7 Multiple integrals

Transcription

1 7 Multiple integrals We have finished our discussion of partial derivatives of functions of more than one variable and we move on to integrals of functions of two or three variables. 7. efinition of double integral Consider the function of two variables f(, ) defined in the bounded region. ivide the region into randoml selected n subregions s, s,..., s k,..., s n where s k, k n, denotes the kth subregion or the area of this subregion. Net we choose a random point in ever subregion P k (ξ k, η k ) s k and multipl the value of the function at the point chosen b the area of the subregion f(p k ) s k. If we assume that f(p k ) then this product equals to the volume of the right prism with the area of base s k and the height f(p k ). The sum n f(p k ) s k k= is called the integral sum of the function f(, ) over the region. The geometric meaning is the sum of the volumes of the right prisms, provided f(, ) in the region. The maimal distance between the points of the subregion s k is called the diameter of this subregion diam s k = ma P Q P,Q s k We have divided the region into subregions randoml. Ever subregion has its own diameter. The greatest diameter of subregions we denote b λ, i.e. λ = ma k n diam s k efinition. If there eists the limit lim λ n f(p k ) s k k=

2 and this limit does not depend of the choice of subregions of and the choice of the points P k in subregions, then this limit is called the double integral of the function of two variables f(, ) over the region and denoted f(, )dd According to this definition n f(, )dd = lim f(p k ) s k (7.) λ If f(, ) in the region then the double integral can be interpreted as the volume of the clinder between the surface z = f(, ) and. k= z z = f(, ) Figure 7.. The clinder between the surface z = f(, ) and There holds the following theorem. Theorem. If f(, ) is continuous in the bounded region then f(, )dd alwas eists. The proof will be omitted. This theorem tells us that for the continuous function f(, ) the limit lim λ n f(p k ) s k k= eists and does not depend of the choice of subregions of and the choice of the points P k in subregions.

3 7. Properties of double integral All of the following properties are reall just etensions of properties of single integrals. Propert. The double integral of the sum of two functions equals to the sum of double integrals of these functions [f(, ) + g(, )]dd = f(, )dd + g(, )dd provided all three double integrals eist. Proof. We use the properties of the sum and the limit. B efinition n [f(, ) + g(, )]dd = lim [f(p k ) + g(p k )] s k λ k= [ n ] n = lim f(p k ) s k + g(p k ) s k λ k= k= n n = lim f(p k ) s k + lim g(p k ) s k. λ λ k= k= B efinition the first limit equals to equals to g(, )dd. f(, )dd and the second limit Propert. If c is a constant then cf(, )dd = c f(, )dd i.e. the constant factor can be carried outside the sign of the double integral. The proof is similar to the proof of Propert. Propert. The double integral of the difference of two functions equals to the difference of double integrals of these functions [f(, ) g(, )]dd = f(, )dd g(, )dd. Propert is the conclusion of the properties and because f(, ) g(, ) = f(, ) + ( )g(, )

4 f(, )dd + Propert 4. If = and the regions and have not common interior points then f(, )dd = f(, )dd Proof. In the definition of the double integral the limit doesn t depend on the division of the region. Therefore, starting the random division of the region we first divide into and. Further random division of the region creates the random divisions into subregions of the regions and. The integral sum we split into two addends n f(p k ) s k = f(p k ) s k + f(p k ) s k (7.) k= where the first addend contains the products, having as one factor the area of the subregions of the region and the second addend contains the products, having as one factor the area of the subregions of the region, The first sum on the right side of this equalit is the integral sum of the function f(, ) over the region and the second over the region. If λ denotes the greatest diameter of the subregions of the region then λ ields that the greatest diameter of the subregions of and approach to zero. We get the assertion of our propert if we find the limits of both sides of the equalit (7.) as λ. 7. Iterated integral. Evaluation of double integral In the previous subsection we have defined the double integral. However, just like with the definition of a definite integral the definition is ver difficult to use in practice and so we need to start looking into how we actuall compute double integrals. In this subsection we assume that the bounded region is closed. There are two tpes of regions that we need to look at. The region is called regular with respect to ais if an straight line parallel to ais passing the interior points of the region cuts the boundar at two points. The regular with respect to ais region can be described b two pairs of inequalities a b and φ () φ (). This notation is a wa of saing we are going to use all the points, (, ), in which both of the coordinates satisf the given inequalities. 4

5 = φ () = φ () a b Figure 7.. Regular region with respect to ais Let the function f(, ) be defined in the region. The iterated integral of the function f(, ) over this region is defined as follows b φ () I = f(, )d d a φ () To compute the iterated integral we integrate first with respect to b holding constant as if this were a definite integral. This is called inner integral and the result of this integration is the function of one variable Φ() = φ () φ () f(, )d Second we multipl the function obtained b d and compute the outer integral b a Φ()d which is another definite integral. So, to compute the iterated integral we have to compute two definite integrals. First we integrate with respect to inner variable and second with respect to outer variable. To avoid the parenthesis we shall further write the iterated integral as I = b d φ () f(, )d (7.) a φ () 5

