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MAT 3100 – Multivariable Calculus Rationale Multivariable functions arise in many real world situations, where physical quantities often depend on two or more variables. This course takes calculus from the two dimensional world of single variable functions into the three dimensional world of multivariable functions which are required to understand and manipulate planes and surfaces, curves in two or three dimensions and scalar-valued and vector-valued functions of several variables.
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MWALE David
∫ (^) d c
{∫ (^) g 2 (y) g 1 (y)^ f(x; y)^ dx
dy: Remember that:
Example 4.1.1. Evaluate
I =
1
2 (x^ + 2y)^ dx dy: Solution
I =
1
2 (x^ + 2y)^ dx
1
[x 2 2 + 2yx
1 (6 + 4y)dy = [(6y + 2y^2 )]^21 = 12:
Theorem 4.1.2. (Fubini’s Theorem for double integrals) If is continuous on the rectangle R = { (x; y) : a ≤ x ≤ b; c ≤ y ≤ d } ; then ∫ (^) b a
∫ (^) d c^ f(x; y)^ dA^ =
∫ (^) b a
∫ (^) d c^ f(x; y)^ dy dx^ =
∫ (^) d c
∫ (^) b a^ f(x; y)^ dx dy:
More generally, this is true if we assume that f is bounded on R; f is discon-
tinuous only on a finite number of smooth curves, and the iterated integrals exist.
Example 4.1.3. Evaluate
I =
1
∫ (^) π 0 (3 + sin^ θ)^ dθ dr: Solution
I =
1
∫ (^) π 0 (3 + sin^ θ)^ dθ dr =
1 [(3θ^ −^ cos^ θ)]
π 0 dr
=
1 (3π^ + 2)dr = [(3π + 2)r]^21 = 3π + 2:
Example 4.1.4. Confirm the Theorem above by showing that
I =
∫ (^) π 0
1 y^ sin(xy)^ dx dy^ =
1
∫ (^) π 0 y^ sin(xy)^ dy dx: Solution Home Work
4.2 Triple Integrals
Triple integrals are of the form ∫ (^) z 2 z 1
∫ (^) y 2 y 1
∫ (^) x 2 x 1 f(x; y)^ dx dy dz:
As in the previous section, we work from the inside out. All variables are regarded as constants for the time being, except for the one variable with respect to which the stage of integration is taking place.
In general, a double integral is denoted by ∫ ∫ Ω^ f(x; y)^ dA;
where Ω is the region over which the integral is to be evaluated and dA = dx dy is an area element. If f(x; y) = 1 for all (x; y) ∈ Ω; then this integral represents the area of the region Ω: That is (^) ∫ ∫
Ω^ dA^ =
Ω^ dx dy^ =^ area of^ Ω: If f(x; y) ≥ 0 for all (x; y) ∈ Ω; then the integral ∫ ∫ Ω^ f(x; y)dA^ =
Ω^ f(x; y)dx dy;
represents the volume of the solid bounded above by the surface z = f(x; y) and below by the region Ω: This leads us to the following discussion
4.3 Application of multiple integration
The following example will be useful to appreciate the notion of area as an application of double integration.
4.3.1 Area rectangular coordinates
Example 4.3.2. Find the area of the region bounded by y = 45 x ; the x-axis and the ordinate at x = 5: Solution We begin by sketching the required region. To find the area, we may, take vertical strips and use the usual formula for area;
Area = δy · δx:
But the total area of the strip is given by y ∑=y 1 y=
δy · δx:
Consequently, the sum of all the strips across the region is given by
A ≈ ∑^ x= x=
y ∑=y 1 y=
δy · δx
∑^ x= x=
y ∑=y 1 y=
δy · δx
It follows that if δx Ï 0 and δy Ï 0 ; we get
Then, as usual, if δx Ï 0 ; δy Ï 0 and δz Ï 0 :
V =
∫ (^) x 2 x 1
∫ (^) y 2 y 1
∫ (^) z 1 0 dx dy dz:
Example 4.3.5. A solid is enclosed by the plane z = 0; the planes x = 1; x = 4; y = 2; y = 5 and the surface z = x + y: Find the volume of the solid. Solution First of all, it is important to give a sketch of the desired solid. As in the description of volume above. It follows that
volume of element ≈ δx · δy · δz volume of column ≈ δx · δy
z= ∑x+y z=
δz
volume of slice ≈ δx ∑^ y= y=
δy
z= ∑x+y z=
δz
volume of total solid ≈ ∑^ y= x=
δx ∑^ y= y=
δy
z= ∑x+y z=
δz
Consequently,
V =
1 dx
2 dy
∫ (^) x+y 0 dz =
1 dx
1 dx
xy + y 2 2
1
3 x +^212
dx =
[ 3 x 2 2 +
21 x 2
1 = 54^ units
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