Lecture Notes on Extrema Review and Double Integration

Extrema Review and D-Test

• Review of extrema process:

  • Find critical points by solving simultaneous partial derivatives with respect to $x$ and $y$ equal to zero.
  • Use $D$.

• $D$ is defined as: D = f_{xx} _{yy} - f_{xy} _{yx}

• Clairaut's theorem implies that $f<em>{xy}$ and $f</em>{yx}$ are almost always equal, so $D$ can be written as: D = f_{xx} _{yy} - (f_{xy})^2

• Rubicon:

  • D < 0 leads to a saddle point.
  • D > 0 leads to a local max or local min.

• When D > 0, the signs of $f<em>{xx}$ and $f</em>{yy}$ can differentiate between local max and local min.

  • If $f<em>{xx}$ and $f</em>{yy}$ are both positive, it indicates a local min.
  • If $f<em>{xx}$ and $f</em>{yy}$ are both negative, it indicates a local max.

Justification of Fxx and Fyy Signs When D > 0

• Recall: $D = f<em>{xx} f</em>{yy} - (f_{xy})^2$
• If D > 0, then f{xx} f{yy} > (f_{xy})^2
• Since $(f<em>{xy})^2$ is always greater than or equal to zero, this implies that f{xx} f_{yy} > 0.
• For the product of $f<em>{xx}$ and $f</em>{yy}$ to be positive, they must have the same sign.
• Therefore, at a critical point (a, b), $f<em>{xx}$ and $f</em>{yy}$ must have the same sign when D > 0.
• Consequence: To differentiate between a local max and local min when D > 0, you can use either $f<em>{xx}$ or $f</em>{yy}$.
  Some strategy can be used to see which is easier to calculate.

Examples

• Two or three examples to come, followed by a discussion of double integration.

Example 1

• From Stacks, page 457, number 327.

• Problem:

  • Given $f(x, y) = 3x^2 - 2xy + y^2 - 8y$
  • Find the critical points and test them using $D$.

• You will know d, which is: $D = f<em>{xx} f</em>{yy} - (f_{xy})^2$

• Step 1: Find the first-order partial derivatives.

  • $f_x = 6x - 2y = 0$. This implies $y = 3x$.
  • $f_y = -2x + 2y - 8 = 0$

• Substitute $y = 3x$ into $f_y = 0$.

  • $-2x + 6x - 8 = 0$
  • $4x = 8$
  • $x = 2$

• Since $y = 3x$, then $y = 6$.

• The critical point is (2, 6).

• Step 2: Calculate $D$.

  • $f_{xx} = 6$
  • $f_{yy} = 2$
  • $f_{xy} = -2$

• $D = (6)(2) - (-2)^2 = 12 - 4 = 8$. $D$ is a constant here.

• Since D > 0 and $f<em>{xx} > 0$ and f{yy} > 0, the critical point (2, 6) is a relative minimum (relmin) or a local minimum (lopmin).

Example 2

• $f(x, y) = arctan(xy)$
• Find the critical points and test them.
• $f_x = \frac{y}{1 + (xy)^2} = \frac{y}{x^2y^2 + 1}$
• $f_y = \frac{x}{1 + (xy)^2} = \frac{x}{x^2y^2 + 1}$
• Setting $f<em>x = 0$ and $f</em>y = 0$ implies that the origin (0, 0) is the only critical point.
• $f_{xx} = \frac{-2xy^3}{(x^2y^2 + 1)^2}$
• $f_{yy} = \frac{-2yx^3}{(x^2y^2 + 1)^2}$
• At the origin (0, 0), both $f<em>{xx} = 0$ and $f</em>{yy} = 0$.
• $f_{xy} = \frac{1 - x^2y^2}{(1 + x^2y^2)^2}$
• At the origin (0, 0), $f_{xy} = 1$.
• $D = f<em>{xx}f</em>{yy} - (f_{xy})^2 = (0)(0) - (1)^2 = -1$.
• Since D < 0, the origin (0, 0) is a saddle point.

