this time...:D

But, now that I've been called out for such laziness (Giggle), here it is...

View attachment 2930

 

[MarkFL]

...Hmm, I wonder if we can create something that looks like a body of revolution. (Wondering)

Yes, you can use the Desmos API to plot solids of revolution...you just have to enter all the correct functions...:D

https://www.desmos.com/calculator/8qdpedyypl

 

[I like Serena]

Yes, you can use the Desmos API to plot solids of revolution...you just have to enter all the correct functions...:D

https://www.desmos.com/calculator/8qdpedyypl

Hmm... let's see... (Sweating)

[desmos="-3,3,-1,1"]x^2-5\left|x\right|+6\le y\le 0; \frac{1}{8}\sqrt{\max \left(4-x^2,0\right)}\le y\le \frac{1}{8}\sqrt{9-x^2}; \frac{1}{8}\sqrt{2.29^2-x^2}-\frac{1}{4}\le y\le \frac{1}{8}\sqrt{\max \left(4-x^2,0\right)}+\min \left(x^2-5\left|x\right|+6,0\right)\left\{\left|x\right|<2.23\right\}; \frac{1}{8}\sqrt{\max \left(2.2^2-x^2,0\right)}\le y\le \frac{1}{8}\sqrt{2.3^2-x^2}; x^2-5\left|x\right|+6\le y\le 0\left\{2.2<\left|x\right|<2.3\right\}[/desmos]

 

[Dethrone]

Wow, and I was begin to wonder whether or not Desmos could plot solids of revolution...

When I do these questions, I do graph them first, then I start with the volume of an element. (I wonder why the call it an 'element'.) I find that it is extremely helpful to sketch the element first, but after a while, you can start to visualize what the element might look like in your head. I actually really enjoy doing these solids of revolution questions... It's really easy once you understand how it works, rather than memorizing the formulas.

 

[I like Serena]

Wow, and I was begin to wonder whether or not Desmos could plot solids of revolution...

When I do these questions, I do graph them first, then I start with the volume of an element. (I wonder why the call it an 'element'.)

From Element | Define Element at Dictionary.com:

el·e·ment [el-uh-muh nt]
noun

1. a component or constituent of a whole or one of the parts into which a whole may be resolved by analysis: Bricks and mortar are elements of every masonry wall.

​

I find that it is extremely helpful to sketch the element first, but after a while, you can start to visualize what the element might look like in your head. I actually really enjoy doing these solids of revolution questions... It's really easy once you understand how it works, rather than memorizing the formulas.

You still have to "memorize" a couple of formulas: (Nerd)

• Circumference of a circle is $2\pi r$.
• Volume of a pillar with constant cross section $A$ is $Ah$.
• Integration goes like $V=\int dV$.

And methods:

• That you can integrate a volume with hollow pipes.
• That you can integrate a volume with disks or washers.

That should be it for this problem. (Angel)

 

[Dethrone]

Yup, by formulas, what I was referring to were these:

$$V=2\pi \int_{a}^{b} x(f(x))\,dx$$
$$V=2\pi \int_{a}^{b} y(f(y))\,dy$$

I don't find them really helpful when you have to rotate around lines that aren't axis, or more complicated revolutions. Since I like to visualize a volume element in my head, I'm already re-deriving the formula anyway. I'm aware of the definition of an element, I've just never read any textbooks that use that word. Well, maybe because I haven't really read any university textbooks... (Wasntme)

What process do you use to solve these problems? To you tend to visualize the volume of an element like what I do, or do you memorize the formulas?

 

[I like Serena]

Yup, by formulas, what I was referring to were these:

$$V=2\pi \int_{a}^{b} x(f(x))\,dx$$
$$V=2\pi \int_{a}^{b} y(f(y))\,dy$$

I don't find them really helpful when you have to rotate around lines that aren't axis, or more complicated revolutions. Since I like to visualize a volume element in my head, I'm already re-deriving the formula anyway. I'm aware of the definition of an element, I've just never read any textbooks that use that word. Well, maybe because I haven't really read any university textbooks... (Wasntme)

What process do you use to solve these problems? To you tend to visualize the volume of an element like what I do, or do you memorize the formulas?

