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Math Help - Triple Integral: Volume problem, Using cylidrical coordinates

  1. #1
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    Triple Integral: Volume problem, Using cylidrical coordinates

    Problem statement:
    A cylindrical drill with radius 1 is used to bore a hole through the center of a sphere of radius 3. Find the volume of the ring shaped solid that remains using cylindrical coordinates.

    My solution: 0=Theta | = integral (Sorry I don't know know Latex. If you can please direct me to a page with syntax and I will happily reformat my solution). Orient the cylindrical hole through the sphere such that the axis of the hole is parallel and on the z axis.

    ||| r dzdrd0

    Limits of integration (left to right): 0 to 2*pi, 1 to 3, -sqrt(9-r^2) to +sqrt(9-r^2)


    I would like to know why the limits of integration of z is -sqrt(9-r^2) to +sqrt(9-r^2). The radius is at most 3 and at least 1. However if r=3, z=0 which is true when r=3 on the xy-plane but a picture can be used to show that we can find a z=/=0 if r=3.

    It seems sqrt(9-r^2)=z doesn't give a correct relation between r and z at all for this problem. I expect to find the max of z (z value when the cylinder intersects the sphere).
    Attached Thumbnails Attached Thumbnails Triple Integral: Volume problem, Using cylidrical coordinates-problem-diag.png  
    Last edited by Elusive1324; December 8th 2013 at 05:30 PM.
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  2. #2
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    Re: Triple Integral: Volume problem, Using cylidrical coordinates

    quickly before I even look at the actual problem

    LaTeX Help
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    Re: Triple Integral: Volume problem, Using cylidrical coordinates

    also, can you not find the volume of the part removed and subtract it from the volume of the sphere? That's got to be an easier integral.
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    Re: Triple Integral: Volume problem, Using cylidrical coordinates

    Quote Originally Posted by Elusive1324 View Post
    Problem statement:
    A cylindrical drill with radius 1 is used to bore a hole through the center of a sphere of radius 3. Find the volume of the ring shaped solid that remains using cylindrical coordinates.

    My solution: 0=Theta | = integral. Orient the cylindrical hole through the sphere such that the axis of the hole is parallel and on the z axis.

    ||| r dzdrd0

    Limits of integration (left to right): 0 to 2*pi, 1 to 3, -sqrt(9-r^2) to +sqrt(9-r^2)
    My solution:
    \int^{2\pi}_{0}\int^{3}_{1}\int^{\sqrt{9-r^2}}_{-\sqrt{9-r^2}} r\,dz\,dr\,d\theta


    Is the expression correct?
    Edited: I used wolfram to do the integration which yields \frac{64\pi\sqrt{2}}{3}. I crossed check this value with the online homework system and it says that this is correct.

    O.K

    Next, I'm going to try to verify this using spherical coordinates.
    Last edited by Elusive1324; December 26th 2013 at 06:15 PM.
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    Re: Triple Integral: Volume problem, Using cylidrical coordinates

    Quote Originally Posted by romsek View Post
    also, can you not find the volume of the part removed and subtract it from the volume of the sphere? That's got to be an easier integral.
    I second this. The material removed consists of a cylinder of radius 1 and length [itex]4\sqrt{2}[/itex] and the two "caps" at each end of the cylinder. Calculating their volume is the only "hard" part.
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    Re: Triple Integral: Volume problem, Using cylidrical coordinates

    Finding the volume of the cylinder: V=\pi(r^2)h=\pi(h) = 4\pi\sqrt{2}. 'r', the radius of the cylinder is 1. 'h' = 4√2 is given in the diagram in the attachment.

    Volume of 1 Cap (Cavalieri's Principle):

    dV=\pi(r^2)*dx and  r=f(x)=\sqrt{9-x^2}
    Vcap = \pi\int^{3}_{2\sqrt{2}} 9-x^2\,dx = 18-\frac{38\sqrt{2}}{3}

    2*Vcap + 4\pi\sqrt{2} \neq \frac{64\pi\sqrt{2}}{3}

    Did I miss something?

    Edited: Oh wait... subtract this from the volume of the sphere.
    Edited: I evaluated (4/3)(pi)(27)-4pi*sqrt(2)-2*[18-38*sqrt(2)/3] == 64pi*sqrt(2)/3 in wolfram alpha to check if the two values are equal but they aren't. So something in my calculation went wrong.
    Attached Thumbnails Attached Thumbnails Triple Integral: Volume problem, Using cylidrical coordinates-diag1.png  
    Last edited by Elusive1324; December 27th 2013 at 02:14 PM.
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    Re: Triple Integral: Volume problem, Using cylidrical coordinates

    Edited: Forgot the pi in front of a term. This expression is true: (4/3)(pi)(27)-4pi*sqrt(2)-2pi*[18-38*sqrt(2)/3] == 64pi*sqrt(2)/3
    On the left hand side it's the volume minus the cylinder and the two caps. The right side is obtained by triple integrate in cylindrical coords.
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