Question re: Limits of Integration in Cylindrical Shell Equation

[HermGnos]

Homework Statement

Use cylindrical shells to find the volume of the solid generated when the region enclosed by the given curves is revolved about the y-axis:

Relevant Equations

x^2 + y^3 = 4, x=0, x=4, y=0

I have managed to get the answer given by the textbook I'm referencing: 3π (∛4) (1 + 3∛3)

However, this took multiple attempts, as I was initially trying to integrate within domain x = 0 - 2. This is the area for the bit that's above the x-axis (y=0 as specified). But the above answer is only gotten when domain x = 2 - 4 is used. This span would seem equally valid, just making y=0 the upper limit of y.

My real question is just: is there something in the question / format of the equations above that should have clued me in to the fact that we're integrating the data under y=0 rather than above it? It seems like every other question in this book with a similar format that only mentioned a single x=0 or y=0 and no other corresponding value for x or y was using the =0 as the lower limit, so I'm wondering if I'm missing something here that should have clued me in.

Apologies also for what's probably very imprecise math language in the question above.

 

[wrobel]

There is a nice theorem: the volume of the body of rotation is equal to the area of the figure in the xy plane multiplied by the length of the circle that the center of mass of the planar figure describes

 

[PeroK]

Homework Statement: Use cylindrical shells to find the volume of the solid generated when the region enclosed by the given curves is revolved about the y-axis:
Relevant Equations: x^2 + y^3 = 4, x=0, x=4, y=0

Are you sure that doesn't say $x =2, x =4$?

I have managed to get the answer given by the textbook I'm referencing: 3π (∛4) (1 + 3∛3)

However, this took multiple attempts, as I was initially trying to integrate within domain x = 0 - 2. This is the area for the bit that's above the x-axis (y=0 as specified). But the above answer is only gotten when domain x = 2 - 4 is used. This span would seem equally valid, just making y=0 the upper limit of y.

I assume the question meant $x =2$ and $y \le 0$.

My real question is just: is there something in the question / format of the equations above that should have clued me in to the fact that we're integrating the data under y=0 rather than above it? It seems like every other question in this book with a similar format that only mentioned a single x=0 or y=0 and no other corresponding value for x or y was using the =0 as the lower limit, so I'm wondering if I'm missing something here that should have clued me in.

Apologies also for what's probably very imprecise math language in the question above.

 

[HermGnos]

Thanks guys. PeroK, can def confirm that the question sets the x boundaries at 0 and 4 and just states the boundary of y=0, pretty much formatted exactly as in the "Relevant Equations" above, carets aside. If it sounds like it might just be a poorly or improperly defined question, I'm happy to hear it; it's the first real one I've hit in a couple hundred pages of this book, though, and I wasn't about to assume that it was potentially faulty and I was right, haha. Definitely wanted to know if I was missing something in the reading.

 

[PeroK]

Thanks guys. PeroK, can def confirm that the question sets the x boundaries at 0 and 4 and just states the boundary of y=0, pretty much formatted exactly as in the "Relevant Equations" above, carets aside. If it sounds like it might just be a poorly or improperly defined question, I'm happy to hear it; it's the first real one I've hit in a couple hundred pages of this book, though, and I wasn't about to assume that it was potentially faulty and I was right, haha. Definitely wanted to know if I was missing something in the reading.

I haven't checked the answer, but if the volume of solid from $x =2$ to $x =4$ is the answer, then it's a simple typo in the question.

 

[vela]

I find you have to integrate from $x=0$ to $x=4$ to get the answer given in the textbook.

 

[PeroK]

I find you have to integrate from $x=0$ to $x=4$ to get the answer given in the textbook.

I found this quite confusing. In terms of the graph, there is an area above the $x$ axis, from $x =0$ to $x =2$. Call this area 1. Then, there is an area below the $x$ axis from $x =0$ to $x =4$. Call this area 2. This contiguous area is not the basis for the volume. Instead, there is a second area below the $x$ axis bounded by the graph and $x =4$. This is the complement of area 2. Call this area 3.