6 Eample. Compute the iterated integral I = d ( + )d The region of integration is described b the inequalities and. The sketch of this region is in Figure 7.. = = Figure 7.. The region of integration in Eamples and First we compute the inner integral b treating as constant Φ() = ( + )d = and then the outer integral ) ( + = = 4 I = 4 d = 5 5 = The region is called regular with respect to ais if an straight line parallel to ais passing the interior points of the region cuts the boundar at two points. The regular with respect to ais region can be described b two pairs of inequalities c d and ψ () ψ (). 6

7 d = ψ () = ψ () c Figure 7.4. Regular region with respect to -ais The iterated integral over the region regular with respect to ais is defined as d ψ () I = d f(, )d. (7.4) c ψ () To compute this iterated integral we have to find two definite integrals again. First we integrate with respect to inner variable b holding constant. The result is a function of one variable Ψ() = ψ () f(, )d ψ () Second we integrate with respect to outer variable I = d c Ψ()d In the iterated integral (7.) the variable is the inner variable and is the outer variable, in the iterated integral (7.4) the situation is vice versa. The conversion of the iterated integral from one order of integration to other order of integration is called the change (or the reverse) of the order of integration. 7

8 Eample. Change the order of integration in the iterated integral I = d f(, )d The region of integration has been sketched in Figure 7.. In the iterated integral given the inner variable is and the outer variable. After changing the order of integration the outer variable has to be and the inner variable. Note that the limits of the outer variable are alwas constants. The limits of the inner variable are in general (but not alwas) the functions of the outer variable. Choosing as the outer variable we can see in Figure 7. that the variable changes between and that is. Solving the equation = for we get = ±. The domain in this eample is bounded b the right branch of the parabola =, thus, the variable in this region is determined b. Changing the order of integration, we obtain the iterated integral I = d f(, )d Let us compute the iterated integral of Eample again, using the reversed order of integration, i.e. compute d ( + )d Here we integrate first with respect to ( ) ( + )d = + = + = + 4 Net we integrate with respect to ( + 4 ) [ d = + 8 ] = 5 The result is, as epected, equal to the result obtained in Eample. 8

9 Eample 4. Sometimes we need to change the order of integration to get a tractable integral. For eample, if we tr to evaluate d e d directl, we shall run into trouble because there is no antiderivative of e, so we get stuck tring to compute the integral with respect to. But, if we change the order of integration, then we can integrate with respect to first, which is doable. And, it turns out that the integral with respect to also becomes possible after we finish integrating with respect to. = = = Figure 7.5. The region of Eample If we choose the reversed order for integration then the region of integration is described b inequalities and, thus d e d = d e d Evaluating the inner integral with respect to we treat as constant, i.e. e is a constant factor and e d = e = e 9

10 Now it is possible to find the outer integral, using the differential d( ) = d e d = e d( ) = e = e Eample 4. Change the order of integration in the iterated integral I = d 6 f(, )d. (7.5) The region of integration is described b inequalities and 6. We sketch in the Figure the lines =, =, = and = 6. = = = 6 6 Figure 7.6. Region of integration of Eample 4 Obviousl in this region 6 and φ(), where {, if φ() = 6, if 6 Changing the order of integration 6 I = d φ() f(, )d

11 B the additivit propert of the definite integral I = d φ() 6 f(, )d + d φ() f(, )d = 6 6 d f(, )d + d f(, )d ividing the region b the line = we obtain two regular regions and. The region is determined b the inequalities and and b inequalities 6 and 6. The result of the Eample 4 we can interpret in two was. First, if we want to change the order of integration in the iterated integral (7.5), we have to divide the region of integration b the line = into two regions and and determine the limits for both regions. Second, if we divide the region of integration b the line parallel to ais into two regions and so that =, then I = I + I The last assertion holds also if we divide the region b the lines parallel to ais into three or more regions. It holds as well if we divide the region b the lines parallel to ais into finite number of subregions. We can conclude the first essential propert of the iterated integral. Propert. If the regular region is divided b the lines parallel to coordinate aes into n subregions then =... n I = n k= We need to prove two more properties of the iterated integral. Suppose the region is determined b inequalities a b and φ () φ (). Propert. If m is the least value and M the greatest value of the function f(, ) in the regular region and S denotes the area of the region then m S I M S (7.6) I k Proof. B the assumption for an (, ) m f(, ) M

12 B the propert of the definite integral φ () φ () md φ () φ () f(, )d φ () φ () Md which ields φ φ () () φ () m f(, )d M φ () φ () φ () or φ () m[φ () φ ()] f(, )d M[φ () φ ()] φ () Using the same propert of the definite integral again, we obtain b b m [φ () φ ()]d d φ () b f(, )d M [φ () φ ()]d a a φ () a The formula of the area of the region S = b a [φ () φ ()]d completes the proof. Propert. If the function f(, ) is continuous in the closed regular region, then there eists P (ξ, η) such that I = f(p )S (7.7) Proof. The continuous in the closed region function f(, ) has the least value m and the greatest value M in this region. Hence, there holds the assertion of the Propert. ividing the inequalities (7.6) b the area of the region S gives m I M S The continuous in the closed region function acquires an value between the least and the greatest, i.e. also the value I. Thus, there eists a S

