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16.1 Double Integrals over Rectangular Regions

If \( y = f(x) \), \( a \leq x \leq b \)

then \( \int_{a}^{b} f(x) dx \) gives us the net area between \( f(x) \) and the x-axis

Graph of a function f(x) over the interval [a, b] showing shaded areas between the curve and the x-axis.

but remember \( \int_{a}^{b} f(x) dx \) really sums up infinitely-many rectangle areas

Riemann sum visualization with rectangles of width delta x approximating the area under a curve f(x).
  • total area = sum of rectangles
  • each rectangle has height \( f(x_i^*) \)
Sample point (left/right/midpoint)

then \( \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x = \int_{a}^{b} f(x) dx \)

Note: \( n \) represents the # of rectangles.

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\( z = f(x, y) \) is a surface

\( [a, b] \times [c, d] \)

if we want the volume under the surface above \( a \leq x \leq b, c \leq y \leq d \)

we can use the same idea as in one-variable case

3D diagram showing a surface f(x,y) above a rectangular region in the xy-plane with a sample volume element.

we want volume under \( f(x, y) \) above the rectangle

subdivide \( x \) into \( n \) parts

subdivide \( y \) into \( m \) parts

use a particular choice of sample points to find height of a rectangular box

the rectangle box has volume \( f(x_i^*, y_j^*) \underbrace{\Delta x \Delta y}_{\Delta A} \)

then sum up all \( \sum_{i=1}^{n} \sum_{j=1}^{m} f(x_i^*, y_j^*) \Delta A \)

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Double Integrals over Rectangular Regions

Now let \( n \to \infty \), \( m \to \infty \), \( \Delta x \to dx \), \( \Delta y \to dy \).

\[ \lim_{n,m \to \infty} \sum_{i=1}^{n} \sum_{j=1}^{m} f(x_i, y_j) \Delta x \Delta y \]

For the region defined by \( a \le x \le b \) and \( c \le y \le d \):

\[ = \int_{a}^{b} \int_{c}^{d} f(x,y) \, dy \, dx = \int_{a}^{b} \int_{c}^{d} f(x,y) \, dA \]

Volume of box: \( f(x,y) \, dy \, dx = f(x,y) \, dx \, dy \)

\[ \int_{c}^{d} \int_{a}^{b} f(x,y) \, dx \, dy \]

If the region is rectangular (\( a \le x \le b \), \( c \le y \le d \)) then the order does NOT matter.

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Example: Evaluating a Double Integral

\[ \int_{0}^{1} \int_{0}^{2} (3 - x - y) \, dy \, dx \]
Coordinate graph showing a rectangle in the first quadrant with vertices at (0,0), (1,0), (1,2), and (0,2).

Region bounds: \( 0 \le y \le 2 \) and \( 0 \le x \le 1 \).

Basic Idea: Work Inside-Out

Inner Integral:

\[ \int_{0}^{2} (3 - x - y) \, dy \]

Note: In this step, \( x \) is treated as a constant, and \( y \) is the "live" variable (others are constants).

\[ = \left[ 3y - xy - \frac{1}{2}y^2 \right]_{y=0}^{y=2} = 6 - 2x - 2 = 4 - 2x \]

Outside Integral:

\[ \int_{0}^{1} (4 - 2x) \, dx = \left[ 4x - x^2 \right]_{x=0}^{x=1} = 3 \]
3
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Double Integrals over Rectangular Regions

This is a rectangular region, order doesn't matter.

\[ \int_{0}^{2} \int_{0}^{1} (3 - x - y) \, dx \, dy \]

Step 1: Inside Integral

Inside: \( \int_{0}^{1} (3 - x - y) \, dx \) others are constants

\[ = 3x - \frac{1}{2}x^2 - yx \Big|_{x=0}^{x=1} = 3 - \frac{1}{2} - y = \frac{5}{2} - y \]

Step 2: Outside Integral

Outside:

\[ \int_{0}^{2} (\frac{5}{2} - y) \, dy = \frac{5}{2}y - \frac{1}{2}y^2 \Big|_{y=0}^{y=2} = 5 - 2 = 3 \]

Interpretation

Interpretation: volume under \( z = 3 - x - y \) above \( 0 \le x \le 1 \), \( 0 \le y \le 2 \).