I get the book answer when I use areas 1 and 3. That was not at all obvious to me. My first thought was to use areas 1 and 2.

 

[PeroK]

PS to be precise:

Area 1 is bounded by $x =0, y =0$ and the graph of the function above the $x$ axis.

Area 2 is bounded by $x=0,y=0$ and the graph below the $x$ axis, until $x =4$.

Area 3 is bounded by $x=4, y =0$ and the graph below the $x$ axis.

 

[HermGnos]

Thank you all. I see and am able to work this out properly with the above. I'm evidently a goofball, because I have no idea now how I got the right answer using only 2 - 4 (twice!). But doing from scratch a few times and keeping the above in mind, I can now see what's going on and at least what's expected here, so thank you again!

 

[vela]

PS to be precise:

Area 1 is bounded by $x =0, y =0$ and the graph of the function above the $x$ axis.

Area 2 is bounded by $x=0,y=0$ and the graph below the $x$ axis, until $x =4$.

Area 3 is bounded by $x=4, y =0$ and the graph below the $x$ axis.

I'm not sure what your Area 2 is. This is the graph of the function I made using Mathematica. Is that what you have?

 

[PeroK]

I'm not sure what your Area 2 is. This is the graph of the function I made using Mathematica. Is that what you have?
View attachment 329374

Yes. Area 2 is everything to the left of that curve below the $x$ axis. It's contiguous with Area 1, which is to the left of that curve above the $x$ axis. The resulting solid would be roughly bell shaped. That was my first thought, but I didn't get the book answer.

The book answer entails Area 3, which is to the right of that curve below the $x$ axis. That fits the problem statement, but seemed unintuitive to me. It's more like two solids joined together.

 

[vanhees71]

I understood the question such that you have to calculate
$$V=\pi \int_{y_1}^{y_2} \mathrm{d} y x^2=\pi \int_{-12^{1/3}}^{4^{1/3}} (4-y^3)=4^{1/3}(3+7\cdot 3^{1/3}).$$
The boundaries are chosen such that $x \in [0,4]$ as given in the question. That, however, differs from the given solution. Are you sure that the question is accurately as given?

 

[PeroK]

I understood the question such that you have to calculate
$$V=\pi \int_{y_1}^{y_2} \mathrm{d} y x^2=\pi \int_{-12^{1/3}}^{4^{1/3}} (4-y^3)=4^{1/3}(3+7\cdot 3^{1/3}).$$
The boundaries are chosen such that $x \in [0,4]$ as given in the question. That, however, differs from the given solution. Are you sure that the question is accurately as given?

That's another option, but doesn't give the book answer. See my posts above.

 

[vanhees71]

Precisely. I don't now, what's the discussion about areas all about. It was asked for the volume of the body defined by revolving the curve around the $y$ axis, and that's what I've calculated. If I didn't make some stupid mistake with the boundaries, that should give the solution, but it differs from the book's answer. So either I've misunderstood the question, made a stupid mistake (although checking with Mathematica), or the answer in the book is wrong...

 

[PeroK]

Precisely. I don't now, what's the discussion about areas all about. It was asked for the volume of the body defined by revolving the curve around the $y$ axis, and that's what I've calculated. If I didn't make some stupid mistake with the boundaries, that should give the solution, but it differs from the book's answer. So either I've misunderstood the question, made a stupid mistake (although checking with Mathematica), or the answer in the book is wrong...

Look at the diagram in post #10. The graph and the x-axis split the plane into four areas. You used the two on the left. That was what I did initially.

The book answer uses the the top left and bottom right. Note that, as I said in a previous post, these two areas are not contiguous.

 

[PeroK]

$$V=\pi \int_{0}^{y_2} \mathrm{d} y x^2+\pi \int_{-12^{1/3}}^{0} (4^2-x^2)^dy$$

Try the above.

 

[vela]

Precisely. I don't now, what's the discussion about areas all about. It was asked for the volume of the body defined by revolving the curve around the $y$ axis, and that's what I've calculated. If I didn't make some stupid mistake with the boundaries, that should give the solution, but it differs from the book's answer. So either I've misunderstood the question, made a stupid mistake (although checking with Mathematica), or the answer in the book is wrong...

