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Triple Integral

Posted on 2006-11-02
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Here is a triple integration problem, where symmetry does not play a role and the order of integration must also be carefully thought out. This example uses Cartesian coordinates.

Find the mass of the volume of the cylinder x²+y²-2ax=0 which lies in the first octant* and under the paraboloid x²+y²=az assuming that the density is constant.

* ie: x,y,z considered positive.
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Question by:BigRat
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by:Infinity08
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Is there an upper limit for z too, otherwise it looks quite infinite to me ...
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by:BigRat
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Yes, the parabaloid cuts across the cylinder.
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by:Infinity08
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We must use different definitions for "under" then. Which axis goes up ? Y or Z ?
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by:BigRat
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Forgetting your 3-D cartesian coords then????

On plain paper x goes to the right, y goes to the top. z goes into the paper away from you.
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by:Infinity08
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>> On plain paper x goes to the right, y goes to the top. z goes into the paper away from you.
Exactly what I thought. So, since the Y axis goes up, that means that "under the paraboloid" refers to the part of the cylinder that goes from the paraboloid to z = +infinity.

Unless you define "under" along the Z axis, which means that the Z axis goes up. And the part of the cylinder we want goes from z = 0 to the paraboloid.

Anyway, since you're saying that it's limited by the paraboloid, I'll assume that you mean the latter. Unless I'm missing something ...
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by:BigRat
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I don't quite understand the "misunderstanding".

Think of a cylinder standing upright. Now slice it at an angle (say 60 degress) off the horizontal. The bit from the floor up to the slice is the volume (ie: mass) to be found. It is just that instead of a plane slice it is a paraboloid.
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by:Infinity08
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I understand that BigRat, but the cylinder is not standing up - it's lying flat on the "ground" (the X-Z plane).
But anyway, I'll just assume the "ground" as the X-Y plane, and Z as the vertical axis, and all is well.
That's just not the way I'm used to do it, that's all :)

Here's what I have :

Int[y = 0 .. 1][x = (a - sqrt(a^2 - y^2)) .. (a + sqrt(a^2 - y^2))][z = 0 .. (x^2 + y^2)/a] dz dx dy

= Int[y = 0 .. 1][x = (a - sqrt(a^2 - y^2)) .. (a + sqrt(a^2 - y^2))] (x^2 + y^2)/a dx dy

= Int[y = 0 .. 1] ((a + sqrt(a^2 - y^2))^3 - (a - sqrt(a^2 - y^2))^3)/(3a) dy
+ Int[y = 0 .. 1] 2y^2 sqrt(a^2 - y^2)/a dy

= Int[y = 0 .. 1] -2sqrt(a^2 - y^2)(4a^2 - y^2)/(3a) dy
+ Int[y = 0 .. 1] 2y^2 sqrt(a^2 - y^2)/a dy

= -8/(3a)  Int[y = 0 .. 1] (a^2 - y^2)^(3/2) dy

But got to go now ... maybe someone else can finish it, or fix errors if there are any :)
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by:jkmyoung
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1. Cylinder with center (a,0,z)
2. Paraboloid starting at origin.

Finding the intersection between the 2, we get 2ax = az -> z = 2x
Since the max value for x is 2a, after z = 2*2a = 4a, the paraboloid will totally encompass the cylinder.

Let us look first at z = 0.
At this point, the paraboloid is nothing, so we have the entire semicircle to add to the volume.
now look at 0 < z <= a
We have 2 circles of cutting across each other, one centered at (0,0,z), the other at (a,0,z). Notice that part of the circle part of our cylinder is above the circle from the paraboloid, therefore, we cannot count it.

Representation of paraboloid when y = 0:
-> Parabola z = x^2/a
-> x = sqrt(az)
So we can sum the area from x = sqrt (az) to 2a

Value of y-coordinate of semicircle given x:
y^2 + (x-a)^2 = a^2
y^2 = a^2 - (x-a)^2
y = (sqrt(a^2 - (x-a)^2))


So the integral is
Int (z = 0 to 4a) Int (x = sqrt(az) to 2a) Int (y = 0 to sqrt(a^2 - (x-a)^2)) 1 dy dx dz
= Int (z = 0 to 4a) Int (x = sqrt(az) to 2a) (sqrt(a^2 - (x-a)^2)) dx dz
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by:BigRat
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The order of integration is correct - dzdydx.