For starters, I would write these formulas differently.
Say like:
$$V=\int_{a}^{b} 2\pi r\,dr \cdot h(r)$$
Then it's easier to recognize the constituents (aka elements ;)) of the formula.

This is a form that I can sort of memorize, or recognize when I see it.
But I don't try too hard.
I'll re-derive it when I need to (or look it up on wiki).

For difficult problems, I always make a drawing.
For just another of the same type of problem, I visualize the element in my mind and write up the formula directly.

As for the word element, whenever a word from natural language is descriptive for what you want to use it for, it's a good word! (Mmm)
Perhaps a high school book might tend to limit its vocabulary a bit to avoid confusion, but in real life that is not the case.

 

[Dethrone]

Integration isn't suppose to be in high school books, unless, I don't know, you live in China. But I do agree that your formula is much better than the ones I have in my textbooks because you can clearly recognize the elements of the formula. Now, if I only I learned your formula first, maybe learning this would have been much easier...but I think I've got the hang of this and is quite fun. Thanks again for the help~ :D

 

[Prove It]

Integration isn't suppose to be in high school books, unless, I don't know, you live in China. But I do agree that your formula is much better than the ones I have in my textbooks because you can clearly recognize the elements of the formula. Now, if I only I learned your formula first, maybe learning this would have been much easier...but I think I've got the hang of this and is quite fun. Thanks again for the help~ :D

Not true, in Year 12 Specialist Mathematics in Australia, volume by revolution is covered - but only using cylindrical shells/washers.

 

[Dethrone]

Obviously it would taught in some places, but I just wanted to stress that it is not normally taught in a regular high school setting. I mean, even in the US and Canada, they're taught maybe if you're in an AP or IB program, but you'd be hard-pressed to find a high school textbook that actually covers it. Most likely they're using a university-level textbook. And my point was that they're not usually included in high school level textbooks, not that they're not taught in high school. But thanks for pointing that out :D

Also, you can calculate solids of revolution with something other than washers/disks and cylindrical shells? You seem to be imply that, and I would like to know more :D

 

[Prove It]

Obviously it would taught in some places, but I just wanted to stress that it is not normally taught in a regular high school setting. I mean, even in the US and Canada, they're taught maybe if you're in an AP or IB program, but you'd be hard-pressed to find a high school textbook that actually covers it. Most likely they're using a university-level textbook. And my point was that they're not usually included in high school level textbooks, not that they're not taught in high school. But thanks for pointing that out :D

Also, you can calculate solids of revolution with something other than washers/disks and cylindrical shells? You seem to be imply that, and I would like to know more :D

Sorry I meant to say discs, not shells.

They only learn the discs/washers method, not the shells method.

 

[Deveno]

Yes, you can use other methods. The most common of these is to integrate the constant function 1 over the 3-dimensional region (or the integral of the characteristic function of the region). Typically this is done with an iterated triple integral:$\displaystyle V = \int_a^b \int_{f_1(x)}^{f_2(x)}\int_{g_1(x,y)}^{g_2(x,y)}\ dz\ dy\ dx$

This may be much harder to do, as it does not leverage the radial symmetry of a solid of revolution.

 

[Dethrone]

Yes, you can use other methods. The most common of these is to integrate the constant function 1 over the 3-dimensional region (or the integral of the characteristic function of the region). Typically this is done with an iterated triple integral:$\displaystyle V = \int_a^b \int_{f_1(x)}^{f_2(x)}\int_{g_1(x,y)}^{g_2(x,y)}\ dz\ dy\ dx$

This may be much harder to do, as it does not leverage the radial symmetry of a solid of revolution.

I can't say I understand this (Wait), but hopefully I will later this month when I work through my book. :D Thanks for the info!

 

[Deveno]

Here is a simple example:

Suppose we have the rectangular region bounded by the planes containing:

(0,0,0), (a,0,0), (0,b,0), (a,b,0), (0,0,c), (a,0,c), (0,b,c), and (a,b,c).

Here, we may take:

$g_1(x,y) = 0, g_2(x,y) = c, f_1(x) = 0, f_2(x) = b$.

The volume of said region is:

$\displaystyle V = \int_0^a \int_0^b \int_0^c dz\ dy\ dx$

$\displaystyle = \int_0^a \int_0^b (c - 0)\ dy\ dx$

$\displaystyle = c \int_0^a \int_0^b\ dy\ dx$

$\displaystyle = c \int_0^a (b - 0)\ dx$

$\displaystyle = bc \int_0^a\ dx = bc(a - 0) = abc$.

