I keep getting negative volumes (volume of sin(x) around y=c, c is in [0,1])

Homework Statement

The arch y= sin(x), x is in [0, pi], is revolved around the line y=c, where c is a constant in [0, 1], to generate a solid...

Anyway, then I have to represent the volume of the solid as a function of c and other stuff.

The Attempt at a Solution

I remember asking in class and the proff said not to find the points of intersection with c.

The volume of the solid would be the integral of sin(x)^2 - c^2 from x=0 to x=1.
OR the integral of c^2 - sin(x)^2, x=0--->1.

Either way, there are values of x for which sin(x) is larger than c, and some where it is smaller. And either way, there are values of c for which I get a negative number, which makes no sense. What am I doing wrong? I can't think of any way of doing this without finding the intercepts, which is said specifically not to do.

Related Calculus and Beyond Homework Help News on Phys.org
LCKurtz
Homework Helper
Gold Member
I think this problem is either not well posed or well understood. You don't rotate "the arch". You rotate an area, presumably the area under the arch, to create a volume. As you have noticed, if c is in (0,1) the area wraps on itself. Your teacher needs to clarify what he wants you to do about that. Does he not want to count the doubled part? It is a poorly phrased problem as you give it.

 Also, did the question ask to rotate the area between the curve and the line?

Last edited:
I think this problem is either not well posed or well understood. You don't rotate "the arch". You rotate an area, presumably the area under the arch, to create a volume. As you have noticed, if c is in (0,1) the area wraps on itself. Your teacher needs to clarify what he wants you to do about that. Does he not want to count the doubled part? It is a poorly phrased problem as you give it.
I copied that from my assignment. When I asked him, he said the volume will look like a candy wrapper. Which is what happens if I find the sin(x) intercepts with c, and take the areas either under c when c > sin(x), or under sin(x) when sin(x) > c. But he said that this is not to be done, that it should be a single integral, (he added that it should be the area under the curve).

This thing is giving me a headache because it looks completely contradictory. How can I take the area under sin(x) without finding the intercepts and not get negative volumes at some values of c?

LCKurtz
Homework Helper
Gold Member
I think this problem is either not well posed or well understood. You don't rotate "the arch". You rotate an area, presumably the area under the arch, to create a volume. As you have noticed, if c is in (0,1) the area wraps on itself. Your teacher needs to clarify what he wants you to do about that. Does he not want to count the doubled part? It is a poorly phrased problem as you give it.

 Also, did the question ask to rotate the area between the curve and the line?
I copied that from my assignment. When I asked him, he said the volume will look like a candy wrapper. Which is what happens if I find the sin(x) intercepts with c, and take the areas either under c when c > sin(x), or under sin(x) when sin(x) > c. But he said that this is not to be done, that it should be a single integral, (he added that it should be the area under the curve).

This thing is giving me a headache because it looks completely contradictory. How can I take the area under sin(x) without finding the intercepts and not get negative volumes at some values of c?
From what you have written I think you understand very well that there is something wrong with the statement of the problem. The only interpretation that makes any sense to me is to rotate the area between the line and the curve about the line. So the integrand on the ends would be π(c2 - sin2(x)) and on the middle part would be π(sin2(x) - c2). And I agree with you, you need the values where sin(x) = c to do the limits.

Last edited:
LCKurtz
Homework Helper
Gold Member
And, for extra credit, find the value of c for which the volume generated between the arch and the line is a minimum.

And, for extra credit, find the value of c for which the volume generated between the arch and the line is a minimum.
That's part (a) (and the max)

I guess I'll go head and do it with the intercepts. An answer is an answer right?

I think I know what I have to do. The radius of the figure will essentially be the absolute value of c-sin(x), so the integral is just pi*(c-sin(x) )^2. That's how you get the shape he drew on the board when I asked. ... I think the question could have been worded differently, or at least included the figure he drew, because without it it can be interpreted in two different ways :/

LCKurtz
Homework Helper
Gold Member
I think I know what I have to do. The radius of the figure will essentially be the absolute value of c-sin(x), so the integral is just pi*(c-sin(x) )^2. That's how you get the shape he drew on the board when I asked. ... I think the question could have been worded differently, or at least included the figure he drew, because without it it can be interpreted in two different ways :/
Careful, you mean π|c2-sin2(x)|, not what you wrote. And you still need those sin-1(c) limits to evaluate it.

Careful, you mean π|c2-sin2(x)|, not what you wrote. And you still need those sin-1(c) limits to evaluate it.
Wouldn't pi*(c-sin(x))^2 work in this case? -- the area is being revolved around c, and the distance from c to the curve of sin(x) is |c-sin(x)|. So choosing |c-sin(x)| to be R, would make the integral pi*R^2.

|c^2-Sin(x)^2| would be the area under sin(x). This is the figure he drew in class when I asked:

|>O<|

like a candy wrapper.

From the figure, I think he wants the area between c and sin(x).

LCKurtz
Homework Helper
Gold Member
Wouldn't pi*(c-sin(x))^2 work in this case? -- the area is being revolved around c, and the distance from c to the curve of sin(x) is |c-sin(x)|. So choosing |c-sin(x)| to be R, would make the integral pi*R^2.

|c^2-Sin(x)^2| would be the area under sin(x). This is the figure he drew in class when I asked:

|>O<|

like a candy wrapper.

From the figure, I think he wants the area between c and sin(x).
Yes, you are correct. I responded too quickly this morning and was thinking revolving about the x axis when I wrote that formula.