13 point P (ξ, η) such that f(ξ, η) = I. Multipling the last equation S b the area S gives us (7.7). Now it turns out wh we have paid so much attention to the iterated integral and what it is useful for. Theorem. If the function f(, ) is continuous in the closed regular region then f(, )dd = I (7.8) Proof. If we divide the region b the lines parallel to coordinate aes into n subregions,,..., n, then b Propert the iterated integral I = n k= I k B Propert in ever subregion k there eists a point P k (ξ k, η k ) k such that I k = f(ξ k, η k )S k Thus, I = n f(ξ k, η k )S k k= enote b λ the greatest diameter of the subregions k (k =,,..., n). Now we find the limits of both sides of the last equalit as λ. On the left side there is an iterated integral, which is a constant and the limit is that constant. On the right side there is the integral sum of the function of two variables f(, ) over the region. B the assumption the function is continuous, hence, the limit of the integral sum eists and equals to the double integral of the function f(, ) over the region. It has turned out that the iterated integral equals to the double integral and so we don t need the term iterated integral an more. It s just a mean to compute the double integral and usuall we shall sa instead of iterated integral double integral. If the region is regular with respect to ais then we compute the double integral b the formula f(, )dd = b d φ () f(, )d (7.9) a φ ()

14 If the region is regular with respect to ais then we compute the double integral b the formula f(, )dd = d d ψ () f(, )d. (7.) c ψ () If the region is regular with respect to either of the coordinate aes then we can choose one of these formulas to compute the double integral. Eample 5. Compute the double integral ( + )dd if is the region bounded b the line + = and parabola =. To sketch the region we find the intersection points of the parabola and line solving the sstem of equations { = + = The second equation gives =. Substituting to the first equation gives the quadratic equation + =, whose roots are = and =. = = Figure 7.7. The region of integration of Eample 5 Using the sketch in Figure 7.7, we determine the limits of integration 4

15 and. B the formula (7.9) Compute the inner integral ( + )d = = ( ) + and the outer integral ( + )dd = ( ) d ) ( + ( + )d 4 = 4 ) ) ( 4 d = ( = 6 4 ( ) = 4, Change of variable in double integral Often the reason for changing variables is to get us an integral that we can do with the new variables. Another reason for changing variables is to convert the region into a nicer region to work with. The following eample gives an idea wh the change of variable can be useful. Eample. Compute the double integral ( 4) dd where is the region bounded b the lines + =, + =, = 6 and =. Notice that the first two lines are parallel and the third and fourth lines are ( parallel. The intersection point of the first and third line is A 9 5 ; 4 ), the ( ) 5 9 intersection point of the first and fourth line is B 5 ; 4, the intersection ( ) 5 point of the second and fourth line is C 5 ; 6 and the intersection point 5 of the second and third line is ( 5 ; 5 5 ).

16 + = + = = 6 A C = B Figure 7.8. The region of integration of Eample To compute this double integral b the formula (7.9), we have to divide the region with two vertical lines passing and B into three subregions, compute this double integral over these three subregions and add the results. The integration demands quite a lot of technical work, which can be avoided if we use the change of variables. Change the variables and b the variables u and v b the equations { = φ(u, v) (7.) = ψ(u, v) We assume that = φ(u, v) and = ψ(u, v) are one-valued continuous functions in the respective region of the uv plane and the have continuous partial derivatives with respect to both variables in that region. In addition we assume that the sstem of equations (7.) can be uniquel solved for the variables u and v. Then to an point of the region in plane there is related one point of the region in uv plane and vice versa. ivide the region in uv plane into subregions with the lines parallel to coordinate aes. Choose the random subregion s of the region The subregion s is a rectangle, whose area is s = u v. On the vertical lines u is constant and the equations (7.) are the parametric equations of some curve in plane and v is the parameter. On the horizontal lines v is constant and the equations (7.) are also the parametric equations of some curve in plane and u is the parameter. Thus, to the vertical lines u = const and u + u = 6

17 v Q 4 Q s Q Q v = v + v v = v u = u u = u + u u Figure 7.9. The region const there correspond two curves in -plane and to the horizontal lines v = const and v + v = const there correspond two another curves in plane. After that, to the point Q (u, v) there is related a point P (φ(u, v), ψ(u, v)), to the point Q (u + u, v) there is related a point P (φ(u + u, v), ψ(u + u, v)), to the point Q (u + u, v + v) there is related a point P (φ(u + u, v + v), ψ(u + u, v + v)) and to the point Q 4 (u, v + v) there is related a point P 4 (φ(u, v + v), ψ(u, v + v)). Therefore, to the rectangle Q Q Q Q 4 there is related quadrangle P P P P 4, whose sides are curves. v + v v P P 4 s P P u + u u Figure 7.. The region We approimate the quadrangle s b the parallelogram built on the vec- 7

