# Thread: Volume - sin(x) and cos(x) quick question

1. ## Volume - sin(x) and cos(x) quick question

Ok so i just took a math test for my calc II class and this question was on there and im just not sure about it.

Find the volume of the region between the curves y=sin(x) and y=cos(x), x > 0 rotated about the y-axis.

So i graphed sin(x) and cos(x) first.

when x > 0 and no boundaries are given (such as x=1) then i thought that this problem could not be calculated because when rotating those curves about the y-axis with no limitation in the x direction then the volume will just be infinitely large.

Does what i said above ^^^^ seem correct?

Thanks for any help

2. Originally Posted by mybrohshi5
Ok so i just took a math test for my calc II class and this question was on there and im just not sure about it.

Find the volume of the region between the curves y=sin(x) and y=cos(x), x > 0 rotated about the y-axis.

So i graphed sin(x) and cos(x) first.

when x > 0 and no boundaries are given (such as x=1) then i thought that this problem could not be calculated because when rotating those curves about the y-axis with no limitation in the x direction then the volume will just be infinitely large.

Does what i said above ^^^^ seem correct?

Thanks for any help
Unfortunately, I think that's not true, because, even though the boundaries are not given, you have to calculate them. On the graph (where x > 0), you will notice that the graphs cross each other quite early. That is right-hand boundary of the region because the area between the curves is the area within the intersection.

You can use pre-calculus to calculate the intersection without a calculator:
lowest value of

$\sin(x) = \cos(x)$

(since the intersections occur when both functions have the same values)

Divide both sides by cos(x):

$\tan(x) = 1 \implies x = tan^{-1}(1)$

You should remember from your precalc days that this is equal to $\frac{\pi}{4}$.

Now, some calc I formulas to calculate the volume...
The y-axis is the line x=0.

$V_{x=0} = \int_0^{\frac{\pi}{4}} 2\pi R H dx = 2\pi \int_0^{\frac{\pi}{4}} R H dx$

R is the distance from the elemental strip to the axis of rotation, which in the case of the y-axis, is simply x.
H is the height of the elemental strip, which is the higher function minus the lower function. You know the cosine functions starts at 1, which is higher than the sine function.
So:
$V_{x=0} = 2\pi \int_0^{\frac{\pi}{4}} x (\cos x - \sin x) dx$

which I'm sure you can evaluate using integration by parts.

Hope I helped

3. Hi. I don't think so. You could have given the volume as a function of x. Look at the plot below and consider the volume using cylindrical shells where the volume is given by $V=\int_a^b 2\pi x f(x)dx$ where $f(x)$ is the height of the shell. In the case below, the height is the difference between the sin and cos function. So we could write for the first red region:

$Vr1(x)=\int_0^x 2\pi u(\cos(u)-\sin(u))du,\quad 0\leq x\leq \pi/4$

Now rotate the first blue section:

$Vb1(x)=\int_{\pi/4}^x 2\pi u(\sin(u)-\cos(u))du,\quad \pi/4\leq x \leq 5\pi/4$

and so on.

4. See the thing is it was a multiple choice question so i dont think it wanted it in terms of x like shawsend did.

I dont remember the exact answers but 4 of them were numbers (like .5 - 2.6 roughly) and 1 was "none of the above".

So mathemagister yours seems like it would be right, but i dont get why it would be that because what about the volume between all of the other regions formed by the two curves? (as shown in shawsend's attached thumbnail). Why arent those parts taken into consideration when finding the volume?

Thanks

5. Originally Posted by mybrohshi5
Find the volume of the region between the curves y=sin(x) and y=cos(x), x > 0 rotated about the y-axis.
In my experience, when a question is worded like this, it probably means the region that borders y=sin x, y=cos x, and x=0. I understand the confusion caused by the question though because there are infinitely many regions to the right of the y-axis. However, only one of those actually "touches" x=0.

So, I would guess that the intent of the problem was to calculate the volume between x=0 and x=pi/4.

But the only way to be sure is to ask your teacher/professor.

6. Thanks drumist.

I get what you are saying and if it would have said x=0 i probably would have evaluated it like that but since it said x>0 i was confused.

If i get it wrong, which i probably will cause im sure thats what it meant, i will talk to my professor and argue my reasoning and maybe she will give me some sympahty

thanks again

7. Originally Posted by mybrohshi5
See the thing is it was a multiple choice question so i dont think it wanted it in terms of x like shawsend did.

I dont remember the exact answers but 4 of them were numbers (like .5 - 2.6 roughly) and 1 was "none of the above".

So mathemagister yours seems like it would be right, but i dont get why it would be that because what about the volume between all of the other regions formed by the two curves? (as shown in shawsend's attached thumbnail). Why arent those parts taken into consideration when finding the volume?

Thanks
I have to agree that you are right in saying that there is some ambiguity in a question, but this is an extremely common type of Calculus (really Calc I) problem (the only REAL Calc II part would be actually evaluating the integral without a calculator if it was a non-calculator question). It is conventionally understood that "the region between the curves" is the area "enclosed" between two curves. That is to say, once the curves intersect, the area on the other side is irrelevant.

Hope this helps