3D coordinate system showing a plane intersecting axes at (3,0,0), (0,3,0), and (0,0,3) with a rectangular region highlighted.
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Example: Double Integral with Substitution

\[ \int_{0}^{1} \int_{0}^{2} y^5 x^2 e^{x^3 y^3} \, dx \, dy \]

Step 1: Inside Integral

Inside: \( \int_{0}^{2} y^5 x^2 e^{x^3 y^3} \, dx \) y is constant

\[ = y^5 \int_{0}^{2} x^2 e^{y^3 x^3} \, dx \]

Substitution:

\( u = y^3 x^3 \)

\( du = 3y^3 x^2 \, dx \)

\( x^2 \, dx = \frac{1}{3y^3} \, du \)

\[ = y^5 \int_{x=0}^{x=2} e^u \cdot \frac{1}{3y^3} \, du = \frac{y^5}{3y^3} \int_{x=0}^{x=2} e^u \, du \]\[ = \frac{1}{3} y^2 (e^u \Big|_{x=0}^{x=2}) = \frac{1}{3} y^2 (e^{y^3 x^3} \Big|_{x=0}^{x=2}) \]\[ = \frac{1}{3} y^2 e^{8y^3} - \frac{1}{3} y^2 \]
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outside:

\[ \int_{0}^{1} (\frac{1}{3} y^2 e^{8y^3} - \frac{1}{3} y^2) dy \]

\[ = \underbrace{\int_{0}^{1} \frac{1}{3} y^2 e^{8y^3} dy}_{\begin{matrix} u = 8y^3 \\ du = 24y^2 dy \\ \vdots \end{matrix}} - \underbrace{\int_{0}^{1} \frac{1}{3} y^2 dy}_{easy} = \dots = \begin{array}{|c|} \hline \frac{1}{72} e^8 - \frac{1}{8} \\ \hline \end{array} \]

what if we had reversed the order?

\[ \int_{0}^{2} \int_{0}^{1} y^5 x^2 e^{x^3 y^3} dy dx \]

inside:

\[ \int_{0}^{1} y^5 x^2 e^{x^3 y^3} dy \]

x is constant

\[ = x^2 \int_{0}^{1} y^5 e^{x^3 y^3} dy \]

by parts

\[ uv - \int v du \]

\[ u = y^5 \quad dv = e^{x^3 y^3} dy \]

etc

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recall if \( y = f(x) \), \( a \le x \le b \)

then the average value of \( f(x) \) over \( a \le x \le b \) is

\[ f_{avg} = \frac{1}{b-a} \int_{a}^{b} f(x) dx \]

because \( (b-a) f_{avg} = \int_{a}^{b} f(x) dx \)

Two coordinate graphs: one showing a rectangle area and one showing the area under a curve f(x) from a to b.

same idea for \( z = f(x,y) \)

\( a \le x \le b \) \( c \le y \le d \)

\[ f_{avg} = \frac{1}{A} \int_{a}^{b} \int_{c}^{d} f(x,y) dy dx = \frac{1}{A} \int_{c}^{d} \int_{a}^{b} f(x,y) dx dy \]
\[ = \frac{1}{A} \iint_{R} f(x,y) dA \]

A = area of \( [a,b] \times [c,d] \)

R = region we integrate over

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Example: Changing Order of Integration

\[ \int_{0}^{1} \int_{0}^{\pi/3} x^2 \cos(xy) \, dx \, dy \]

Order as given, inside:

\[ \int_{0}^{\pi/3} x^2 \cos(xy) \, dx \]

Note: \( y \) is constant

This requires two rounds of integration by parts. It is not "easy".

Try Switching Order

\[ \int_{0}^{\pi/3} \int_{0}^{1} x^2 \cos(xy) \, dy \, dx \]

Inside:

\[ \int_{0}^{1} x^2 \cos(xy) \, dy \]

Note: \( x \) is constant

\[ = x^2 \sin(xy) \cdot \frac{1}{x} \Big|_{y=0}^{y=1} = x \sin x \]

Outside:

\[ \int_{0}^{\pi/3} x \sin x \, dx \]

This requires only one round of integration by parts.