 So, we have started looking at double integrals, integral of functions of two variables, so over a domain. So, we had a function f defined in a domain d in R2 to R. So, we looked at double integral over d of f xy dxy. So, that was the double integral. So, whenever it exists, we said there was a theorem called Fubini's theorem that this double integral under suitable conditions, namely the function exists, the double integral exists. For example, when d is closed bounded and f is a continuous function, this will exist and can be given as integral over, okay. So, let us write case 1 is equal to x is equal to a to x is equal to b integral y goes from some function eta x to phi x of f xy d by dx whenever the domain d can be written as all points x y such that x lies between a and b and y lies between some function eta x to phi of x. So, this was type 1. So, these are type 1 domains. So, how does one go ahead? Page done, more pages, okay. You have to add a page, okay, right. And similarly for type 2 domain, if d is of type 2, namely x y such that y is between c and d and x is between some function say eta y to phi of y, then the double integral can be computed as iterated integral. So, y is going from c to dx goes from eta y to phi of y f xy sorry dx. So, meaning that the double integral for a function of two variables if the double integral exists can be computed by computing one variable integral at a time. So, these were called the iterated integrals. So, that gave us lot of maneuverability of computing double integrals. Let us look at what is a situation in three variables. So, let us take a domain d in R 3. So, a domain d in R 3 can be quite complicated, right. So, we are going to look at one possibility which says d is projectable onto x y plane. So, we want to look at d is a domain which is projectable into x y plane. So, there is think of the domain. So, that is z this is x this is y. So, this is a domain which can be projected onto x y plane. So, what does it mean? That means I can look at the projection of this. So, that will be something. So, this is my domain d let us call it as region R. So, that means what? So, mathematically means that d can be written as x y and z such that x y belongs to a region R contained in R 2 and z. So, at any point here so that is x y in the projection my z starts somewhere and goes up to somewhere. Let us make this as a. So, top is some surface. So, let us call it as the top one is z is equal to a function of x and y. So, let us call it phi x y and the bottom this one is z is equal to something eta of x y. So, the domain looks like this it can be projected onto the x y plane projection is a region in R in x y plane. So, how does the points in the domain look like? They look like x and y belong to R that is a projection and for every point x and y the z starts at this surface that is eta of x y and goes up to the top surface that is phi of x y. So, it is a kind of a cylinder you can think of some kind of a cylinder you can think of in R 3. So, that can be projected onto the it has a projection onto x y plane. So, when you project it onto x y plane that is the region R. So, how are the points described for a projection that is projection meaning z coordinate becomes 0. So, for any point x and y in the projected region R how high you will go so that you are in the domain. So, when you want to go up you start at this surface and end up this surface. So, this is the value of z so that you are inside the region D. So, we start at R is x y belong to R. So, point is here and you go along z x is how how high you have to go you have to start at the surface eta x y and go up to the surface of phi x y. So, this is a domain which is projectable onto x y plane. So, what is the advantage of this advantage of this is that Fubini's theorem states for R 3 the computation says the following. So, Fubini's theorem for D projectable onto x y onto x y plane that means D is written as x y and z where x y belongs to the region R that is inside the plane R 2 and z goes from some function we said say eta of x y to phi of x y. So, with that description of D as projectable onto x y plane the triple integral of F D x y z. Can be computed as x y belong to R and z. So, double integral over R D x y and integral of F x y z with respect to a variable z z goes from eta of x y to phi of x y. So, that is what we are saying the Fubini's theorem says if the triple integral exists then and the domain is of this type projectable onto x y plane then I can split the triple integral into two parts integrate the variable z from the bottom surface to the top surface. And so that is this from the bottom surface in the domain is x y in R. So, you start here at this surface and end up at this surface. So, from this surface so integrate the variable z once you have integrated the variable z it is a function of two variables. So, it is a function of two variables. So, that says that this as a function of two variables can be integrated over R with respect to x and y and that will give you the triple integral clear. So, we are splitting the triple integral into two iterated integrals. So, one is a double integral other is the ordinary integral of one variable. Now, suppose we further this region R which is a projection into x y plane also is of either type 1 or type 2. So, let us write further. So, that will be using the further if this R the region R which is in x y plane x y belonging to R2 is of type 1. So, whatever is type 1 in the plane x lies between A and B y lies between some function. So, let us call that as xi of y theta of x sorry theta of x that was type 1. X lies between A and B and y goes from one lower curve to the upper curve. So, lower curve is ix upper curve is theta x. Then what does this triple integral look like? So, this triple integral over d f x y z dx y z can be computed. Now, this double integral when R is of this type I can write as integral x goes from A to