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Double integral problem for finding volume

  • Thread starter arl146
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Homework Statement


A cylindrical drill with radius r1 is used to bore a hole throught the center of a sphere of radius r2. Find the volume of the ring shaped solid that remains.


Homework Equations


x=r*cos(theta)
y=r*sin(theta)


The Attempt at a Solution


i know that the boundaries are:
for theta: 0 to 2*pi
for r: r1 to r2

i just dont know the equation to use in the double integral.. the equation of the cylinder is x^2+y^2= r1^2 i know that much.
 

Answers and Replies

  • #2
Simon Bridge
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Volume integrals are triple integrals (3D right?)
The cartesian equation for a sphere is

[itex]x^2+y^2+z^2=r^2[/itex]
You're equation for the bit drilled out fails to account for the cylinder end-caps, which are sections of the sphere.

Why not use the solid of rotation method? It reduces to an area integral - which is probably what you are looking for.

Sketch the situation - I'm a fan of this and you'll find that sketch is often worth a lot of marks so it's worth taking a bit of trouble.
 
  • #3
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i sketched it on like the x-y axis ... yea its triple integrals i guess.... but we're still in the double integrals section of the book...
so i figured that was the equation for a sphere but that still got me no where haha
 
  • #4
Simon Bridge
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Look up "solid of rotation".

Can you find the area between [itex]y=\sqrt{r^2-x^2}[/itex] and the line [itex]y=a: a<r[/itex] ?

Rotate that about the x axis and you have your volume with only a double-integral.
 
  • #5
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im confused
 
  • #6
HallsofIvy
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First, rather than directly calculating the volume of wood left, I would find the volume of wood removed and subtract from the volume of the sphere.

There are really three parts of the hole that you need to consider. First, there is a cylinder of raidius [itex]r_1[/itex] and height h. Note that h is NOT [itex]2r_2[/itex], the diameter of the sphere, because the drill will also take of two "caps" at each end. Those are the other three parts.

Draw a picture, a circle with a rectangle through it center. Draw a line from the center of the circle to the point where the rectangle cuts the circle. Draw a second line from the center of the circle perpendicular to the rectangle and the line from the point where that line cuts the rectangle to meet the first line. You should see a right triangle. The hypotenuse is a radius of the sphere and its length is [itex]r_2[/itex]. The second line is a raidus of the hole and is [itex]r_1[/itex]. The third line is half the length of the cylinder and its length is [itex]h/2= \sqrt{r_2^2- r_1^2}[/itex]. So the volume removed in that cylinder is [itex]\pi hr^2= \pi r_1^2\sqrt{r_2^2- r_1^2}[/itex].

Now, calculate the volume of the two "caps" removed at each end of the cylinder. Those are the same, of course, so you really only have to find one, then double. This can be done as a "volume of revolution". Take the portion of the circle [itex]x^2+ y^2= r_2^2[/itex] above the line [itex]y= \sqrt{r_2^2- r_1^2}[/itex] and rotate around the y-axis.
 
  • #7
HallsofIvy
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I have added this as a separate post since it is not directly relevant.

I have seen a similar problem in which we are told that "a hole four inches long" is drilled through a sphere and we are asked to find the volume left. Neither the radius of the hole nor the radius of the sphere is given! Now by doing the calculation as I outlined, you can show that they are, in fact, not relevant. As long as the relation between the two radii is such that the hole is 4 inches long, you get the same volume left.

But a clever way of doing that problem is to argue that, since the two radii are not given, they are not relevant to the problem! So take a hole of "radius" 0. Now the diameter of the sphere must be 4 inches and no wood is removed. Find the volume of a sphere of diameter 4 and you have the correct answer to the problem!
 
  • #8
Simon Bridge
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@Hallsofivy - he does not need to consider the end-caps of the cylinder if he carefully chooses the limits of his integration.

@arl146: cannot help you unless you tell me what you are confused about.