18 tors P P and P P 4. The area of the parallelogram built on two vectors equals to the absolute value of the determinant, whose rows are the components of these vectors. Since P P = (φ(u + u, v) φ(u, v), ψ(u + u, v) ψ(u, v) and P P 4 = (φ(u, v + v) φ(u, v), ψ(u, v + v ψ(u, v)), then s = φ(u + u, v) φ(u, v) ψ(u + u, v) ψ(u, v) φ(u, v + v) φ(u, v) ψ(u, v + v ψ(u, v) B the assumptions made the total increments of the functions = φ(u, v) and = ψ(u, v) can be written and = u + u v v + ε u + ε v = u + u v v + ε u + ε 4 v where ε, ε, ε and ε 4 are infinitesimals as ( u, v) (; ). In the first row of the determinant of s there are the increments of the functions and with respect to u and in the second row with respect to v. In the first row v = and in the second row u =. Hence, φ(u + u, v) φ(u, v) = u u + ε u ψ(u + u, v) ψ(u, v) = u u + ε u φ(u, v + v) φ(u, v) = v v + ε v and ψ(u, v + v) ψ(u, v) = v v + ε 4 v Ignoring the infinitesimals of a higher order ε u, ε u, ε v and ε 4 v with respect to u and v, we get the approimate formula to compute the area s u u u u u u s = u v (7.) v v v v v v The functional determinant in the formula (7.) is called jacobian and denoted u u J = (7.) v v 8

19 Now we have the approimate equalit between the areas of subregions and s J s (7.4) which attains more accurac as u and v are getting smaller (since the functions are continuous, and are also getting smaller). The double integral is defined as the limit of the integral sum as the greatest diameter of the subregions approaches zero, i.e. (not writing the indees) f(, )dd = lim f(ξ, η) s λ where P (ξ, η) is a point in the subregion s chosen randoml. Let (u, v) be a point in the subregion s related to the point P (ξ, η) s. B our assumptions this is uniquel determined. Using the equalit (7.4), we obtain f(ξ, η) s = f(φ(u, v), ψ(u, v)) J s (7.5) where the last summation is over all subregions s of the region. Let λ denotes the greatest diameter of the subregions of the region. Then lim f(φ(u, v), ψ(u, v)) J s = f(φ(u, v), ψ(u, v)) J dudv λ If λ, then the continuit of the functions = φ(u, v) and = ψ(u, v) ields λ. Finding the limit as λ of both sides of the equalit (7.5) gives us the formula of the change of variable in the double integral f(, )dd = f(φ(u, v), ψ(u, v)) J dudv (7.6) Now return to Eample of this subsection at let s change the variable setting { u = + (7.7) v = This wa the parallelogram in plane converts to the rectangle in uv plane determined b the inequalities u and 6 v.the integrand converts to ( 4) = (v 4). To compute the jacobian (7.) we solve the sstem of equations (7.7) for the variables and = 5 u + 5 v = 5 u 5 v 9

20 and find the partial derivatives Thus, the jacobian u = 5, v = 5, u = 5 v = J = = = 5 5 and b the formula of change of variable (7.6) we find ( 4) dd = (v 4) 5 dudv = 5 du (v 4) dv = ouble integral in polar coordinates If we substitute the Cartesian coordinates and b the polar coordinates φ and ρ, we use the formulas { = ρ cos φ (7.8) = ρ sin φ Recall that φ denotes the polar angle and ρ the polar radius. To the constant polar angles there correspond the straight lines passing the origin in plane and to the constant polar radius there correspond the circles centered at the origin in plane. Therefore, first of all we use the change of variable (7.8) if the region of integration is a disk or the part of disk. To use the formula (7.6) we find f(, ) = f(ρ cos φ, ρ sin φ) and the jacobian φ φ J = = ρ sin φ ρ cos φ cos φ sin φ = ρ ρ ρ Since ρ is the polar radius, which is non-negative, we have J = ρ. Suppose the region in plane converts to the region in φρ plane. Then the general formula (7.6) gives us the formula to convert the double

21 integral into polar coordinates f(, )dd = f(ρ cos φ, ρ sin φ)ρdφdϱ (7.9) Eample. Convert to polar coordinates the double integral f(, )dd if the region of integration is the disk + 4. The region is bounded b the circle + = 4. Converting this equation, we have + 4 =, i.e = 4 or +( ) = 4. This is the equation of the circle with radius centered at the point (; ). Figure 7.. The disk + 4 The ais is the tangent line of this circle, hence, the polar angle changes from to, i.e. φ. The least value of the polar radius is for an polar angle. The greatest value of the polar radius depends on the polar angle. To get this dependence, we convert b (7.8) the equation of the circle + = 4 into polar coordinates which ields ρ cos φ + ρ sin φ = 4ρ sin φ ρ = 4 sin φ Thus in the disk given, the polar radius satisfies the inequalities ρ 4 sin φ.