B y goes from xi x to theta x. So, this was the double integral this was a function. So, what is that function? So, that function is z equal to eta x y to phi of x y f x y z dz right. Here is with respect to dy and then with respect to right. So, what we are saying is the triple integral if your domain is special. First it is projectable on to x y plane. Then I can split the triple integral into two parts. One is the double integral over the projected region R and the third variable that is z that is the ordinary integration of one variable. Further if the projected part looks like domain of type 1 in the plane then you can use the earlier Fibonius theorem. The double integral will look like x goes from A to B iterated integral x A to B y goes from some function psi x to phi x theta x and the double integral. So, this means that if the surface if the domain d is projectable on to x y plane and the projection itself is of type 1 then the triple integral becomes integration of one variable at a time right. So, it will be three integrations which will be coming with respect to x y and z iteratively. So, that is if the projection R is of type 1 supposing it was of type 2 that is also possible. Then what will happen then this one will split according to the type 2 right will be integrating y first and then with respect to x. So, this is when the domain is projectable on to x y plane. So, two possibilities right projection on to x y plane is possible and the projection is either type 1 or type 2. So, you get two iterated integrals possibly if you are the if your surface if your region R is projectable on to say x z plane. Then again two possibilities will arise right and similarly on to say the y z plane. So, in all there are six possible iterated integrals possible for a function of three variables on a domain d right possible you can rise. So, let us try to do some examples on this. So, this is what it says. So, this is just recalling what we have said. So, this is a domain which is projectable on to x y plane. So, this is the region R which is the projection. So, every point you go from this lower surface right that is written here as h and g. So, from here to here right. So, can you see the pointer moving you cannot see the pointer moving. But I do not know whether you can write on top of it or not let us just see. So, if you take upon it good additional thing. So, if you go up you will start here and you will go up to here. So, that will be your z. So, for a point x y in the region R right you want to go it is projectable on to x y plane. So, you are moving along z axis. So, z goes from this region to this region. So, this is for the triple integral. So, it says if you can write it this way right projectable on to the x y plane. So, we will call it as type 1 elementary region. So, this is what x goes from A to B projection is again of type 1 right. So, x goes from A to B y goes from phi 1 to phi 2 and z goes from lower surface to the upper surface dx dy dz right. So, suitable conditions have to be put that namely this phi 1 phi 2 have to be continuous g is to be continuous and so on. So, we will assume all those nice conditions hold that it all the iterated integrals exist and R equal to the double integral. So, let us see. So, similarly if this projection is of type 2 that will be y going from C to D right x going from lower curve to the upper curve psi 1 to psi 2 and then z going from lower surface to the upper surface right. So, let us do one example. So, it is this example. So, we want to compute triple integral of the function 1 where D is the solid in R 3 bounded by the ellipsoid right. So, this is ellipsoid. So, ellipsoid is 4 x square plus 4 y square plus z square equal to 16. What does that ellipsoid look like? What is the ellipsoid? How do you visualize the ellipsoid? If you take z is equal to 0 what will that give me? z equal to 0 will give me the projection of this solid onto x y plane right and that looks like 4 x square plus 4 y square right. So, that is 4 x square plus 4 y square equal to 16 right okay. And similarly if you want projection on to other we have put y equal to 0 and projectible. So, this is a solid which is projectible on to all 3 coordinate planes x y, y z and z x. So, let us project it on to the x y plane that means z is equal to 0 you get 4 x square plus 4 y square equal to 16. So, what is that? So, that is a region R that is the projection okay. So, R is 4 x square plus 4 y square equal to 16. So, that gives me okay. So, I can write it as x y belonging to the region R the region R was the region R that is 4 x square plus 4 y square less than or equal to 16 right inside of that and that itself I can write it as there is a circle I can write it as a type 1 or type 2 right 4 x square plus 4 y square equal to 16. So, what will be that? So, x goes from so this is a region R okay and that itself I can write it. So, that will be the region R in terms of so that that is a circle. So, you can write type 1 or type 2 right 4 x square plus 4 y square equal to 16 that is x square plus y square equal to 4 right. So, type 1 where does x vary from that circle minus 2 to plus 2 lower part of the circle to the upper part of the circle or if you like you can write it also as type 2 where y goes from minus 2 to 2 x goes from the left side of the circle to the right side of the circle right. So, if you do that then you can write your so x goes from minus 2 to 2 that is a circle and where does y go from upper to the lower. So, what is the upper circle? So, upper circle is 4 minus x square positive square root upper part lower part is minus of that right. So, what does your domain look like? Domain looks like x goes from minus 2 to 2 right y goes from here to here and where does z vary from when you project you get the circle. So, what is the lower part of the circle that is a lower part of the ellipsoid to the upper part of the ellipsoid right. In the projection