sql server reporting services barcode font Integrate[f[x, y], {x, xmin, xmax}, {y, ymin, ymax}] evaluates the double integral in Software

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Integrate[f[x, y], {x, xmin, xmax}, {y, ymin, ymax}] evaluates the double integral
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Integrate[f[x, y, z], {x, xmin, xmax}, {y, ymin, ymax}, {z, zmin, zmax}] xmax ymax zmax evaluates the triple integral f ( x , y, z) dz dy dx.
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Higher order iterated integrals are evaluated in a similar manner. Note that the rightmost variable of integration in the Integrate command is the variable that is evaluated first. As an alternative, the integral symbol from the Basic Math Input palette may be used repeatedly for the evaluation of multiple integrals.
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Multivariate Calculus
EXAMPLE 13 To evaluate the triple integral
xyz dz dy dx, we can type either
Integrate[x y z, {x, 0, 2}, {y, 0, x}, {z, 0, x y}] 4 or
x y z z y x
If the integral is such that it s exact value cannot be evaluated, numerical integration can be used instead.
NIntegrate[f[x, y], {x, xmin, xmax}, {y, ymin, ymax}] returns a numerical approxixmax ymax mation of the value of the double integral xmin ymin f ( x , y) dy dx. NIntegrate[f[x, y, z], {x, xmin, xmax}, {y, ymin, ymax}, {z, zmin, zmax}] xmax ymax zmax returns a numerical approximation of the value of the triple integral xmin ymin zmin f ( x , y, z ) dz dy dx.
Higher-order iterated integrals are approximated in a similar manner. If the Basic Math Input palette is used, the N command (or //N to the right of the integral) may be used. All of the options for Nintegrate as applied to single integrals apply to multiple integrals.
EXAMPLE 14
NIntegrate[Exp[x^2 y^2], {x, 0, 1}, {y, 0, 1}] 1.1351
2 y2
y x //N
SOLVED PROBLEMS
10.16 Use a double integral to compute the area bounded by the parabola y = x2 + 2x + 3 and the line y = x + 1.
SOLUTION
f[x_] = x2 2 x + 2; g[x_] = x + 1; Plot[{f[x], g[x]}, {x, 1, 3}]
5 4 3 2 1
It is always a good idea to sketch the region under consideration before integrating.
intersections = Solve[f[x] g[x]]
{{x 1 (3 5)}, {x 1(3 + 5)}} 2 2
Multivariate Calculus
xvalues = x /.intersections 1 (3 5 ), 1 (3 + 5 ) 2 2
a = xvalues[[1]]; b = xvalues[[2]];
g[x]
f[x]
5 5 6
10.17 Find the center of mass of the lamina bounded by the parabola y = 9 x2 and the x-axis if the density at each point is proportional to its distance from the x-axis.
SOLUTION
Let R be the region bounded by y = 9 x2 and the x-axis. Plot[9 x2, {x, 3, 3}]
8 6 4 2
M M The graph intersects the x-axis at 3 and 3. The coordinates of the center of mass are y , x where M M M y = moment about the y-axis = M x = moment about the x -axis = M = mass of lamina = The density function r (x,y) = ky. q[x_, y_]= k y; my = 0 mx =
3 3 0 3 3 0
x (x , y)dA
y (x , y)dA
(x , y)dA
9 x2
xq [x, y] y x
9 x2
[x, y] y x yq
23 328 k 35 m=
3 3 0
9 x2
q [x, y] y x
648 k 5 my mx , m m 36 0, 7
Multivariate Calculus
10.18 Compute the shaded area. The curve shown is the Spiral of Archimedes and has polar equation r = q. It is shown for 0 q 6p.
SOLUTION
The area inside a polar region, R, is
r dr d . The smaller arc
R 5 / 2
of the shaded region is described by r = q, 2 p q 5 p/2 and the larger arc may be represented by r = q + 2 p, 2 p q 5 p/2. The enclosed area can be expressed as
+ 2
r dr d .
5 5
5 /2 +2 2
r p r
13 3 4
10.19 Compute the volume under the paraboloid z = x2 + y2, above the region bounded by y = x2 and y = x + 1 .
SOLUTION
The volume bounded by a surface z = f(x, y) and the x-y plane, above a region R, is First let us look at R. plot[{x2, x + 1},{x,1,2}]
f ( x , y) dA.
1.0
0.5
Next we find the points of intersection. Because of the complicated nature of the solution, we will obtain a numerical approximation. NSolve[x2 x + 1] {{x 1.22074}, {x 0.724492} Now we can express the volume as a double integral. Two solutions are shown.
.724492 x 2
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