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Multivariable Calculus

Introduction to Multivariable Calculus

These notes review integral calculus in one dimension and expand the discussion to multiple dimensions, with particular emphasis on two-dimensional integrals.

1: Introduction to Multivariable Calculus


Extending the concepts learned in introductory calculus into multiple dimensions, particularly focusing on 3D spaces. Revisits the basics of single integrals, extending these ideas to double and volume integrals, essential for formulating physical laws in three-dimensional space.

1. Review of Single Variable Integrals

An integral in one dimension can be viewed as calculating the area under a curve. For a function f(x)f(x) defined on the interval [a,b][a, b], the integral can be expressed informally as:

abf(x)dx\int_a^b f(x) \, dx

Integral Approximation As The Sum Of Rectangles The process involves subdividing the interval [a,b][a, b] into mm subintervals, each of width Δx\Delta x. If x1,x2,...,xmx_1, x_2, ..., x_m are points in these subintervals, the integral is approximately the sum of the areas of rectangles under the curve, and mathematically described by the limit:

abf(x)dx=limΔx0r=1mf(xr)Δx\int_a^b f(x) \, dx = \lim_{\Delta x \to 0} \sum_{r=1}^m f(x_r) \Delta x

This approximation becomes exact as Δx\Delta x approaches zero, provided f(x)f(x) is continuous over [a,b][a, b].

2. Introduction to Double Integrals

Extending the concept to two dimensions, consider a scalar field S(x,y)S(x, y) defined over a region RR in the xyxy-plane. The double integral is used to compute the volume under the surface described by S(x,y)S(x, y) over the region RR:

RS(x,y)dxdy\iint_R S(x, y) \, dx \, dy

This is achieved by partitioning the plane into small rectangular elements of area ΔA=ΔxΔy\Delta A = \Delta x \Delta y, and summing the volumes corresponding to these elements. The process is similar to the line integral, but in two dimensions:

RS(x,y)dxdy=limΔA0i=1nSiΔA\iint_R S(x, y) \, dx \, dy = \lim_{\Delta A \to 0} \sum_{i=1}^n S_i \Delta A

Where SiS_i is the value of the scalar field at the center of the ii-th subregion.

3. Properties of Double Integrals

Double integrals inherit several properties from single integrals, such as linearity and the ability to split the domain of integration. These properties simplify calculations and are particularly useful when dealing with complex regions or functions.

i. Linearity:
R(af(x,y)+bg(x,y))dA=aRf(x,y)dA+bRg(x,y)dA\iint_R (a f(x, y) + b g(x, y)) \, dA = a \iint_R f(x, y) \, dA + b \iint_R g(x, y) \, dA
ii. Order: If f(x,y)g(x,y)f(x, y) \geq g(x, y) for all (x,y)R(x, y) \in R, then
Rf(x,y)dARg(x,y)dA\iint_R f(x, y) \, dA \geq \iint_R g(x, y) \, dA
iii. Domain Splitting: If R=R1R2R = R_1 \cup R_2 with R1R2=R_1 \cap R_2 = \emptyset,
Rf(x,y)dA=R1f(x,y)dA+R2f(x,y)dA\iint_R f(x, y) \, dA = \iint_{R_1} f(x, y) \, dA + \iint_{R_2} f(x, y) \, dA

Example: Double Integral Over a Rectangular Domain

  • Region: R=[1,2]×[1,3]R = [1, 2] \times [1, 3]
  • Integral:
Rx2y3dxdy\iint_R x^2 y^3 \, dx \, dy

Method 1: Integrate with respect to xx first

  • Integral:
13(12x2y3dx)dy\int_1^3 \left( \int_1^2 x^2 y^3 \, dx \right) dy
  • Inner Integral:
12x2y3dx=x33y312=8y33y33=7y33\int_1^2 x^2 y^3 \, dx = \left. \frac{x^3}{3} y^3 \right|_1^2 = \frac{8y^3}{3} - \frac{y^3}{3} = \frac{7y^3}{3}
  • Outer Integral:
137y33dy=73y4413=73(81414)=1403\int_1^3 \frac{7y^3}{3} \, dy = \frac{7}{3} \left. \frac{y^4}{4} \right|_1^3 = \frac{7}{3} \left( \frac{81}{4} - \frac{1}{4} \right) = \frac{140}{3}

Method 2: Integrate with respect to yy first

  • Integral:
12(13x2y3dy)dx\int_1^2 \left( \int_1^3 x^2 y^3 \, dy \right) dx
  • Inner Integral:
13x2y3dy=x2y4413=x2(81414)=20x2\int_1^3 x^2 y^3 \, dy = x^2 \left. \frac{y^4}{4} \right|_1^3 = x^2 \left( \frac{81}{4} - \frac{1}{4} \right) = 20 x^2
  • Outer Integral:
1220x2dx=20x3312=20(8313)=1403\int_1^2 20 x^2 \, dx = 20 \left. \frac{x^3}{3} \right|_1^2 = 20 \left( \frac{8}{3} - \frac{1}{3} \right) = \frac{140}{3}

Simplified Method

  • Integral:
(12x2dx)(13y3dy)\left( \int_1^2 x^2 \, dx \right) \left( \int_1^3 y^3 \, dy \right)
  • Results:
(x3312)(y4413)=73804=1403\left( \left. \frac{x^3}{3} \right|_1^2 \right) \left( \left. \frac{y^4}{4} \right|_1^3 \right) = \frac{7}{3} \cdot \frac{80}{4} = \frac{140}{3}

General Rectangular Domain

  • Continuous Bounded Function f(x,y)f(x, y) on [a1,b1]×[a2,b2][a_1, b_1] \times [a_2, b_2]:
a1b1(a2b2f(x,y)dy)dx=a2b2(a1b1f(x,y)dx)dy\int_{a_1}^{b_1} \left( \int_{a_2}^{b_2} f(x, y) \, dy \right) dx = \int_{a_2}^{b_2} \left( \int_{a_1}^{b_1} f(x, y) \, dx \right) dy
  • Condition: Function ff must be continuous and bounded.

Example: Area of a Region Bounded by a Parabola

  • Region: Bounded by y=xy = \sqrt{x}, xx-axis, and y=x2y = x - 2

Area of a Region Bounded by a Parabola

  • Area:
Rdxdy\iint_R dx \, dy

Integration Order: xx first, then yy

  • Limits: xx between y2y^2 and y+2y + 2
  • Integrals:
02(y2y+21dx)dy=02((y+2)y2)dy=10/3\int_0^2 \left( \int_{y^2}^{y+2} 1 \, dx \right) dy = \int_0^2 ((y + 2) - y^2) \, dy = 10/3

Integration Order: yy first, then xx

  • Split Domain:
    • 0x20 \leq x \leq 2: yy between 0 and x\sqrt{x}
    • 2x42 \leq x \leq 4: yy between x2x - 2 and x\sqrt{x}
  • Integrals:
02(0x1dy)dx+24(x2x1dy)dx=10/3\int_0^2 \left( \int_0^{\sqrt{x}} 1 \, dy \right) dx + \int_2^4 \left( \int_{x-2}^{\sqrt{x}} 1 \, dy \right) dx = 10/3
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