to be inside the domain D you will start with the lower part of the ellipsoid to the upper part of the ellipsoid. So, what will be the lower part of the ellipsoid? z in terms of y x and y. So, z square equal to so what are the equation of that ellipsoid? So, 16 right. So, you divide it by that. So, we will get z in terms of so I think it is quite clear. There is a lower part of the ellipsoid okay z square is it okay and that is upper part of the ellipsoid. So, this is a region ellipsoid when projected on to x y plane the points inside the ellipsoid can be described as x goes from minus 2 to plus 2. So, this is x goes from minus 2 to plus 2 and where does y vary from for every point x y varies from minus 4 minus x square to 4 minus x square plus square root. So, lower part of the circle to the upper part of the circle that is the region R. So, this is described in the region R the projection on to x y plane okay and then looking at to be z where in the region if you take a point in the region how high or how much high or how much low you should go to be inside. So, we will start at the lower part go to the upper part. So, this is a lower part and this is a upper part. So, what we have done is a function of three variables instead of integrating it three variable at a time right all together we have split integrate with respect to z integrate with respect to y and then integrate with respect to x okay. So, this is integrating one variable at a time. So, that is illustration of Fubini's theorem for okay. So, you can integrate all that and finally your answer comes out something. So, let us not bother about okay sign inverse and all that thing okay. Let us look at this. Let us find out what is what is integrate z dv say dx yz right that is also written as in short as d of v volume basically it comes from that. So, this dv is same as d of x y and z okay where d is the solid wedge in the first octant. So, your imagination also will come into picture in the first octant cut from the cylindrical solid by the planes y equal to x and x is equal to 0. So, how do we visualize that this is solid wedge in the first octant cut from the cylindrical solid. So, there is a cylinder y square plus z square less than or equal to 1 what is that cylinder if you want to just look at the cylinder y square plus z square less than or equal to 1 it is a three dimensional solid right cylinder three dimensional solid cylindrical solid. So, how do we visualize it is a cylinder with x is going through x axis it is symmetrical around x axis right x is independent. So, the x is goes minus infinity to plus infinity right and if you look at any point y square plus z square less than or equal to 1 at any point x how does y and z vary is a circle right. So, sections of the solid are circles y square plus z square less than or equal to 1 is that okay cylinder you are able to visualize okay in that cylinder we are looking at first octant that means we are looking at only part of the cylinder when x is bigger than 0 y is bigger than 0 and z is bigger than 0 on that part only right. So, you can see here. So, there is upper part only we are looking at in this picture where it is bounded it is bounded by the planes y equal to x that we do not bind it then it will be infinite right. So, y equal to x what is the plane y equal to x y equal to x is a plane right z is independent here that means what it is a plane which meets x y plane in the line y equal to x. So, in the in this line. So, this is a plane going up okay y equal to x how able to visualize y equal to x is a plane in R3 in right in coordinate plane x y it is a line but if you visualize it as a object as a set in R3 then it is a plane and what does it mean it is cutting the line it is cutting x y plane at the line y equal to x. So, it is going through that so it is a plane okay. So, that is one side of the plane bottom is x y plane anyway we are going up only right and then and x is equal to 0. So, what is x is equal to 0 that is a y z plane right x is equal to 0 the y z plane y and z vary. So, this is the wedge. So, this wedge is predictable on to x y plane right this wedge is predictable on to x y plane and what is the projection projection is this triangle okay bounded by y equal to x. So, this is the projection. So, that is visualized here in this side on the right hand side that is a projection okay. Is it clear the wedge when it is projected on to the x y plane that will give you a triangle. So, how do you write that now? So, x y z x and y z x and y in the projection z goes from bottom that is x y plane 0 how high it will go up to the part of the cylinder. So, that is z less than or equal to 1 minus x square okay. So, and the projection that are that is that triangle. So, let us write down the x between 0 and 1 y goes from the line y equal to x right that is y equal to x that line up to y equal to 1. So, it starts and that triangle it goes up to there or you can write other way around if you like the projection here is both type 1 and type 2 or you can write y equal to 0 to 1 x goes between 0 and x less than or equal to y either way it is okay. So, you can write down the integral by using Fubini's theorem it is a double integral over the region R which is a projection of z the function z integrated 0 to 1 minus y square right. So, that is the inside thing that is a projection is R which is type 1 or type 2. So, you can further write it as okay you can further compute 0 to 1 0 to y that integrated z integrated out gives you this and you integrate all. So, basically the idea is the region in R 3 you have to visualize whether you want to project it on to the x y plane or y z plane or z x plane which one it is easier to visualize okay the same way if you try to project it on to the y z plane that looks like very weird one and you cannot describe it as going from one point to another right. So, that is where your visualization power will come into picture.