Calculus/Multiple integration

Directional derivatives and the gradient vector Change of variables __TOC__ Double integration thumb|Illustration of partitioning interval For (Riemann) integrals, we consider the Riemann sum. Recall in the one-variable case, we partition an interval into more and more subintervals with smaller and smaller width, and we are integrating over the interval by summing the areas of corresponding rectangles for each subinterval. For the multivariable case, we need to do something similar, but the problem arises when we need to partition 'interval' in or in general. (Actually, we only have the term green interval in .) In multivariable case, we need to consider not just 'interval' itself (which is undefined in multivariable case), but green Cartesian product over intervals for , and more generally green n-ary Cartesian product over intervals for . thumb|Illustration of Cartesian product by an example. darkblue Definition. (n-ary Cartesian product) The green n-ary Cartesian product over sets is the set of -tuples (or vectors). rosybrown Remark. recall that green interval is essentially a set, e.g. . 2-ary Cartesian product is simply called Cartesian product, e.g. is green Cartesian product over two intervals, and also a green rectangle with side lengths 1 and 2 geometrically. green special case: is called green n-ary Cartesian power and is denoted by , e.g. . Area (for ), volume (for ) or measure (for each positive number ) of geometric objects (e.g. rectangles in and cubes in ) in is the product of the lengths of all its sides (in different dimensions). red Example. is the Cartesian product over two intervals (It is square with side length in geometrically) is the 3-ary Cartesian product over three intervals (It is a rectangular cuboid in geometrically) is the 4-ary Cartesian product over four intervals, or 4-ary Cartesian power (it is denoted as ) (It is a 4-dimensional cube in geometrically) Now, we are ready to define multiple integral in an analogous way compared with single integral. For simplicity, let us first discuss double integral, and then generalize it to multiple integral in an analogous way. thumb|Illustration of partitioning rectangle in . def:double integrals darkblue Definition. (Double integrals) Let be a function defined on a rectangle in . Consider a partition of into small rectangles with areas respectively. Choose an green arbitrary point in the th rectangle. The function is green integrable over if exists. In that case, we denote this limit by ( is a mnemonic of area), and call it the green double integral of over . rosybrown Remark. the limit is the Riemann sum over with the partition of into small rectangles as we partition more and more small rectangles (with smaller and smaller area) from the rectangle , green which arbitrary point we choose in each rectangle becomes less and less important, because different points in each small rectangle become 'closer and closer' together, and thus the position of different points are more and more 'similar' we may assume that nice functions (functions given in the questions here are nice, unless stated otherwise) to be integrable, so we do not need to check integrability of every function we encounter here (Checking of integrability involves the formal definition of limit, and is out of scope here) the process of computing green double integral is called green double integration A physical meaning of double integration is computing volume. prp:volume given by double integration purple Proposition. (Volume given by double integration) Let be an green integrable function defined on a rectangle in . Suppose for each . Then, the volume under the graph of over is rosybrown Remark. if is green not always , the region below the -plane has negative volume, so this negative volume will cancel out the positive volume, which may or may not be desired if is always negative, then the volume computed by this formula is negative, and we usually take its absolute value to get the volume, because volume is usually defined to be nonnegative Let's also introduce some properties of double integral to ease computation of double integral. purple Proposition. (Properties of double integral) Let and be green integrable functions defined on a rectangle in . Then, the following properties hold. (Linearity) is also integrable over for each real number and , and (Monotonicity) if for each , then (Triangle inequality) is integrable over and purple Proposition. (Continuity implies integrability) Let be a green continuous function defined on a rectangle in . Then, the function is integrable. rosybrown Remark. Because nice functions are often continuous, most nice functions are integrable. Iterated integrals Thankfully, we need not always work with Riemann sums every time we want to calculate an integral in more than one variable. There are some results that make life a bit easier for us. Before stating the result, we need to define green iterated integral, which is used in the results. darkblue Definition. (Iterated integral) Let be a continuous function defined on a rectangle in . The iterated integral is defined by and rosybrown Remark. e.g., for the integral , we first compute the definite integral with respect to by treating as a green constant. Then, we compute the remaining integral (in terms of ) with respect to . Computation of iterated integrals is generally much easier than computing the double integral directly using Riemann sum. So, it will be nice if we have some relationships between iterated integral and double integral for us to compute double integral with the help of iterated integral. It is indeed the case and the following theorem is the bridge between iterated integral and double integral. thm:Fubini's theorem tomato Theorem. (Fubini's theorem) Let be a continuous function defined on a rectangle in . Then, rosybrown Remark. i.e., we can use iterated integral in green either order to compute the corresponding double integral we should notice the change of bounds for each integral after changing the integration order { red Example. } {Choose the correct expression(s) for the integral . |type="[]"} + || This is correct by Fubini's theorem. + || This is correct by Fubini's theorem. - || Be careful about the change of bounds. - || Be careful about the change of bounds. {Choose the correct expression(s) for the integral . |type="[]"} + - - - + red Example. (Volume of rectangular cuboid) (i) Prove that the volume under the graph of over the rectangle is . (ii) Hence, prove that volume of rectangular cuboid with length , width and height is . Proof. (i) Because for each , by the proposition of volume given by double integration, the volume is . By Fubini's theorem, this equals rosybrown Remark. Geometrically, the graph is a rectangular cuboid, and its volume is given by product of each side length, namely , which matches with our answer. (ii) The desired volume is given by (double integral of a constant function over the rectangle with length and width ). (We may also express the integral as without affecting the result.) Then, by Fubini's theorem, Double integrals over more general regions in R2 We have defined double integrals over rectangles in . However, we often want to compute double integral over regions with shape other than rectangle, e.g. circle, triangle, etc. Therefore, we will discuss an approach to compute double integrals over more general regions reasonably, without altering the definition of double integrals. Consider a function in which is a general region. To apply the definition of double integrals, we need to transform the general region to a rectangle (say ). An approach is finding a rectangle green containing (i.e., ), and let for each lying green outside (i.e., for each ). Because the value of the function is zero outside the region we are integrating over, this green does not change the volume under the graph of over , so this way is a good way to define such double integrals. Let's define such double integrals formally in the following. def:double integrals over general regions darkblue Definition. (Double integrals over general regions) Let be a function defined on a region (of arbitrary shape) . Then, we define for each lying outside (i.e., for each ), and define the green double integral of function over the region by if the latter integral exists. rosybrown Remark. Then, we may compute double integrals over general regions by computing the corresponding Riemann sum for the latter integral. However, this way of computation (by computing Riemann sum) is generally very difficult, and usually we use a generalized version of Fubini's theorem to compute such integrals. It will be discussed in the following. tomato Theorem. (Generalized Fubini's theorem) Let be a continuous function defined on a region (with arbitrary shape) . Then, the following hold. (i) If in which functions and are continuous, then (ii) If in which functions and are continuous, then in which are real numbers satisfying the above conditions. violet Proof. We can prove this theorem by Fubini's theorem (ungeneralized version) and applying the definition of double integrals over general regions. (We can use Fubini's theorem because we assume that functions and are continuous.) Part (i): () Take arbitrary rectangle containing (i.e. and ). Then, define if . After that, The result follows. (We say that whether equality holds in the above inequalities does not matter, because the definite integral over a point equals zero anyway, so it does not affect the result.) Part (ii): () Similarly, take arbitrary rectangle containing (i.e. and ). Then, define if . After that, The result follows. rosybrown Remark. Usually, finding the bounds for and is the most difficult step when we compute such double integrals. red Example. Let be the triangle in with vertices and . Prove that . Proof. thumb|440x440px|Illustration of approach 1 thumb|440x440px|Illustration of approach 2 Approach 1: The bound for is . Given a fixed , the bound for is . Thus, the integral is Approach 2: The bound for is . Given a fixed , the bound for is . Thus, the integral is red Example. (Volume of tetrahedron) Consider a tetrahedron in with vertices and in which and are positive numbers. Define vectors and . Prove that volume of the tetrahedron is . (i.e., of the volume of parallelepiped spanned by the vectors and ) Proof. thumb|330x330px|Illustration of finding volume of tetrahedron by double integration Let the plane containing and be . To find the equation of , consider its normal vector. A normal vector of is Therefore, the equation of is The desired volume is the volume under the graph of over a region , and the region is the