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16.2 Double Integrals over General Regions

last time: over rectangular regions \( R = \{ (x,y) : a \leq x \leq b, \ c \leq y \leq d \} \)

the order of integration doesn't matter

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

but switching order arbitrarily is only ok for rectangular regions for general regions, order is very crucial

example

\( f(x,y) = xy^2 \)

\( R = \{ (x,y) : 0 \leq x \leq \sqrt{2}, \ x^2 \leq y \leq 2 \} \)

  • top curve: \( y = 2 \)
  • bottom curve: \( y = x^2 \)
Coordinate graph showing region R bounded by y=2, y=x^2, and the y-axis from x=0 to x=sqrt(2).

Basic rule: integrate the constant-bounded variable LAST (outside)

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so, here we integrate \( x \) last, \( y \) first

\[ \int_{0}^{\sqrt{2}} \int_{x^2}^{2} xy^2 \, dy \, dx \]

then proceed as before: inside-out

\[ = \int_{0}^{\sqrt{2}} \left( \frac{1}{3}xy^3 \Big|_{y=x^2}^{y=2} \right) \, dx = \int_{0}^{\sqrt{2}} \left( \frac{8}{3}x - \frac{1}{3}x^7 \right) \, dx \]
\[ = \frac{4}{3}x^2 - \frac{1}{24}x^8 \Big|_{0}^{\sqrt{2}} = 2 \]

what if we integrated using the wrong order?

\[ \int_{x^2}^{2} \int_{0}^{\sqrt{2}} xy^2 \, dx \, dy = \int_{x^2}^{2} \frac{1}{2}x^2y^2 \Big|_{x=0}^{x=\sqrt{2}} \, dy \]
\[ = \int_{x^2}^{2} y^2 \, dy = \frac{1}{3}y^3 \Big|_{x^2}^{2} = \frac{8}{3} - \frac{x^6}{3} \]

what do we do with this? (pointing to the variable result \( \frac{x^6}{3} \))

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Switching the Order of Integration

But, we still can switch order, but must be done carefully.

\[ R = \{ (x, y) : 0 \le x \le \sqrt{2}, \quad x^2 \le y \le 2 \} \]
Graph of a Type I region bounded by y=2 and y=x^2 from x=0 to sqrt(2) with vertical rectangles.

height = \( 2 - x^2 \)

This is sometimes called a Type I region.

  • Bunch of vertical rectangles
  • \( x \): bounded by constants

Formulating as a Type II Region

To switch order, we need to formulate \( R \) as a Type II region.

Graph of a Type II region with horizontal rectangles bounded by x=0 and x=sqrt(y).

Horizontal rectangles: right curve − left curve

right: \( x = \sqrt{y} \)

left: \( x = 0 \)

\( 0 \le x \le \sqrt{y} \)

\( 0 \le y \le 2 \)

\[ R = \{ (x, y) : 0 \le x \le \sqrt{y}, \quad 0 \le y \le 2 \} \]
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Integrate using the new order

\[ \int_{0}^{2} \int_{0}^{\sqrt{y}} xy^2 \, dx \, dy = \int_{0}^{2} \left( \frac{1}{2}x^2y^2 \Big|_{x=0}^{x=\sqrt{y}} \right) \, dy \]
\[ = \int_{0}^{2} \frac{1}{2}y^3 \, dy = \frac{1}{8} y^4 \Big|_{0}^{2} = \boxed{2} \]
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Example: Changing Order of Integration

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

Region \( R \) as given: \( R = \{ (x, y) : y \leq x \leq 1, \, 0 \leq y \leq 1 \} \) Type II

For this Type II region, the boundaries are defined as:

  • Left boundary: \( x = y \)
  • Right boundary: \( x = 1 \)
Graph of triangular region R with vertices (0,0), (1,0), (1,1) and a horizontal representative rectangle.

Integrate using the given order:

Inside integral:

\[ \int_{y}^{1} e^{x^2} dx \]

is not something we can do by hand

We must switch to a different type of \( R \) to integrate.

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Switch to Type I

Graph of region R with a vertical representative rectangle bounded by y=0 and y=x from x=0 to x=1.

Boundaries: \( y = x \) (top) and \( y = 0 \) (bottom)

\( 0 \leq x \leq 1 \)

\( 0 \leq y \leq x \)

\( R = \{ (x, y) : 0 \leq x \leq 1, \, 0 \leq y \leq x \} \)

\[ \int_{0}^{1} \int_{0}^{x} e^{x^2} dy \, dx \]
\[ = \int_{0}^{1} x e^{x^2} dx \]

Substitution:

\( u = x^2 \)

\( du = 2x \, dx \)

\[ = \dots = \frac{1}{2}(e - 1) \]

Inside integral calculation:

\[ \int_{0}^{x} e^{x^2} dy \]

Note: \( x \) is constant with respect to \( y \)

\[ = e^{x^2} y \Big|_{y=0}^{y=x} = x e^{x^2} \]
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Example: Switching the Order of Integration

\[ \int_{e^{-2}}^{1} \int_{-\ln y}^{2} f(x,y) \, dx \, dy + \int_{1}^{e^{2}} \int_{\ln y}^{2} f(x,y) \, dx \, dy \]

Rewrite the integral by switching order.

First Integral (Type II)

\( e^{-2} \le y \le 1 \)

\( -\ln y \le x \le 2 \)

  • Left: \( x = -\ln y \)
  • Right: \( x = 2 \)

Second Integral (Type II)

\( 1 \le y \le e^{2} \)

\( \ln y \le x \le 2 \)

  • Left: \( x = \ln y \)
  • Right: \( x = 2 \)

Type II Region

The region is split into two parts along the horizontal line \( y = 1 \). Horizontal rectangles are used to represent the Type II integration.

Coordinate graph showing a region bounded by x=2 and curves x=ln(y) and x=-ln(y) with horizontal test strips.
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Change to Type I (Vertical Rectangles)

Graph of the region from x=0 to x=2, bounded by y=e^x and y=e^-x, showing a vertical rectangle.

One integral is enough

Since top and bottom curves don't change.

\[ e^{-x} \le y \le e^{x} \]\[ 0 \le x \le 2 \]

Boundary Conversions

\( x = -\ln y \implies \ln y = -x \implies y = e^{-x} \)
\( x = \ln y \implies y = e^{x} \)

Resulting Region and Integral

\[ R = \{ (x,y) : 0 \le x \le 2, e^{-x} \le y \le e^{x} \} \]
Integral:\[ \int_{0}^{2} \int_{e^{-x}}^{e^{x}} f(x,y) \, dy \, dx \]