The cylinder is standing in the first octant, so the lower limits are?????

These sort of questions are cleverly constructed (if I may be so bold) so that no "funny" multiples are present. Hence what is the diameter of the cylinder, and therefore what is the radius?
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by:jkmyoung
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Using http://integrals.wolfram.com/index.jsp
Int (a^2 - y^2)^(3/2) dy =
1/8 [3 invtan(y / sqrt(a^2 - y^2)) a^4 + y (5a^2  - 2y^2) sqrt(a^2 - y^2))
not very pretty.

Mine is even worse.

Instead I suggest, try to calculate y in terms of angles,
draw the angle with points (0, 0), (a, 0), (x, y), let this be p

a cos (p) = a - x
a sin (p) = y
p = invcos((a-x)/a)
now x starts at sqrt(az)
p = invcos ((a - sqrt(az))/a)
p = invcos (1 - sqrt(z/a))

and p ends at pi


= Int [z = 0 to 4a] Int [p = invcos (1 - sqrt(z/a)) to pi]  a sin p dp dz
= Int [z = 0 to 4a] (- a cos p) | [p = invcos (1 - sqrt(z/a)) to pi]  dz
= Int [z = 0 to 4a] (-a cos (invcos (1-sqrt(z/a))) - (-a  cos pi) dz      cos and invcos cancel:
= Int [z = 0 to 4a] (-a (1-sqrt(z/a)) - a  dz
= Int [z = 0 to 4a] -a (2 - sqrt(z/a))  dz
This nicely works out to: (used http://integrals.wolfram.com/index.jsp)
= -2/3 a z (sqrt (z/a) - 3)  [z = 0 to 4a]
= -2/3 a 4a (sqrt (4a/a) - 3) - (-2/3 a * 0 (sqrt (0/a) - 3))
= -2/3 a 4a (2 - 3)
= 8/3 a^2

I hope I have not made any mistakes anywhere.
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by:jkmyoung
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Sorry, I forgot to mention,
Cylinder:
x²+y²-2ax = 0
<=> x²-2ax+a² -a² + y² = 0
<=> (x-a)² + y² = a²
Cylinder has center (a,0,z) with radius a.

If we look at any x-y plane, with constant z, we see a circle with center (a,0) and radius a. eg. left border on the origin for this plane (0,0,z), right border on (2a, 0, z), and top is (a, a, z)
=====

Paraboloid:
x²+y² = za
z = (x² + y²)/sqrt(a)²
radius is sqrt(a)
Looking at the paraboloid on any x-y plane with constant z, we see a circle with center (0,0) and radius sqrt(a)
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by:BigRat
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jkmyoung:

I don't see how you even started????

Basically it is a triple integral in the order dz, dy and dx. The order has been established but so far without justification.

The next problem is to determine the limits of the three integrations. To do this knowing the radius of the cylinder is a good idea, which will enable us to determine the limits of the dx-integration.

In any event your answer is wrong. A quick ball-park check says that, like the volume of a sphere, pi should turn up somewhere.
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by:jkmyoung
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What I did is merely a form of substitution, but I guess I'll have to explain it better.
The reason I tried to use angles is because I wanted a pi somewhere.

Correction:
Let us take any z-slice, eg. one x-y plane.
For the cylinder: we will always see the circle centered at (a,0) with radius 0.

Paraboloid:
x²+y² = za            *correction here.
x²+y² = sqrt(za)
For the paraboloid we will see a circle centered at (0,0) with radius sqrt (za)
This circle's rightmost point is (sqrt(0), za))

So, for this particular z plane, we are looking to find the portion of the circle where x >= za
Using pythagorus, we get y = sqrt(a^2 - (x-a)^2)

Now for z, z goes from 0, to at most 4a, since at 4a and larger,
x²+y² = sqrt(za) = radius of parabaloid's cirlce <= sqrt(4a²) = 2a
After z = 4a, all of the cylinder lies within the paraboloid.