Did you look up "solid of rotation" as suggested?
[itex]V=\pi \int_p^q |f(x)-g(x)|dx[/itex]

Draw a picture of the crossection of the solid (srsly: do it!) - center the hole along the x axis, then you will see :

f(x) is the equation of the positive y half of the circle in x-y radius r,
g(x) is the horizontal line at y=[radius of the hole]
choose limits p and q to be the intersection of g(x) with f(x) - so the end-bits will get excluded automatically.
 
  • #9
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I just had that exact same question in my book. The equation for a sphere is
R^2=x^2+y^2+z^2 so,
z=sqrrt(R^2-x^2-y^2)

I found that the easiest way to do that is to convert it into polar coordinates, which gives you....
z=sqrrt(R^2-r^2), R being the constant value of the radius, and r being the variable value of the radius.


V=∫[itex]^{2π}_{0}[/itex] ∫[itex]^{R}_{b}[/itex] r*(R[itex]^{2}[/itex]-r[itex]^{2}[/itex])[itex]^{\frac{1}{2}}[/itex], b being the cylindrical radius.

And then solve. I think that should give you it.

EDIT: Although... now that I look at it... I'm kinda retarded. I'll be right back.
 
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  • #10
Simon Bridge
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@Kingptka: yep - polar co-ords as what the solid of rotation does, Since the function does not depend on the angular component, you just get a pi out the front. You forgot to the z integral.

You sketched the z-r graph right?
 
  • #11
Simon Bridge
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Here's what I'm talking about:
attachment.php?attachmentid=40768&stc=1&d=1320736627.png

The ball is, here, approximated by a polyhedron.
A wedge has been cut out of the ball for clarity.
a and b are limits of integration chosen to exclude the end-caps.
Notice that z=a is not the same as z=r2?

I've suggested solid of revolution method, but you can also slice this parallel to the x-y plane - each slice will be a disk with a hole in it with thickness dz.

Each disk has inner radius r1 and an outer radius r(z) and thickness dz.
The volume of each disk is [itex]\pi (r^2(z)-r_1^2)dz[/itex] - the total volume is the sum of all the disks. Just stick an integration sign in front of the expression :)

[note: you may use the pic under creative commons, attribution, no-deriveratives 3.0 (nz)]
 

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  • #12
ehild
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One more picture...The left figure is the side view, it is sliced, every slice has the thickness Δz. The right figure is the horizontal the cross section of the green slice: two concentric circles, with radii r and a. The inner circle is the hole, you need to find the area A between the circles, as the volume of the slice is ΔV=AΔz. r is obtained from the position z of this slice and the radius of the sphere, R:
[tex]r^2=R^2-z^2[/tex]
The area of the slice at height z is
[tex]A(z) = (r^2-a^2)˛\pi[/tex].
You can get it by integrating in polar coordinates, but I hope you are allowed to use the formula for the area of a circle. To get the volume of the whole thing, you need to sum of the volume elements, ΔV=AΔz, that is, calculating the integral
[tex]V=\int_{-Z}^Z{A(z)dz}[/tex]
You have to integrate from -Z to Z: Z is the height where the cylinder crosses the sphere [itex]Z=\sqrt{R^2-a^2}[/itex].

ehild
 

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  • #13
Simon Bridge
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Don't think we can do better than literally drawing OP a picture :)
 
  • #14
ehild
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Don't think we can do better than literally drawing OP a picture :)
Well, the OP seems to be scared away, I think. A simple picture can help more than a clever explanation, but students usually do not draw anything. My sketch is illustration to your explanation, Simon.


ehild
 
  • #15
Simon Bridge
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I think the main gotcha in this problem is picking the limits.

Students are very reluctant to draw pictures - you are right. I suspect this is because you have to draw the pic before you know what the right answer is and students are scared of writing/drawing the wrong thing.

We tend to train them into thinking in terms of putting numbers into equations too ... so they end up with an algorithmic approach to problem solving... plugging numbers into equations. But it's this fear of being wrong that holds so many back.

I see it all through into senior college too. But the post-grads seem to have overcome it (or developed astounding memories.)

It's a tough problem.
 

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