22 B the formula (7.9) we get f(, )dd = f(ρ cos φ, ρ sin φ)ρdφdρ where is determined b the inequalities φ and ρ 4 sin φ. Using the iterated integral, we obtain f(, )dd = dφ 4 sin φ f(ρ cos φ, ρ sin φ)ρdρ Eample. Using the polar coordinates, compute the double integral dd + + if the region of integration is bounded b = and =. Since + + = ρ cos φ + ρ sin φ + = ρ +, then dd + + = ρdφdρ ρ + The region is bounded b ais and the upper half of the circle + =. Figure 7.. The half-disk bounded b -ais and half-circle = The region in Cartesian coordinates converts to the region in polar coordinates, determined b the inequalities φ and ρ. Thus, ρdφdρ ρ + = dφ ρdρ ρ +

23 First we find the inside integral. Since the differential of the denominator d(ρ + ) = ρdρ, then Now, the outside integral ρdρ ρ + = d(ρ + ) ρ + ln dφ = ln = ln(ρ + ) dφ = ln = ln 7.6 Computation of areas and volumes b double integrals While defining the double integral we got the geometrical meaning of this. Assuming that f(, ) in the region, the double integral f(, )dd is the volume of the solid enclosed b the region in the plane, the graph of the function z = f(, ) and the clinder surface, whose generatri is parallel to z ais. This idea can be etended to more general regions. Suppose that in the region for two functions f(, ) and g(, ) there holds f(, ) g(, ). The propert of the double integral gives [f(, ) g(, )]dd = f(, )dd g(, )dd Geometricall both integrals on the right mean the volumes of the solids. The first integral equals to the volume of the solid that lies below the surface z = f(, ) and above the region in the plane. The second integral is the volume of the solid that lies between the surface z = g(, ) and the region. Thus, the volume of the solid sketched in Figure 7. is computed b the formula V = [f(, ) g(, )]dd (7.) This is the volume of the solid enclosed b the surface z = f(, ) from the top, b surface z = g(, ) from the bottom and the clinder surface,

24 z z = f(, ) z = g(, ) Figure 7.. whose directri is the boundar of the region and generatri is parallel to z ais. Eample. Compute the volume of the solid enclosed b the planes =, =, z = and + + z =. The solid enclosed b these planes is the pramid sketched in Figure 7.4. The pramid is bounded b the plane z = from the top and b -plane z = from the bottom. Using the formula (7.) f(, ) = and g(, ) =. The volume of the pramid is V = ( )dd The region is determined b inequalities and, thus, V = d ( )d First we compute the inside integral ( )d = ( )d( ) = ( ) = ( ) 4

25 z Figure 7.4. The solid in Eample and second the outside integral V = ( ) d = ( ) d( ) = ( ) 6 = 6 If the height of the solid f(, ) = at an point of the region, then the volume of this solid V = S, where S is the area of the bottom (the region ). So, in this case the area of the bottom and the volume of the solid are numericall equal. Substituting the function f(, ) = into the formula V = f(, )dd we get the formula to compute the area of the plane region S = dd (7.) Eample. Compute the area of the region bounded b the lemniscate ( + ) = a ( ). Converting the equation of the lemniscate into polar coordinates, we obtain ρ = a cos φ. Hence, φ, which ields 4 φ 4 or 4 φ 5 4. The lemniscate is smmetrical with respect to the ais and ais. Therefore we compute the area of the region bounded b the lemniscate 5

26 a a Figure 7.5. The lemniscate in the first quadrant of the coordinate plane and multipl the result b 4. Converting the the formula (7.) to polar coordinates gives S = 4 dd = 4 ρdφdρ The region of integration is determined b the inequalities φ 4 and ρ a cos φ. Hence, S = 4 4 dφ a cos φ ρdρ Computing the inside integral gives and the outside integral S = 4 4 a a cos φ ρdρ = ρ cos φdφ = a 4 a cos φ = a cos φ cos φd(φ) = a sin φ 4 = a 7.7 efinition and properties of triple integral We used a double integral to integrate over a two-dimensional region and so it s natural that we ll use a triple integral to integrate over a three dimensional region. Suppose the function of three variables f(,, z) is defined in 6

27 the three dimensional region V. Choose the whatever partition of the region V into n subregions v, v,..., v k,..., v n where v k denotes the kth subregion, as well the volume of this subregion. For each subregion, we pick a random point P k (ξ k, η k, ζ k ) v k to represent that subregion and find the product of the value of the function at that point and the volume of subregion f(p k ) v k. Adding all these products, we obtain the sum n f(p k ) v k, (7.) k= which is called the integral sum of the function f(,, z) over the region V. Let diam v k = ma P Q P,Q v k be the diameter of the subregion v k and λ the greatest diameter of the subregions, i.e. λ = ma k n diam v k efinition. If there eists the limit lim λ n f(p k ) v k k= and this limit doesn t depend on the partition of the region V and the choice of the points P k in the subregions, then this limit is called the triple integral of the function f(,, z) over the region V and denoted f(,, z)dddz Thus, b this definition V V f(,, z)dddz = lim λ n f(p k ) v k (7.) If the function f(,, z) is continuous in the closed region V, then the triple integral (7.) alwas eists. The properties of the triple integral are quite similar to the properties of the double integral. 7 k=