With a solid of revolution, these integrals will be much more complicated.

 

[I like Serena]

Not true, in Year 12 Specialist Mathematics in Australia, volume by revolution is covered - but only using cylindrical shells/washers.

Actually, in my country, I have recently tutored someone in grade 11 on the disks method.
It was explained in his regular math textbook, and there was a problem like that in a previous final exam.
I have also tutored someone in grade 10 on integration.

I can't say I understand this (Wait), but hopefully I will later this month when I work through my book. :D Thanks for the info!

Just to give you a heads up. (Wink)

Suppose you have a cube with sides 1.
Then the cube can be divided into small blocks with size $dx,dy,dz$.
Such a small block has volume:
$$dV=dxdydz$$

Integrating it, we get:
$$V=\int dV = \int_0^1 \int_0^1 \int_0^1 dxdydz
= \int_0^1 \int_0^1 [x]_0^1\,dydz
= \int_0^1 \int_0^1 1\,dydz
= \int_0^1 1\,dz
= 1$$

Edit: I'm only seeing just now that Deveno came up with a similar explanation.
I missed it, because it was on the next page. Now suppose you would want to find the volume below
$$z=x^2+y^2$$
bounded by $z=0$, $x=\pm 1$, and $y=\pm 1$... (Evilgrin)

 

[I like Serena]

Oh, and if you want something to look forward to... (Wasntme)

In some cases you can use Gauss's divergence theorem to find a volume.
$$\iiint \nabla \cdot \mathbf f \, dV = \bigcirc \kern{-26mu} \iint \, \mathbf f \cdot d\mathbf S$$

Applied to our cube, we get with $\mathbf f(x,y,z) = x \mathbf{\hat \imath}$:
$$\iiint 1 \, dV = \bigcirc \kern{-26mu} \iint \mathbf f \cdot d\mathbf S$$
The left hand side corresponds to the volume integral.
This means that we can calculate the volume by calculating how much effect $x \mathbf{\hat \imath}$ has on each face of the cube.

In this case it only has a non-zero contribution on the $x=1$ face.
Its contribution is the area of the face ($1 \times 1$) times the magnitude of $x \mathbf{\hat \imath}$, which is $1$.
That brings the volume to $1$ as expected. (Wink)

 

[Dethrone]

Actually, in my country, I have recently tutored someone in grade 11 on the disks method.
It was explained in his regular math textbook, and there was a problem like that in a previous final exam.
I have also tutored someone in grade 10 on integration.

I'll read this when I get home later today, but I do want to say that I'm not as fortunate as those who can learn calculus early... Though, at my age in the country that I'm in, I still have a huge head start
But calculus is so strangely fun :D

 

[Dethrone]

Here, we may take:

$g_1(x,y) = 0, g_2(x,y) = c, f_1(x) = 0, f_2(x) = b$.

Hi Deveno, I understand your example, but I don't understand what I have quoted and how it fits in with the example. i.e what does $g_1, g_2, f_1, f_2$ refer to?

Now suppose you would want to find the volume below
$$z=x^2+y^2$$
bounded by $z=0$, $x=\pm 1$, and $y=\pm 1$... (Evilgrin)

Hi 'I like Serena'!

Spoiler

Is there a better way you want me to address you? :D

I don't like that evil grin you have there, but I will attempt this now! (Devil) Wait...does this require spherical coordinates?

 

[I like Serena]

Hi 'I like Serena'!

Spoiler

Is there a better way you want me to address you? :D

I'm used to ILS by now. (Happy)

I don't like that evil grin you have there, but I will attempt this now! (Devil) Wait...does this require spherical coordinates?

Nope. Just normal coordinates.
With the given boundaries, spherical coordinates don't work very well.
And due to the lack of rotational symmetry, the shells and disks methods don't work either.

 

[Dethrone]

Using the knowledge I have learned from this thread, I will try my best. (Nod)

$$V=\int_{-1}^{1} \int_{-1}^{1} \int_{0}^{0} x^2+y^2\,dz\,dx\,dy$$

I have no idea if that's correct, and I'm not sure about the z-bounds.

