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# Chapter 14: Multiple Integrals ## 14.1 Double Integrals over Rectangles ### Review of Single Integrals Suppose $f(x)$ is defined for $a \leq x \leq b$. We divide the interval $[a, b]$ into $n$ subintervals of equal width $\Delta x = \frac{b-a}{n}$ and choose sample points $x_i^*$ in these subin...

# Chapter 14: Multiple Integrals ## 14.1 Double Integrals over Rectangles ### Review of Single Integrals Suppose $f(x)$ is defined for $a \leq x \leq b$. We divide the interval $[a, b]$ into $n$ subintervals of equal width $\Delta x = \frac{b-a}{n}$ and choose sample points $x_i^*$ in these subintervals. Then the definite integral of $f$ from $a$ to $b$ is $\qquad \int_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i^*) \, \Delta x$ Geometrically, this represents the area under the curve $y = f(x)$ from $a$ to $b$. ### Double Integrals In double integrals, instead of integrating over an interval, we integrate over a region in the $xy$-plane. We start with a rectangular region $R$: $\qquad R = \{(x, y) \mid a \leq x \leq b, c \leq y \leq d\}$ We divide $R$ into subrectangles $R_{ij}$ with dimensions $\Delta x$ and $\Delta y$ and area $\Delta A = \Delta x \Delta y$. We choose a sample point $(x_{ij}^*, y_{ij}^*)$ in each $R_{ij}$. Then, the double integral of $f$ over $R$ is $\qquad \iint_R f(x, y) \, dA = \lim_{m, n \to \infty} \sum_{i=1}^m \sum_{j=1}^n f(x_{ij}^*, y_{ij}^*) \, \Delta A$ This represents the volume under the surface $z = f(x, y)$ and above the region $R$ in the $xy$-plane. ### Iterated Integrals If $f$ is continuous on the rectangle $R = [a, b] \times [c, d]$, then $\qquad \iint_R f(x, y) \, dA = \int_a^b \left( \int_c^d f(x, y) \, dy \right) \, dx = \int_c^d \left( \int_a^b f(x, y) \, dx \right) \, dy$ **Note:** The inner integral is performed first, treating the other variable as a constant. ### Example 1 Evaluate $\iint_R (x - 3y^2) \, dA$, where $R = \{(x, y) \mid 0 \leq x \leq 2, 1 \leq y \leq 2\}$. **Solution:** $\qquad \iint_R (x - 3y^2) \, dA = \int_0^2 \int_1^2 (x - 3y^2) \, dy \, dx$ First, integrate with respect to $y$: $\qquad \int_1^2 (x - 3y^2) \, dy = \left[ xy - y^3 \right]_1^2 = (2x - 8) - (x - 1) = x - 7$ Now, integrate with respect to $x$: $\qquad \int_0^2 (x - 7) \, dx = \left[ \frac{1}{2}x^2 - 7x \right]_0^2 = (2 - 14) - (0) = -12$ Therefore, $\iint_R (x - 3y^2) \, dA = -12$. ### Fubini's Theorem Fubini's Theorem states that if $f(x, y)$ is continuous on the rectangle $R = [a, b] \times [c, d]$, then $\qquad \iint_R f(x, y) \, dA = \int_a^b \int_c^d f(x, y) \, dy \, dx = \int_c^d \int_a^b f(x, y) \, dx \, dy$ ### Example 2 Evaluate $\iint_R y \sin(xy) \, dA$, where $R = [1, 2] \times [0, \pi]$. **Solution:** $\qquad \iint_R y \sin(xy) \, dA = \int_0^\pi \int_1^2 y \sin(xy) \, dx \, dy$ First, integrate with respect to $x$: $\qquad \int_1^2 y \sin(xy) \, dx = \left[ -\cos(xy) \right]_1^2 = -\cos(2y) + \cos(y)$ Now, integrate with respect to $y$: $\qquad \int_0^\pi (-\cos(2y) + \cos(y)) \, dy = \left[ -\frac{1}{2} \sin(2y) + \sin(y) \right]_0^\pi = (0 + 0) - (0 + 0) = 0$ Therefore, $\iint_R y \sin(xy) \, dA = 0$.