In all:
Int [z = 0 to 4a] Int [x = sqrt(za) to 2a] Int [y = 0 to sqrt(a^2 - (x-a)^2)] dy dx dz
the y integral is easily simplifiable:
Int [z = 0 to 4a] Int [x = sqrt(za) to 2a] sqrt(a^2 - (x-a)^2) dx dz

Now is the hard part, finding the correct substitution. In order to explain why I'm doing this particular substitution, I will post a simpler example in the next post.





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by:jkmyoung
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Example: Find the area of a circle with radius 1 in the first quadrant
http://celtickane.com/school/math_circle.php explains a little more thoroughly.
x²+y² = 1
y = sqrt(1-x²)

A = Int [y = 0 to 1] sqrt(1-y²) dx
The classic substitution is x = sin(u), u = invsin(x)
dx = cos(u)

invsin 0 = 0
invsin 1 = pi/2

So, we get:
A = Int[u = 0 to pi/2] sqrt (1-sin(u)²) cos(u) du
= Int[u = 0 to pi/2] cos(u) cos(u) du

Now by the double angle formula, we have cos(2u) = 2 cos²(u) - 1
cos²(u) = (cos(2u) + 1) / 2
Subbing this back in, we have
A = Int[u = 0 to pi/2] (cos(2u) + 1) / 2 du
= -sin(2u)/4 +  u/2   [u = 0 to pi/2]
= (0 + pi/2/2)  (0 + 0/2)
= 0 + pi/4
= pi/4
=============================
My mistake was I forgot to calculate dp in terms of dx. Corrected post to follow.
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by:BigRat
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jkmyoung :

The order of integration has been established as dz, dy, dx, yet you are trying dy, dx, dz. Why?

The clue to the order was given by the word "under", which got queried for the wrong reasons, but Infinity08 did start off right. Now why that order of integration?  Because a line parallel to the y-axis gives different limits according as it cuts the parabaloid and cylinder or the xz plane and the cylinder. There is a similar difficulty when first integrating w.r.t. x, since in the cylinder x and y are "symmetric".
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jkmyoung earned 450 total points
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There's more than one way to do it; dz dy dx it is then.
Assuming the definition of 'under' is that of  'in front of'. that is, (x, y, z) is 'under' (x, y, z1) if z < z1.

Note: solve for y first.
(x-a)² + y² = a²
y² = a² - (x-a)²
y = sqrt(a² - (x-a)²)

Int [x = 0 to 2a] Int[y = 0 to sqrt(a² - (x-a)²)] Int [z = 0 to (x²+y²)/a] dz dy dx
Int [x = 0 to 2a] Int[y = 0 to sqrt(a² - (x-a)²)] (x²+y²)/a dy dx
Int [x = 0 to 2a]  yx²/a + y^3/3a  dx | [y = 0 to sqrt(a² - (x-a)²)]
Int [x = 0 to 2a]  sqrt(a² - (x-a)²)x²/a + sqrt(a² - (x-a)²)^3/3a  dx
Int [x = 0 to 2a]  sqrt(a² - (x-a)²)x²/a + sqrt(a² - (x-a)²)(a² - (x-a)²)/3a  dx
1/3a Int [x = 0 to 2a]  3x² sqrt(a² - (x-a)²) + sqrt(a² - (x-a)²)(a² - (x-a)²)  dx
1/3a Int [x = 0 to 2a]  sqrt(a² - (x-a)²) (3x² + a² - (x-a)²)  dx
1/3a Int [x = 0 to 2a]  sqrt(a² - (x-a)²) (2x² +  2xa)  dx
2/3a Int [x = 0 to 2a]  sqrt(a² - (x-a)²) (x² +  xa)  dx
Let u = a - x,
du = -dx
x = a - u
2/3a Int [u = a to -a]  sqrt(a² - u²) ((a-u)² + (a-u)a)  -du
2/3a Int [u = -a to a]  sqrt(a² - u²) (a²-2au + u² + a²-ua)  du
2/3a Int [u = -a to a]  sqrt(a² - u²) (2a² - 3au + u²)  du

