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Math Help - Volume Integral - Cylinder & Cone

  1. #1
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    Volume Integral - Cylinder & Cone

    Hello all,

    In mathematical terms, I've got a problem where I'd like to calculate the volume of a cylinder whose upper surface is bounded by a right circular cone. A brief discription of the geometries follow, but it's probably best to look at the attachments.

    In the real world, a description of the problem follows: consider the cylinder an inground tank with radius 7.5 m and height 8.1 m. The tank is used to collect sand form the top edge [along the edge of the tank] such that the sand entering the tank creates the shape of half of a right circular cone of radius 14 m [and height 8.1 m]. Notice that the cone's base does not touch the far edge of the tank [in fact, it is 1 m away from the edge].

    What is the volume of sand in the tank?

    MY Attempt:
    1. I decided to use a triple integral in cylindrical coordinates.
    2. I defined my cylinder equation
    3. I defined my cone equation
    4. I identified the angles to integrate through
    5. Compute the triple integral

    But my answer is obviously wrong since the volume that I'm calculating is larger than the total volume of the tank.

    Can anyone offer help? I've checked and re-checked. What am I not seeing? Is my method wrong? Or is my arithmetic wrong? Any help would be greatly appreciated!!

    Also, I would have loved to type out the question in the thread -- but sadly, i dont know how to.

    Thanks in advance!

    Juggernaut.
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  2. #2
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    Re: Volume Integral - Cylinder & Cone

    First off, why are you using the entire base of the cylinder when there's no sand in the cone from 14 to 15.
    Second, really a question, is the base of the cone entirely inside the circular base of the cylinder?
    Next, why are you using

    z = h + \frac{r}{c} instead of z = h - \frac{r}{c}

    Your formula says that when r = 0 then z = h (correct) but as you increase r, z increases instead of decreases.
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    Re: Volume Integral - Cylinder & Cone

    Thanks a lot Danny! Responding to your questions...

    1. "why are you using the entire base of the cylinder when there's no sand in the cone from 14 to 15. "
    - Shouldn't I? Wont the surface component of the integral (i.e. the cone's surface) take this into account? If not, how would I qo about account for this?

    2. "is the base of the cone entirely inside the circular base of the cylinder?"
    - No, it cant be, the radius of the cone is far bigger than that of the cylinder. Am I missing something here?

    3. "Your formula says that when r = 0 then z = h (correct) but as you increase r, z increases instead of decreases. "
    - Great catch!! I should have seen that! Thanks!

    However, based on comments 1 and 2, I'm guessing that I'm missing more to this question. Any further insights?

    Thanks so much!
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    Re: Volume Integral - Cylinder & Cone

    Ya, here's some more. Since you are looking for volume then what I would do is

    \iint \limits_R z dA

    where z = h - \dfrac{h}{a}\sqrt{x^2+y^2} (the cone with height h and radius a) and R the region, in the xy plane, when the part of the cone inside the cylinder is projected down into the xy plane. If we view this downward what I see is two circles - one is the base of the cone and the other is the base of the cylinder. The region of integration in the area inside both circles. What I'm saying is that you're going to need two integrals to do this problem!
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    Re: Volume Integral - Cylinder & Cone

    So, what you're saying is:

    1. I need to do 2 integrals instead of 3?
    2. Forget about cylindrical coordinates
    3. Find the area of the base of the sand -- the area formed by the two circles you've described
    4. Use this area to compute the integral you specified above


    I'm tripping up over the dA in your integral above...How would I define it? Is it simply dxdy? I know it's the differential area element of the area in the two circles, but in my mind, I'd calculate this area and use it as a bound for my integral :S

    Thanks so much for your help!!!
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    Re: Volume Integral - Cylinder & Cone

    What you want to calculate is

    \iint \limits_R \left( h - \frac{h}{a}r\right) r dr d\theta

    where R is the region inside both circles
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    Re: Volume Integral - Cylinder & Cone

    Quote Originally Posted by Danny View Post
    What you want to calculate is

    \iint \limits_R \left( h - \frac{h}{a}r\right) r dr d\theta

    where R is the region inside both circles

    I think I get it! My integral will look like this...correct? :

    \int_{0}^{\pi}\int_{15\sin\theta}^{14} \left( h - \frac{h}{a}r\right) r dr d\theta

    Where R is defined by the area inside r = 14 and r = 15\sin\theta, and ranging from  {0} to {\pi} Correct?
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    Re: Volume Integral - Cylinder & Cone

    Not quite. First, there's some symmetry so integrate \theta from 0 to \pi/2 and multiply by two. Next, there's a point where these two circles intersect, call this \theta^*.  Then what you have is


    \int_0^{\theta ^*}{\int_0^{15 \sin \theta} \left(h - \frac{h}{a} r\right) r dr d \theta + \int_{\theta ^*}^{\pi/2}{\int_0^{14} \left(h - \frac{h}{a} r\right)r dr d \theta
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    Re: Volume Integral - Cylinder & Cone

    Amazing!! I see it!! I'll give it a shot, and I'll post back to let you know how it goes...Thank you so much for your help!!
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    Re: Volume Integral - Cylinder & Cone

    So I've computed the integral, and I'm getting a volume of about 871 cubic meters. This seems like it's in the right ball park.

    As a thought experiment though, lets assume that instead of a surface of a cone, it was a simple plane, defined by  z = h -\frac{h}{a}y that cut the cylinder...what scenario whould have more volume?

    My thought was that the volume under the plane would be larger.

    I did the calc explicitly, and found that my guess was wrong -- it was in fact he cone scenario.

    Does this make sense? I can post my work if you'd like to see -- but what makes sense in your mind?

    Thanks again in advance!!
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  11. #11
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    Re: Volume Integral - Cylinder & Cone

    I'm not sure what you did but I agree that I think the volume under the plane is more. In fact it is easy to show that (of course for h,a >0)

    h - \frac{h}{a}y \ge h - \frac{h}{a} \sqrt{x^2+y^2} for all (x,y).
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