28 Propert. [f(,, z)±g(,, z)]dddz = f(,, z)dddz± g(,, z)dddz V Propert. If c is a constant then cf(,, z)dddz = c V V f(,, z)dddz V Propert. If V = V V and the regions V and V have no common interior point then f(,, z)dddz = f(,, z)dddz + f(,, z)dddz V V Let f(,, z) be the densit of a three-dimensional solid V at the point (,, z) inside the solid. B picking a point P k to represent the subregion v k we treat the densit f(p k ) constant in the subregion v k and the product f(p k ) v k is the approimate mass of the subregion v k. The approimate mass because we have substituted the variable densit f(,, z) b the constant densit f(p k ). The integral sum is the sum of the approimate masses of the subregions, i.e. the approimate mass of the region V. The limiting process λ means that all diameters of the subregions are infinitesimals. The densit at the point P k represents the densit of the subregion v k with the greater accurac and the integral sum will approach to the total mass of the region V. Therefore, if the function f(,, z) is the densit of a three-dimensional solid V then the triple integral equals to the mass of the solid V m = f(,, z)dddz V If the region V has the uniform densit, then the mass and volume are numericall equal, i.e. if f(,, z), then the volume of the region V is computable b the formula V = dddz (7.4) An eample how to use this formula we have later. V 8 V V

29 7.8 Evaluation of triple integral The region V in the space is called regular in direction of z ais if there are satisfied three conditions.. An line parallel to the z ais passing the interior point of this region cuts the boundar surface at two points.. The projection of the region onto plane is a regular plain region.. Cutting the region b the plane parallel to some coordinate plane creates two regions satisfing the conditions. and. If those conditions are fulfilled, then the region V is determined b inequalities a b, φ () φ () and ψ (, ) z ψ (, ). We can define the iterated integral I V = b d φ () d ψ (,) f(,, z)dz a φ () ψ (,) To compute this iterated integral we have to compute three definite integrals. First we integrate with respect to the variable z holding and constant Ψ(, ) = ψ (,) f(,, z)dz ψ (,) We call this inside integral and the result is a function of two variables Ψ(, ). Net we integrate with respect to the intermediate variable holding constant φ () Φ() = Ψ(, )d φ () The result is a function of the one variable Φ(). Finall we compute the outside integral b a Φ()d Notice that the limits of the outside variable a and b are alwas constants. The limits of the intermediate variable φ () and φ () depend in general on the outside variable. The limits of the inside variable ψ (, ) and ψ (, ) 9

30 depend in general on the outside variable and on the intermediate variable. Eample. Compute the iterated integral d d ( + )dz First we integrate with respect to the inner variable z. Since + is constant, then ( + )dz = ( + ) z = + This result we integrate with respect to intermediate variable ( + )d = and finall with respect to + = = d = = 6 Since we have assumed that the projection of the region V onto plane is a regular plane region, then the region V can be determined b inequalities c d, φ () φ () and ψ (, ) z ψ (, ) and the iterated integral can be defined as I V = d d φ () d ψ (,) f(,, z)dz c φ () ψ (,) Just like we have defined the regular region in direction of z ais, we can define the regular region in direction of ais and the regular region in direction of ais. In the first case it is possible to define the iterated integrals b φ () ψ (,z) I V = d dz f(,, z)d or a φ () ψ (,z) b φ (z) I V = dz d ψ (,z) f(,, z)d a φ (z) ψ (,)

31 and in the second case I V = b d φ () dz ψ (,z) f(,, z)d a φ () ψ (,z) or I V = b dz φ (z) d ψ (,z) f(,, z)d a φ (z) ψ (,z) So, if the region V is regular in direction of all coordinate aes, si orders of integration are possible. The conversion of the iterated integral for one order of integration to the iterated integral for another order of integration is called the change of the order of integration. The iterated integral has the most simple limits, if the region of integration is a rectangular bo defined b a b, c d and p z q. All the faces of that bo are parallel to one of three coordinate planes. z q p a c d b Figure 7.6. Rectangular bo If we choose the outer variable, the intermediate variable and z the inner variable, we compute I V = b d d q d f(,, z)d a c p and, of course, five more orders of integration are possible.

32 The iterated integral is the appropriate tool to compute the triple integral. Theorem. If the function f(,, z) is continuous in the closed regular region V, then f(,, z)dddz = b d φ () d ψ (,) f(,, z)dz (7.5) V a φ () ψ (,) Eample. Compute the triple integral zdddz V if the region V is bounded b the planes =, =, z = and ++z =. First three planes are the coordinate planes. The fourth plane passes three points (; ; ), (; ; ) and (; ; ). The intersection line of this plane and plane z = is + =. The region of integration is sketched in Figure 7.7. z + = Figure 7.7. Region of integration of Eample The projection of the region of integration onto the plane is the triangle, which is determined b equalities and. Since the region of integration is bounded b the plane z = on the bottom and b z = on the top, the region of integration is determined b, and z. B the formula (7.5) zdddz = d d zdz V

33 First we compute the inside integral zdz = z ( ) = Net we integrate with respect to and finall ( ) d = [( ) ( ) + ]d = ] [( ) ( ) = [ ( ) 4 ( ) 4 d = 4 4 ( )4 + ( )4 4 ( )( ) 4 d ] = ( )4 4 = ( )( ) 4 d = [( ) 5 ( ) 4 ]d( ) 4 4 = [ ] ( ) 6 ( )5 = ( ) = Change of variable in triple integral Changing variables in triple integrals is nearl identical to changing variables in double integrals. We are going to change the variables in the triple integral f(,, z)dddz V over the region V in the z space. We use the transformation = φ(u, v, w) = ψ(u, v, w) z = χ(u, v, w) (7.6)

