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10.24 A silo is formed above the x-y plane by the intersection of a right circular cylinder of radius 3 and a sphere of radius 5. Compute its volume.
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It is easiest to work in cylindrical coordinates. The equation of the spherical cap is z = 25 r 2 . It will intersect the cylinder when r = 3. The height of the cylinder will be 4.
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cylinder = Graphics3D[Cylinder[{{0, 0, 0}, {0, 0, 4}}, 3]];
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cap = RevolutionPlot3D[ 25 r2 , {r,0,3}, {p, 0,2 o} ] ;
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Show[cylinder, cap, Boxed False, PlotRange {0,5}]
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The projection of the solid is a circle of radius 3, centered at the origin. volume = 122 3
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25 r2
r p z r
10.25 Find the center of mass of a solid hemisphere of radius a if its density at each point is proportional to its distance above the x-y plane.
SOLUTION
The center of mass has coordinates ( x , y, z ) where
x (x ,y,z) dV
(x ,y,z) dV
, y=
y (x ,y,z) dV
(x ,y,z) dV
, z=
z (x ,y,z) dV
(x ,y,z) dV
Multivariate Calculus
The density function, ( x , y, z ) = k z, where k is the constant of proportionality. The problem is most conveniently solved by using spherical coordinates: x = sin cos , y = sin sin , z = cos x = q Sin[e]Cos[p]; y = q Sin[e]Sin[p]; z = q Cos[e]; r = k z; mass = 1 a4 k 4 2 /2 a x r q2 Sin[e] q e , centerofmass = 0 0 0 mass
2o 0
o /2
q2 Sin[e] p q e
q e y r q2 Sin[e] , mass
zrq
0 0 0
/2
{0,0, 8 a } 15
SOLUTION
q e Sin[e] mass
10.26 Find the moment of inertia of a solid hemisphere of radius a about its axis if its density is equal to the distance from the center of its base.
The moment of inertia about the z-axis is
(d ( x , y, z))
( x , y, z ) dV where d (x, y, z) is the distance from the
point (x, y, z) to the z-axis and (x, y, z) is the density at the point (x, y, z). In this problem we should use a spherical coordinate system: x = r sin f cosq , y = r sin f sin q , z = r cos f d ( x , y, z ) = s ( x , y, z ) = c = q Sin e] [ r=q x 2 + y 2 = (r sin f cosq )2 + (r sin f sin q )2 = r sinf x 2 + y2 + z 2 = r
/2
c 2 r q2 Sin[e] q e
2 a6 9
CH AP TE R 11
Ordinary Differential Equations
11.1 Analytical Solutions
Simply put, a differential equation is an equation expressing a relationship between a function and one or more of its derivatives. A function that satisfies a differential equation is called a solution. The Mathematica command DSolve is used to solve differential equations. As with algebraic or transcendental equations, a double equal sign, , is used to separate the two sides of the equation.
DSolve[equation, y[x], x] gives the general solution, y[x] of the differential equation, equation, , whose independent variable is x. DSolve[equation, y, x] gives the general solution, y, of the differential equation expressed as a pure function (see appendix) within a list. ReplaceAll (/.) may then be used to evaluate the solution. Alternatively, one may use Part or [[ ]] to extract the solution from the list.
dy = x + y , we simply type dx
EXAMPLE 1 To solve the first-order differential equation
DSolve[y'[x] x + y[x], y[x], x]
{{y[x]
1 x + x C[1]}}
EXAMPLE 2 To obtain the solution of
{{y
dy = x + y as a pure function (see appendix, Section A.1), we enter dx solution = DSolve[y'[x] x + y[x], y, x]
Function[{x}, 1 x + x C[1]]}}
If we wish to evaluate the solution, we can type y[x] /.solution
{{y[x]
1 x + x C[1]}}
Using the pure function, we can evaluate the derivatives of the solution. This would be clumsy using the solution of Example 1. y'[x]/.solution
+ x C[1]} + 2 x C[1]}
y''[x]+ y'[x] /.solution
We can define a function, f, representing the solution: f = solution[[1, 1, 2]] Function[{x}, 1 x + x C[1]]
Ordinary Differential Equations
We can then directly evaluate f or any of its derivatives. f[x]
1 x + x C[1]
f'[x]
1 + x C[1]
f''[x] x C[1]
It is extremely important that the unknown function be represented y[x], not y, within the differential equation. Similarly, its derivatives must be represented y'[x] y''[x], etc. The next example illustrates , some common errors.
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