Here's a plot with the four bounds given in the problem. What @PeroK called areas 1 and 3 are the only two areas that are bounded. The problem was to use cylindrical shells to calculate the volume when those areas are rotated about the y-axis.

 

[vanhees71]

Yes, and that's what I calculated. So where is the mistake? In my calculation or is the book's solution simply wrong?

 

[PeroK]

Yes, and that's what I calculated. So where is the mistake? In my calculation or is the book's solution simply wrong?

That's not what you calculated according to what you posted. You used the areas adjacent to the y axis.

 

[vanhees71]

Which areas? I just calculated the volume of the body I get when rotating the blue curve around the $y$-axis. The question is pretty ambigously stated. Here's another interpretation of the text:
$$V'=\pi \int_0^{4^{1/3}}\mathrm{d} y (4-y^3)+\pi \int_{-12^{1/3}}^0 \mathrm{d} y (12+y^3),$$
and this indeed gives the answer in #1.

It's pretty tricky, how to interpret the question.

 

[vanhees71]

Which areas? I just calculated the volume of the body I get when rotating the blue curve around the $y$-axis. The question is pretty ambigously stated. Here's another interpretation of the text:
$$V'=\pi \int_0^{4^{1/3}}\mathrm{d} y (4-y^3) + \pi \int_{-12^{1/3}}^0 \mathrm{d} y (12+y^3).$$

 

[PeroK]

Which areas? I just calculated the volume of the body I get when rotating the blue curve around the $y$-axis. The question is pretty ambigously stated. Here's another interpretation of the text:
$$V'=\pi \int_0^{4^{1/3}}\mathrm{d} y (4-y^3)+\pi \int_{-12^{1/3}}^0 \mathrm{d} y (12+y^3),$$
and this indeed gives the answer in #1.

It's pretty tricky, how to interpret the question.

You got there in the end.

 

[PhDeezNutz]

I'd just like to add that the question states we should use the method of cylindrical shells but i do agree the question was poorly stated by the book. I think that $y = 0$ was not needed/incorrect.

$V = 2 \pi \int x y\left(x\right) \, dx$

Where of course $x$ is the radius. $y\left(x\right)$ is the height. $dx$ being the thickness of the shell.

If $y \left( x \right)$ is the height we need to make sure it's positive.

$y(x) = \begin{cases} \left(4 - x^2 \right)^{\frac{1}{3}}, & 0 \leq x \leq 2 \\ -\left(4 - x^2 \right)^{\frac{1}{3}}, & 2 \leq x \leq 4 \\ \end{cases}$

If I evaluate

$V = 2 \pi \int_{0}^{2} x \left(4 - x^2 \right)^{\frac{1}{3}} \, dx - 2 \pi \int_{2}^{4} x \left(4 - x^2 \right)^{\frac{1}{3}} \, dx$

I get the same answer in the OP.

 

[HermGnos]

Not to resurrect the dead but haven't hopped on here in a bit and just wanted to say thanks all again, really. I see what I was doing wrong, but also see that it was a confusingly phrased question, which definitely gives some comfort. I very much appreciate how the discussion's helped break things down further to provide some fodder for playing with different interpretations, honestly. Probably straightforward for most of you, but as a middle-aged dude trying to learn math without a teacher or mathematically inclined people in his circle, definitely cool.

 

[PhDeezNutz]

Not to resurrect the dead but haven't hopped on here in a bit and just wanted to say thanks all again, really. I see what I was doing wrong, but also see that it was a confusingly phrased question, which definitely gives some comfort. I very much appreciate how the discussion's helped break things down further to provide some fodder for playing with different interpretations, honestly. Probably straightforward for most of you, but as a middle-aged dude trying to learn math without a teacher or mathematically inclined people in his circle, definitely cool.

Hey man that’s what this place is all about! Glad you gained some clarity. And I commend you for learning math later in life.