Example 3

• $f(x, y) = (x^2 + y^2)e^{-x} = x^2e^{-x} + y^2e^{-x}$
• $f_y = 2ye^{-x} = 0$. Since $e^{-x}$ is never zero, this implies $y = 0$.
• $f_x = 2xe^{-x} - x^2e^{-x} - y^2e^{-x} = e^{-x}(2x - x^2 - y^2)$
• Since $y = 0$, $f_x = e^{-x}(2x - x^2) = 0$.
• $e^{-x}(2x - x^2) = 0$. Dividing by $e^{-x}$, $2x - x^2 = 0$.
• $x(2 - x) = 0$, so $x = 0$ or $x = 2$.
• The critical points are (0, 0) and (2, 0).
• $f_{yy} = 2e^{-x}$
• $f_{yx} = -2ye^{-x}$
• $f_{xx} = e^{-x}(x^2 + y^2 - 4x + 2)$
• $D = f<em>{xx} f</em>{yy} - (f_{xy})^2 = (2e^{-x})(e^{-x}(x^2 + y^2 - 4x + 2)) - (-2ye^{-x})^2$
• $D = 2e^{-2x}(x^2 + y^2 - 4x + 2) - 4y^2e^{-2x} = e^{-2x}(2x^2 + 2y^2 - 8x + 4 - 4y^2) = e^{-2x}(2x^2 - 2y^2 - 8x + 4)$
• At (0, 0), D = e^0 (0 + 0 - 0 + 4) = 4 > 0.
• Also, f_{yy} = 2e^0 = 2 > 0. Therefore, (0, 0) is a relative minimum.
• At (2, 0), D = e^{-4}(2(4) - 0 - 8(2) + 4) = e^{-4} (8 - 16 + 4) = -4e^{-4} < 0.
• Therefore, (2, 0) is a saddle point.

Comment on Calculating D

• Alternative to setting up $D$ algebraically: Evaluate $f<em>{xx}$, $f</em>{yy}$, and $f_{xy}$ separately at each critical point.
• Determine the sign of $D$ based on the signs of its components.
• This piecemeal approach can sometimes be easier.
• The algebra of D was calculated algebraically so:
  • $D = f<em>{xx} f</em>{yy} - (f_{xy})^2$
• You don't have to have an algebra expression for it.
• It works just as well to figure out, basically, you know, sign of $D$ by looking at the signs of the components of D, namely the two pure partials and the mixed partial partials, which are the same.

Double Integration

• Generalizes single variable integration.

• Review of single variable integration (Calc 1):

  • Partition the interval [a, b] into n subintervals of length $\Delta x = \frac{b - a}{n}$.
  • Each subinterval has the form $[x<em>{i-1}, x</em>i]$.
  • Choose a sample point $x_i^*$ from each interval.
  • Goal: Find the area under the curve $y = f(x)$ between $y = 0$, $x = a$, and $x = b$.
  • Approximate the area using rectangles with width $\Delta x$ and height $f(x_i^*)$.
  • The area of the i-th rectangle is $f(x_i^*) \Delta x$.
  • Riemann sum: $\sum<em>{i=1}^{n} f(x</em>i^*) \Delta x$
  • Definite integral: $\int<em>{a}^{b} f(x) dx = \lim</em>{n \to \infty} \sum<em>{i=1}^{n} f(x</em>i^*) \Delta x$

• Double integration involves a double Riemann sum set up on each axis.

Double Integration on a Rectangle

• Consider a function $f(x, y)$ on a rectangle with sides $b - a$ and $d - c$.
• The rectangle is defined by $a \le x \le b$ and $c \le y \le d$.
• Partition the intervals [a, b] and [c, d] into subintervals of length $\Delta x$ and $\Delta y$ respectively.
• Form sub-rectangles $R_{ij}$, each with area $\Delta A = \Delta x \Delta y$.
• Pick a sample point $(x<em>{ij}^<em>, y</em>{ij}^</em>)$ from each sub-rectangle $R_{ij}$.
• The volume of the box is \Delta A (x{ij}^, y{ij}^)
• Double Riemann sum: Sum with respect to both variables. The volume of boxes, many boxes, create a double Riemann sum.

Double Riemann Sum and Double Integral

• Double Riemann sum: $\sum<em>{i=1}^{m} \sum</em>{j=1}^{n} f(x<em>{ij}^<em>, y</em>{ij}^</em>) \Delta A$
• Let $m, n \to \infty$. If $f(x, y)$ is continuous on the rectangle, then the limit exists.
• Double integral: $\iint<em>{R} f(x, y) dA = \lim</em>{m, n \to \infty} \sum<em>{i=1}^{m} \sum</em>{j=1}^{n} f(x<em>{ij}^<em>, y</em>{ij}^</em>) \Delta A$
• The double integral represents the volume under the surface $z = f(x, y)$ and above the rectangle R in the xy-plane.
• The proof setting up a double Riemann sum, letting, m and n go to infinity, assuming that f is continuous.
• If asking about what problem we're looking for, we're really looking at what? What's the corresponding problem? Not area under a curve, but volume basically under a surface.
• This step is existence only; it does not provide a method for computation.