Let u = a sin p
du = a cos p dp

2/3a Int [p = -pi/2 to pi/2]  sqrt(a² - a²sin²p) (2a² - 3a² sin p + (a sin p)²)  a cos p dp
2/3a Int [p = -pi/2 to pi/2]  (a cos p)²a²(2 - 3 sin p + sin²p) dp
2a^3/3 Int [p = -pi/2 to pi/2]  cos²p (2 - 3 sin p + sin²p) dp
2a^3/3 Int [p = -pi/2 to pi/2]  2cos²p - 3 sinpcos²p + cos²psin²p dp
2a^3/3 Int [p = -pi/2 to pi/2]  2(1 + cos (2p))/2 - 3 sin(p)cos²(p) + (sin(2p)/2)² dp
2a^3/3 Int [p = -pi/2 to pi/2]  1 + cos (2p) - 3 sin(p)cos²(p) + sin²(2p)/4 dp
2a^3/3 Int [p = -pi/2 to pi/2]  1 + cos (2p) - 3 sin(p)cos²(p) + (1 - cos²(2p))/4 dp
2a^3/3 Int [p = -pi/2 to pi/2]  1 + cos (2p) - 3 sin(p)cos²(p) + 1/4 - cos²(2p)/4 dp
2a^3/3 Int [p = -pi/2 to pi/2]  5/4 + cos (2p) - 3 sin(p)cos²(p) - ((1 + cos(4p))/2)/4 dp
2a^3/3 Int [p = -pi/2 to pi/2]  5/4 + cos (2p) - 3 sin(p)cos²(p) - 1/8 + cos(4p)/8 dp
2a^3/3 Int [p = -pi/2 to pi/2]  9/8 + cos (2p) - 3 sin(p)cos²(p) + cos(4p)/8 dp
2a^3/3 [9p/8 + sin(2p)/2 + cos(p)^3 + sin(4p)/32]  | Int [p = -pi/2 to pi/2]
2a^3/3 [9pi/2/8 + sin(2pi/2)/2 + cos(pi/2)^3 + sin(4pi/2)/32  - 9*-pi/2/8 - sin(-2pi/2)/2 - cos(-pi/2)^3 - sin(-4pi/2)/32]
2a^3/3 [9pi/16 + sin(pi)/2 + cos(pi/2)^3 + sin(2pi)/32  + 9pi/16 - sin(-pi)/2 - cos(-pi/2)^3 - sin(-2pi)/32]
2a^3/3 [9pi/16 + 0 + 0 + 0  + 9pi/16 - 0 - 0 - 0]
2a^3/3 [9pi/8]

3/4  pi a^3

I believe you would come up with the same answer using infinity08's integral, changing y = 0 to 1 to y = 0 to a, and using a similar trigonometric substitution.
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by:BigRat
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Well, it seemed so easy but it took a time to get there!

Taking the coords I gave and rotating the z-axis into the vertical, making the y-axis go off to the south west of the paper, we see that the cylinder, standing upright, is sliced off by the paraboloid. "Under" was a clue to consider an infinitesimal area on the surface of the paraboloid multiplied by z and summed to give the volume. Therefore dz is the first integral summed from (obviously) z=0 to the surface where z=(x²+y²)/a, directly from the paraboloid equation. It is quite obvious that, at the base of the cylinder x and y are interchangable, but since the origin of the center of the cylinder is at x=a, the limits for x are 0 and 2a and hence this should be done last. That gives limits for y as 0 and sqrt(2ax-x²), directly from the cylinder equation. Getting the limits right is half the work and having the lower bound as zero is an added bonus - it greatly simplifies the work.

Now onto the integration. The first part, dz, trivially gives z which then gives (x²+y²)/a. The second part results in x²*sqrt(2ax-x²) + 1/3*(2ax-x²)^3/2.

The substitution x=a(1-cost), sqrt(2ax-x²)=asint, dx = asintdt produces the integral (from zero to pi) of

    (1-cost)²*sin²t + 1/3*(sin²t)²

which produces 3/4*pi*a³.

Well done jkmyoung!
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