EDIT: On second though, should I start with this?

$$V=\int_{-1}^{1} \int_{-1}^{1} \int_{0}^{0} z \,dz\,dx\,dy$$

 

[I like Serena]

Using the knowledge I have learned from this thread, I will try my best. (Nod)

$$V=\int_{-1}^{1} \int_{-1}^{1} \int_{0}^{0} x^2+y^2\,dz\,dx\,dy$$

I have no idea if that's correct, and I'm not sure about the z-bounds.

The formula for the volume should be:
$$V = \iiint dV = \iiint dz\,dy\,dx$$

The bounds specify through which range each of the coordinates goes.
You have the right range for the x- and y-coordinates.
However, the z-coordinate has the range $0$ to $x^2+y^2$.
The latter is $g_2(x,y)$ in Deveno's explanation.

 

[Dethrone]

The formula for the volume should be:
$$V = \iiint dV = \iiint dz\,dy\,dx$$

The bounds specify through which range each of the coordinates goes.
You have the right range for the x- and y-coordinates.
However, the z-coordinate has the range $0$ to $x^2+y^2$.
The latter is $g_2(x,y)$ in Deveno's explanation.

Ah! I understand! :D

$$V = \iiint dV$$
$$= \int_{-1}^{1} \int_{-1}^{1} \int_{0}^{x^2+y^2} \,dz\,dx\,dy$$
$$= \int_{-1}^{1} \int_{-1}^{1} x^2+y^2\,dx\,dy$$
$$= \int_{-1}^{1} \frac{2}{3}dy$$
$$= \frac{4}{3}$$

Not sure if this is right. (Sweating)

 

[I like Serena]

Ah! I understand! :D

$$V = \iiint dV$$
$$= \int_{-1}^{1} \int_{-1}^{1} \int_{0}^{x^2+y^2} \,dz\,dx\,dy$$
$$= \int_{-1}^{1} \int_{-1}^{1} x^2+y^2\,dx\,dy$$
$$= \int_{-1}^{1} \frac{2}{3}dy$$
$$= \frac{4}{3}$$

Not sure if this is right. (Sweating)

Hmm... can you redo the following integral with a couple more steps? (Wondering)
$$\int\limits_{-1}^{1} (x^2+y^2)dx$$

 

[Dethrone]

Hmm... can you redo the following integral with a couple more steps? (Wondering)
$$\int_{-1}^{1} (x^2+y^2)dx$$

I had no idea what to do, but I thought of y^2 as a constant since we're integrating with respect to x.

$$= [\frac{x^3}{3}]_{-1}^1=\frac{2}{3}$$

EDIT: Somehow, I thought the integral of a constant was 0. Let me re-work this.

 

[I like Serena]

I had no idea what to do, but I thought of y^2 as a constant since we're integrating with respect to x.

$$= [\frac{x^3}{3}]_{-1}^1$$

Yep! Let's treat $y^2$ as a constant.
Let's suppose $y^2=2$ for a minute, so it really looks like a constant.
What is:
$$\int 2\,dx$$

 

[Dethrone]

Yep! Let's treat $y^2$ as a constant.
Let's suppose $y^2=2$ for a minute, so it really looks like a constant.
What is:
$$\int 2\,dx$$

It's $2x$. I think I can do this now. I knew it was a constant, so I got excited when I was doing it, but I really have no clue why I thought the integral of a constant was 0...

 

[I like Serena]

It's $2x$. I think I can do this now. I knew it was a constant, so I got excited when I was doing it, but I really have no clue why I thought the integral of a constant was 0...

Oh, I know.
The derivative of a constant is 0.
It means that the anti-derivative of 0 is a constant, which is the other way around.
I suspect you were just mixing them up, which is a pretty normal thing to do.

 

[Dethrone]

$$V=\int_{-1}^{1}\int_{-1}^{1} x^2+y^2 \,dx \,dy$$
$$=\int_{-1}^{1} [\frac{x^3}{3}+xy^2]_{-1}^1\,dy$$
$$=\int_{-1}^{1}\frac{2}{3}+2y^2\,dy$$
$$=\frac{2}{3}[y+y^3]_{-1}^1$$
$$=\frac{8}{3} units^3$$

:D :D :D