34 to transform the region V into the new region V in the uvw space. We assume that the functions, and z of the variables u, v and w are onevalued and the sstem of equations (7.6) has unique solution for u, v and w. Then to an point in the region V there is related one point in the region V and vice versa. In addition we assume that the functions (7.6) are continuous and the have continuous partial derivatives with respect to all three variables in the region V. The jacobian of this change of variables is the determinant u u z u J = v v z v (7.7) w w z w and we can transform the triple integral over the region V into the triple integral over the region V b the formula f(,, z)dddz = f(φ(u, v, w), ψ(u, v, w), χ(u, v, w)) J dudvdw V V (7.8) 7. Triple integral in clindrical coordinates The clindrical coordinates are reall nothing more than an etension of polar coordinates into three dimensions leaving the z coordinate unchanged. For the given point P (,, z) in the z space we denote P the projection of this point onto plane. enote b ρ the distance of P from the origin and b φ the angle between the segment P O and ais. Those φ and ρ are eactl the same as the polar coordinates in the two-dimensional case. efinition. The clindrical coordinates of the point P are called φ, ρ and z. Since φ and ρ have in the plane the same meaning as the polar coordinates then the conversion formulas from the Cartesian coordinates into clindrical coordinates are = ρ cos φ = ρ sin φ z = z (7.9) Find the jacobian of this change of variables. B the formula (7.7) we get φ φ z φ J = ρ ρ z ρ z z z z 4

35 z z P O φ ϱ P Figure 7.8. The clindrical coordinates φ, ρ and z of the point P The variable z does not depend on φ and ρ, hence, z φ = and z ρ =. The variables and does not depend on z, i.e. z = and z =. Consequentl, ρ sin φ ρ cos φ J = cos φ sin φ Epanding this determinant b the last column gives J = ρ sin φ ρ cos φ cos φ sin φ = ρ sin φ ρ cos φ = ρ. Since ρ is a distance J = ρ. Let V be the region in clindrical coordinates, which corresponds to the region V in Cartesian coordinates. B the general formula for change of variables in the triple integral (7.8) we obtain the formula to convert the triple integral in Cartesian coordinates into the triple integral in clindrical coordinates f(,, z)dddz = f(ρ cos φ, ρ sin φ, z)ρdφdρdz (7.) V V Supposing that the region V in clindrical coordinates is given b the inequalities α φ β, ρ (φ) ρ ρ (φ) and z (φ, ρ) z z (φ, ρ), we can write b the formula (7.5) f(,, z)dddz = b dφ ρ (φ) dρ z (φ,ρ) f(ρ cos φ, ρ sin φ, z)ρdz (7.) V α ρ (φ) z (φ,ρ) 5

36 Eample. Convert d d + f(,, z)dz into an integral + in clindrical coordinates. The ranges of the variables in Cartesian coordinates from this iterated integral are + z + The first two inequalities define the projection of this region onto plane, which is the half of the disk of radius centered at the origin. The third equalit determines that the region of integration is bounded b the paraboloid of rotation z = + on the bottom and b the cone z = + on the top. The region of integration and its projection onto plane are in Figure 7.9. z Figure 7.9. Region of integration of Eample In clindrical coordinates the equation on the paraboloid of rotation converts to z = ρ and the equation of the cone to z = ρ. So, the ranges for the region of integration in clindrical coordinates are, Now, b the formula (7.) we write + d d + f(,, z)dz = φ ρ ρ z ρ 6 dφ ρ dρ ρ f(ρ cos ρ, ρ sin φ, z)ρdz

37 Notice that the limits of integration are simpler in the clindrical coordinates. Eample. Using the clindrical coordinate, compute the triple integral d d a z + dz In Cartesian coordinates the region of integration is defined b the inequalities, and z a, i.e. bounded b the planes =, =, =, z = and z = a and b the clinder =. The generatri of the clinder is parallel to the z ais and the projection onto plane is the half circle =. This is the upper half of the circle = or + =, i.e. ( ) + = which is the circle of radius centered at (; ). The region of integration is sketched in Figure 7.. a z Figure 7.. The region of integration of Eample Convert the integral given into the integral in clindrical coordinates. The range of the angle φ in the projection of this region onto plane is φ. Converting the equation of the clinder + = into clindrical coordinates gives ρ cos φ + ρ sin φ = ρ cos φ or ρ = cos φ. Hence, the range for ρ is ρ cos φ. We didn t convert the third coordinate z, thus, z a. Converting the integrand into clindrical coordinates gives z ρ cos φ + ρ sin φ = zρ 7

