[b][size=150]Volumes of solids[/size][/b][br][br]For a function of one variable with positive values i.e. the graph of the function is above the x-axis, its definite integral over an interval is the area under the graph on such interval. We can generalize the definition of definite integral to functions of two variables as follows:[br][br]Let [math]z=f(x,y)[/math] be a function of two variables such that [math]f(x,y)\geq 0[/math] on the domain [math]R=[a,b]\times[c,d]=\left\{(x,y) \ | \ x\in [a,b], \ y\in [c,d]\right\}[/math] i.e. a [b]rectangular region[/b] in [math]\mathbb{R}^2[/math]. Then we divide [math]R[/math] into [math]N[/math] rectangular subregions using lines parallel to x and y axes. [br][br]For [math]k=1,\ldots, N[/math], let [math]\Delta x_k[/math] and [math]\Delta y_k[/math] be the length and width of the [math]k^{\text{th}}[/math] rectangular subregion [math]R_k[/math] respectively. Then the area of [math]R_k[/math] is [math]\Delta A_k=\Delta x_k \Delta y_k [/math]. Take any point [math](x_k^*,y_k^*)[/math] in [math]R_k[/math] and consider the rectangular box with base [math]R_k[/math] and height [math]f(x_k^*,y_k^*)[/math]. Then its volume is [br][br][math]f(x_k^*,y_k^*)\Delta A_k [/math][br][br]And we can approximate the volume [math]V[/math] under the graph of [math]z=f(x,y)[/math] over [math]R[/math] by the sum of volumes of all these rectangular boxes/prisms:[br][br][math]V \approx \sum_{k=1}^{N}f(x_k^*,y_k^*)\Delta A_k [/math][br][br]As [math]N\to \infty[/math], [math]\Delta A_k \to 0[/math] and if limit of the above "Riemann sum" exists for all partitions of [math]R[/math] and for all choices of [math](x_k^*,y_k^*)[/math] within those partitions, then the limit is the volume under the graph over [math]R[/math] and is called the [b]double integral[/b] of [math]f[/math] over [math]R[/math], denoted as follows:[br][br][math]\iint_R f(x,y) \ dA=\lim_{N\to\infty}\sum_{k=1}^{N}f(x_k^*,y_k^*)\Delta A_k [/math][br][br]In the following applet, we cut [math]R=[a,b]\times [c,d][/math] evenly into [math]n\times n[/math] identical rectangular regions i.e. [math]N=n^2[/math] and [math](x_k^*,y_k^*)[/math] is chosen to be the lower left corner of the rectangular region. As [math]n[/math] get larger and larger, the sum of the rectangular boxes/prisms will get closer and closer to the volume under the graph over [math]R[/math].[br]

As we know, the definite integral [math]\int_a^b f(x) \ dx[/math] is still well-defined even when the graph of [math]y=f(x)[/math] is not entirely above the x-axis. The area below the x-axis can be regarded as the "negative area". Similarly, the double integral [math]\iint_R f(x,y) \ dA[/math] is well-defined for any function [math]z=f(x,y)[/math]. The portion of volume below the xy-plane can be regarded as the "negative volume".[br][br]

[b][size=150]Iterated integrals[/size][/b][br][br]How can we compute a double integral? It turns out that we have the following extremely theorem to help us computing any double integral over a rectangular region:[br][br][u]Fubini's theorem[/u]: Suppose [math]z=f(x,y)[/math] on [math]R=[a,b]\times [c,d][/math]. Then[br][br][math]\iint_R f(x,y) \ dA = \int_c^d \left(\int_a^b f(x,y) \ dx \right) \ dy = \int_a^b \left(\int_c^d f(x,y) \ dy\right) \ dx[/math][br][br] The integrals on the right hand side are called [b]iterated integrals[/b]. The integral [math]\int_a^b f(x,y) \ dx[/math] means we integrate the function [math]f(x,y)[/math] with respect to [math]x[/math] i.e. [math]y[/math] is regarded as a fixed value. What you obtain after integrating is a function of [math]y[/math] and we then integrate it with respect to [math]y[/math].[br][br][br][u]Example[/u]: Compute [math]\iint_R (4-x-y) \ dA[/math], where [math]R=[0,1]\times [0,2][/math].[br][br][u]Answer[/u]:[br][br]By Fubini's theorem, we have [br][math]\iint_R (4-x-y) \ dA = \int_0^2 \left(\int_0^1(4-x-y) \ dx \right) \ dy[/math][br][math]=\int_0^2\left(\left[4x-\frac{x^2}2-yx\right]_0^1\right) \ dy[/math][br][math]=\int_0^2\left(\frac 72-y\right) \ dy = \left[\frac 72y-\frac{y^2}2\right]_0^2=5[/math][br][br]Alternatively, we compute the following iterated integral:[br][math]\iint_R (4-x-y) \ dA = \int_0^1 \left(\int_0^2(4-x-y) \ dy \right) \ dx[/math][br][math]=\int_0^1 \left(\left[4y-xy-\frac{y^2}2\right]_0^2\right) \ dx[/math][br][math]=\int_0^1\left(6-2x\right) \ dy = \left[6x-x^2\right]_0^1=5[/math][br][br][br][br][br]

[b][size=150]More about double integrals[/size][/b][br][br]The following are some nice properties of double integrals: [br][br][list=1][*][math]\iint_R cf(x,y) \ dA=c\iint_R f(x,y) \ dA[/math], for any real number [math]c[/math][/*][*][math]\iint_R \left(f(x,y)\pm g(x,y)\right) \ dA=\iint_R f(x,y) \ dA \pm \iint_R g(x,y) \ dA [/math][/*][*][math]\iint_R f(x,y) \ dA=\iint_{R_1} f(x,y) \ dA + \iint_{R_2} f(x,y) \ dA [/math], where [math]R[/math] is the disjointed union of [math]R_1[/math] and [math]R_2[/math].[/*][*]If [math]f(x,y)\leq g(x,y)[/math] on [math]R[/math], then [math]\iint_R f(x,y) \ dA \leq \iint_R g(x,y) \ dA[/math].[/*][*][math]\left|\iint_R f(x,y) \ dA\right|\leq \iint_R \left|f(x,y)\right| \ dA[/math].[/*][*]If [math]f(x,y)=g(x)h(y)[/math], then [math]\iint_R f(x,y) \ dA = \left(\int_a^b g(x) \ dx\right)\left(\int_c^d h(y) \ dy\right)[/math], where [math]R=[a,b]\times[c,d][/math].[/*] [/list][br][br][br]For a function of one variable [math]g(x)[/math], we define its average value over [math][a,b][/math] by[br][br][math]\overline{g}=\frac{1}{b-a}\int_a^b g(x) \ dx[/math][br][br]We have an analogous definition for function of two variables: Suppose [math]f(x,y)[/math] is defined on a domain [math]R[/math], then its average value over [math]R[/math] is[br][br][math]\overline{f}=\frac 1{\text{area of }R}\iint_R f(x,y) \ dA[/math][br][br][u]Example[/u]: Find the average value of [math]f(x,y)=4-x-y[/math] over [math][0,1]\times[0,2][/math].[br][br][u]Answer[/u]:[br][br][math]\overline{f}=\frac 1{1\times 2} \iint_R \left(4-x-y\right)\ dA = \frac 12 \cdot 5 = \frac 52[/math] (Note: We already computed the double integral in the previous example.)[br][br][br][br][br][br][br]