Fubini's Theorem (Computation of Double Integrals)

• If $f(x, y)$ is continuous on the rectangle $R = {(x, y) | a \le x \le b, c \le y \le d }$, then
  • $\iint<em>{R} f(x, y) dA = \int</em>{a}^{b} \int<em>{c}^{d} f(x, y) dy dx = \int</em>{c}^{d} \int_{a}^{b} f(x, y) dx dy$
• These are called iterated integrals.
• The inner integral is evaluated first, then the outer integral.
• The order of integration can be reversed (Fubini's Theorem).
• What's really going on is partial integration.

Examples of Double Integration

• Want to do it may be you know, we want to do a couple examples where you do both ways, and we see that the answer comes out the same as it should.

Example 1

• Compute $\int<em>{0}^{3} \int</em>{1}^{2} x^2y dy dx$
  • The rectangle is defined by $0 \le x \le 3$ and $1 \le y \le 2$.
• 1. Evaluate the inner integral:
  • $\int<em>{1}^{2} x^2y dy = x^2 \int</em>{1}^{2} y dy = x^2 \Big[ \frac{1}{2}y^2 \Big]_{1}^{2} = x^2 \Big( \frac{1}{2}(4) - \frac{1}{2}(1) \Big) = \frac{3}{2} x^2$
• 2. Evaluate the outer integral:
  • $\int<em>{0}^{3} \frac{3}{2} x^2 dx = \frac{3}{2} \int</em>{0}^{3} x^2 dx= \frac{3}{2} \Big[ \frac{1}{3}x^3 \Big]_{0}^{3} = \frac{3}{3 \times 2} 3^3 = \frac{27}{2}$
  • The final answer is $\frac{27}{2}$.
• Alternate Setup:
  • Integrate with respect to x first.
  • $\int<em>{1}^{2} \int</em>{0}^{3} x^2 y dx dy = \int<em>{1}^{2} y \int</em>{0}^{3} x^2 dx dy = \int<em>{1}^{2} y \Big[ \frac{1}{3}x^3 \Big]</em>{0}^{3} dy = \int<em>{1}^{2} 9y dy = \Big[ \frac{9}{2}y^2 \Big]</em>{1}^{2} = \frac{9}{2} \times 4 - \frac{9}{2} \times 1 = \frac{36 - 9}{2} = \frac{27}{2}$

Example 2

• Evaluate $\iint_{R} (x - 3y^2) dA$ where $R = {(x, y) | 0 \le x \le 2, 1 \le y \le 2 }$.
• Method 1: Integrate with respect to y first.
  • Set up the iterated integral: $\int<em>{0}^{2} \int</em>{1}^{2} (x - 3y^2) dy dx$
  • Evaluate the inner integral: $\int<em>{1}^{2} (x - 3y^2) dy = \Big[ xy - y^3 \Big]</em>{1}^{2} = (2x - 8) - (x - 1) = x - 7$
  • Evaluate the outer integral: $\int<em>{0}^{2} (x - 7) dx = \Big[ \frac{1}{2} x^2 - 7x \Big]</em>{0}^{2} = (2 - 14) - (0) = -12$
  • The answer is -12.
• Method 2: Integrate with respect to x first.
  • Set up the iterated integral: $\int<em>{1}^{2} \int</em>{0}^{2} (x - 3y^2) dx dy$
  • Evaluate the inner integral: $\int<em>{0}^{2} (x - 3y^2) dx = \Big[ \frac{1}{2} x^2 - 3xy^2 \Big]</em>{0}^{2} = (2 - 6y^2) - (0) = 2 - 6y^2$
  • Evaluate the outer integral: $\int<em>{1}^{2} (2 - 6y^2) dy = \Big[ 2y - 2y^3 \Big]</em>{1}^{2} = (4 - 16) - (2 - 2) = -12$

Special Case

• Calculate $\iint_{R} sin(x)cos(y) dA$ where $R$ is defined by $0 \le x \le \frac{\pi}{2}$, and $0 \le y \le \frac{\pi}{2}$.
• Observe that the integrand is a factorable product of a function of x and a function of y.
• This means that you're basically going to be able to factor this as a product of two calc 1 integrals.