38 Now, b the formula b (7.) d d a z + dz = dφ cos φ a dρ zρ ρdz The integration with respect to z gives a zρ dz = ρ z the integration with respect to ρ gives a cos φ ρ dϱ = a ρ a cos φ Finall, integrating with respect to φ, we get 4a cos φdφ = 4a = a ρ ( sin φ)d(sin φ) = 4a = 4a cos φ ( ) sin φ sin φ 7. Triple integral in spherical coordinates = 8a 9. In the previous subsection we looked at computing integrals in terms of clindrical coordinates and we now take a look at computing integrals in terms of spherical coordinates. First, we have to define the spherical coordinates. The following sketch shows the relationship between the Cartesian and spherical coordinate sstems. For a point P (,, z) in Cartesian coordinates given denote b P the projection of this point onto plane. enote b φ the angle between the segment OP and ais, b θ the angle between OP and z ais and b r the distance of the point P from the origin, i.e. the length of segment OP. An point in the space is uniquel determine b three parameters φ, θ and r. In Figure 7. the point P has been chosen in the first octant, hence all three coordinates are positive. The abscissa of the point P is the length of segment OR, the ordinate the length of segment RP and third coordinate z the length of segment OQ. Using the right triangle OQP, we get sin θ = QP r 8

39 z z Q r P R O φ θ P Figure 7.. The spherical coordinates φ, θ and r of the point P and cos θ = z r Since OP = QP, then the first equalit gives OP = r sin θ. The second equalit gives z = r cos θ. From the right triangle ORP we get cos φ = which ield and = OP cos φ = OP sin φ and sin φ = OP OP Substituting OP into the formulas for and, we obtain the conversion formulas from Cartesian coordinates into spherical coordinates = r cos φ sin θ = r sin φ sin θ z = r cos θ (7.) Find the jacobian for this change of variable. B the (7.7) we obtain φ φ z φ J = θ θ z θ r r z r = r sin φ sin θ r cos φ sin θ r cos φ cos θ r sin φ cos θ r sin θ cos φ sin θ sin φ sin θ cos θ Epanding this determinant, we get J = r sin φ sin θ cos θ r cos φ sin θ r sin φ sin θ r cos φ sin θ cos θ 9

40 Adding the st and 4th and the nd and rd term gives J = r sin θ cos θ(sin φ + cos φ) r sin θ(cos φ + sin φ) = r sin θ(cos θ + sin θ) = r sin θ In spherical coordinates the angle θ is measured with respect to z ais. Hence, we have the restriction θ. Then sin θ and the absolute value of jacobian is J = r sin θ Let V denotes the region in spherical coordinates corresponding to the region V in Cartesian coordinates. B the formula (7.8) we get the formula to convert the triple integral into the spherical coordinates f(,, z)dddz = f(r cos φ sin θ, r sin φ sin θ, r cos θ)r sin θdφdθdr. V V The ranges for the variables φ, θ and r are α φ β γ φ δ r (φ, θ) r r (φ, θ) (7.) Therefore the integral (7.) will become, f(,, z)dddz = β δ dφ dθ r (φ,θ) f(r cos φ sin θ, r sin φ sin θ, r cos θ)r sin θdr. V α γ r (φ,θ) (7.4) First of all we use the spherical coordinates to compute the triple integral, if the region of integration is ball or some portion of the ball. Eample. Compute the volume of the intersection of balls bounded b the spheres + + z = R and + + z = Rz. We are going to use the formula V = dddz V The first sphere is the sphere of radius R centered at the origin. Converting the equation of the second sphere to + + (z R) = R, we see that this is the sphere of radius R centered at B(; ; R). The intersection of the two balls is the solid of rotation, with z ais as the ais of rotation. Thus, the angle φ makes the full rotation, i.e. φ. 4

41 z B C A O Figure 7.. The intersection of the balls Converting the sum of the squares + + z into spherical coordinates, we obtain r cos φ sin θ + r sin φ sin θ + r cos θ = = r sin θ(cos φ + sin φ) + r cos θ = = r sin θ + r cos θ = r (sin θ + cos θ) = r Consequentl, the equation of the sphere + + z = R is in the spherical coordinates r = R and the equation of the sphere + + z = Rz r = R cos θ Since the maimal distance is different on these spheres, we have to divide the region of integration V into two subregions V = V V. In the first subregion θ rotates from the position OB to the position OA, hence, θ. In the second subregion θ rotates from the position OA to the ais, hence, θ. In the first subregion V the limits for the variables are, φ θ 4

42 and in the second subregion V, r R φ θ r R cos θ The volume of the intersection of two balls is V = dddz = dddz + dddz = V V V = dφ dθ R r sin θdr + dφ dθ R cos θ r sin θdr We start the computation of the first addend. The integration with respect to r gives R r sin θdr = sin θ r the integration with respect to θ, R = R sin θ R sin θ = R cos θ = R ( ) = R 6 and the integration with respect to φ, R R dφ = 6 6 φ = R Integrating the second addend with resect to r, we get R cos θ r sin θdr = sin θ r R cos θ = 8R sin θ cos θ 4

43 integrating with respect to θ, we get 8R = 8R sin θ cos θdθ = 8R cos 4 θ 4 = 8R an integrating with respect to φ gives ( 6 4 cos θd(cos θ) ) = R 4 R R dφ = 4 4 φ = R If we add these two results, we obtain the volume of this solid of rotation V = R + R = 5R 4