projection of the tetrahedron on the xy-plane, which is the triangle with vertices in . Because the line passing through and has equation , the bound for is , and the bound for is given a fixed . Thus, the desired volume is On the other hand, which equals the desired volume. red Example. (Switching integration order) Prove that is approximately . (correct to two decimal places) Proof. thumb|440x440px|Illustration of the bounds of the region for this integral (region ABC). If we integrate in this order, the computation will be very tedious. Thus, we will interchange the integration order using generalized Fubini's theorem to ease the computation. Originally, the bounds are and (given a fixed ). If we integrate over first instead, the suitable expressions of bounds are and (given a fixed ). Therefore, prp:union of regions for double integration purple Proposition. (Union of regions for double integration) Let be a continuous function defined on a bounded set in . If is the union of and , in which the green overlapping set of and (in ) has green zero area, then if the integrals exist. rosybrown Remark. Example of analogous case in : (the overlapping set of and has green zero length) green Curves and green points have zero area in . crimson Corollary. (Subtraction of regions for double integration) Let be a continuous function defined on a bounded set in . If is the union of and , in which ,then if the integrals exist. violet Proof. First, the overlapping set of and is , which has zero area. Next, because , The result follows. purple Proposition. (Area given by double integration) Let be a bounded region in . Then, the area of is violet Proof. The volume under the graph of the constant function over equals the base area (which is the area of ) times the height (the height is one in this case). Therefore, the area of is the volume under the graph of over , which is by proposition about volume given by double integration (because for each ) rosybrown Remark. Recall that area of a bounded region can also be found by single integration. In some cases, using this proposition instead is more convenient. red Example. Let be a region bounded by the curves and , in which is a positive number. Prove that the area of bounded region is . Proof. Solving and , we get . Because , intersection points of these two curves are and . Therefore, the bound for is , and given a fixed , the bound for is . Thus, the desired area is rosybrown Remark. Geometrically, the bounded region is a disk (region in a plane bounded by a circle) of radius . Triple integration The concepts in the section of double integrals apply to green triple integrals (and also multiple integrals generally) analogously. We will give several examples for triple integrals in this section. darkblue Definition. (Triple integrals) Let be a function defined on a rectangular box (or rectangular cuboid) in . Consider a partition of into small boxes with volumes respectively. Choose an green arbitrary point in the th box. The function is green integrable over if exists. In that case, we denote this limit by ( is a mnemonic of volume), and call it the green triple integral of over . rosybrown Remark. The process of computing green triple integral is called green triple integration thm:generalized Fubini's theorem (triple integrals version) tomato Theorem. (Generalized Fubini's theorem (triple integrals version)) Let be a continuous function defined on a region (with arbitrary shape) . Then, the following hold. (i) if , then (ii) if , then (iii) if , then (iv) if , then (v) if , then (vi) if , then in which each function involved is continuous. That is, we can use either one of all possible integration orders for iterated integrals to compute triple integrals, with suitable bounds. purple Proposition. (4-dimensional volume given by triple integration) Let be an green integrable function defined on a rectangular box in . Suppose for each . Then, the green 4-dimensional volume under the graph of over is rosybrown Remark. It is just a theoretical result, and is difficult to be visualized. prp:volume given by triple integration purple Proposition. (Volume given by triple integration) Let be a bounded region in . Then, the green volume of is red Example. Consider a region that is bounded by a sphere (with radius , which is positive) that lies in the octant (+,+,+) (i.e., the octant in which and are positive). (i) Prove that the volume of is . (ii) Hence, prove that the volume of the region that is bounded by the sphere (in ) is . Proof. (i) The given bounds for is and . We can express the bounds as follows: given a fixed , aim: finding bounds for in the form of . Steps: given fixed and ( is selected from its bound, namely , and is selected from its bound, which depends on the fixed ), aim: finding bounds for in the form of . Steps: Therefore, by generalized Fubini's theorem (triple integrals version) (i) and proposition about volume given by triple integration, the desired volume is (ii) Because there are eight octants (in ), and each octant is symmetric to each other. There are seven more regions that are symmetric to in octants other than octant (+,+,+). Thus, the desired volume is rosybrown Remark. The region mentioned in (ii) is a ball (solid figure bounded by a sphere) of radius . Directional derivatives and the gradient vector Change of variables