LAB 8: INTGRATION The purpose of this lab is to give intuition about integration. It will hopefully complement the, rather-dry, section of the lab manual and the, rather-too-rigorous-and-unreadable, section of your textbook.. Recollections The integral is most concretely defined as a limit of a sequence of summations. If you have a function f : D R from some domain D, then you can approximate the volume under the graph of f by objects that you already know the volume of (i.e. if f is a function from R R then its graph is some surface in R 3 ; to approximate the volume under this graph it might be easiest to try to use rectangular prisms, or cylinders, or another object, to build the region you want to know the volume of). It s easiest to see this idea by using pictures. Figure. Approximating volume: the left by cubes, the right by cylinders Now, even with this definition in place, it shouldn t be clear how one can calculate the integral of a function on a given domain. There are only a few instances which are amenable to direct computation via the definition. ven in low dimension, computing an integral using this definitions is rather difficult. xample.. We can calculate the integral of f(x) x on the domain [, ] directly: first we subdivide our domain, say by n regions of length /n. Approximating the area under the graph of f gives us a sum x dx ( n n + ) ( (n ) + + n n n n ) 3 i. i
Taking the limit of the sequence defined by these sums as we get better and better approximations (i.e. as n goes to infinity), we can find ( n ) x dx lim n n 3 i lim n n 3 lim n n 3 i ( ) (n )n(n ) ( n 3 3n ) + n 3 Here we used the sum of squares identity to simplify the sum into a usable form. Similarly, one can compute other simple integrals by using more complicated sums; in particular, the above method will work for any function f(x) x n. But, this is effectively useless if we want to do serious computations. It really outlines the problem of finding an efficient method to do computations since, most sums one will encounter are virtually impossible to give an exact form for. This should really outline how amaing the next theorem is: Theorem.. If f is a nice-enough function, then the integral of f on a interval [a, b] can be computed by the formula b a f dx F (a) F (b) where F is an anti-derivative of f (i.e. F is a function whose derivative is f). The proof of this theorem isn t too difficult it s just abstract. To prove it, you first have to find a general-enough function F to serve as the anti-derivative of f. Then all you do is calculate the integral via the definition. The hardest part of the proof is in proving the derivative of F is f. But the beauty in the theorem is that we know a lot of derivatives. This means, by doing the reverse process, we also know a lot of anti-derivatives and can therefore calculate a lot of integrals. Remark.3. Notice that if one took the naive definition of the integral defined as the volume between the graph of a function f and the domain, then we wouldn t be able to use Theorem. to compute the integral. This is because what Theorem. computes is the signed-volume. This is a crucial, and subtle, difference when it comes to our intuition. Of course, negative volume doesn t make much sense but, negative signed-volume is just fine. This was something that was utilied in the last qui fairly often in order to quickly compute some integrals knowing only the shape of the domain and the symmetry of the function one integrated.. higher-dimensions We ve spent some time already talking about double integrals. The difficult part when computing a double integral is usually in setting up the bounds for integration. Here, we ll focus on triple integrals. Again, the most difficult part will be setting up the bounds for integration. On the other hand, the most tedious part will be evaluating a triple integral. There are three pieces of information one needs to set up an integral: a function, a domain, and a choice of volume used to approximate the domain. The way one writes an integral is supposed to capture this. If we write f dv then from dv we should know how we re approximating our domain (e.g. by boxes, by spheres, by cylinders, etc.), from and dv we should be able to find out the domain we re integrating over, and from f what we are integrating. Although knowing your function and how to find out bounds for your domain might be easy to understand conceptually, it might seem mysterious why one can or should use various different choices of volume in
order to specify dv. Thinking about it another way, we could say that the choice of dv is a way to linearie a different choice of coordinates. For instance, if we were trying to approximate the volume of an object in R 3 by using boxes, then we d know a given box has volume its length times its width times its height. If these three values are chosen to be with respect to the x, y, -axis, then we could approximate dxdyd (or in any other order, since multiplying one way or the other is independent of order). If instead of boxes one wanted to use cylinders, then one would need to know the volume of a cylinder: its the square of the radius times times the height of the cylinder. Note then that a cylinder has volume dv dad where A should be related to the area of a circle. If we took a small sliver from a circle, it s area would be given by its ratio to the whole circle times r where r is the radius. Again writing da d( θ r ) rdθdr we find that using cylinders, or sections of cylinders, our volume should be measured in dv rdrdθd. Using the definition of the integral, it isn t too hard to show that this intuitionistic-way of thinking can actually be made precise. We re just summing different objects. This is also what allows one to use the same formula we ve always used in order to compute the integral. Once it s set up, we can just apply Theorem. to the innermost integral, and then repeat. xample.. This example will use Cartesian coordinates. Using triple integration, we can find volumes of three-dimensional objects by integrating the constant function. This is easiest seen in lower-dimensions: if you integrate along an interval the function, you ll get the length of the interval; if you integrate the function along a region in R, you ll get the area of that region; similarly, integrating along a higher-dimensional object will give you the higher-dimensional volume of that object. Let s compute the volume of the tetrahedron of.aa) in Qui 3; I ll call it for short. More explicitly, this tetrahedron can be described by the region bounded by the planes y x, x,, and x. Once we find bounds for this region, we ll be left with an integral that looks like dv. From Qui 3 we also know our integral should equate to. How I like to set up triple integrals is like this: first, I choose an axis to ignore for the moment. I ll pick the -axis for this example. Then, I look at sections which are orthogonal to the -axis. These sections should be regions which are smaller in dimension and, most of the time, they ll be -dimensional. This is me just saying I m going to think about the triple integral as a single integral of a double integral: ( ) dv da d. bounds D Geometrically, I am trying to keep this picture in mind: In our integral, D should be a region similar to the ones appearing in the right diagram of Figure below. Notice that in order to get a definite integral, we ll need to use the variable to describe D. This corresponds to the fact that as we change our value, the region changes as well. First, let s set up the double integral of one of these regions with respect to some fixed (but arbitrary) -value. We ll need to pick either da to be dxdy or dydx. I ll pick the first one for no particular reason. Well, integrating with respect to x first means our lower bound is the left-most vertical line on one of these triangular regions. Since this value depends on, and is equal to, this will be our lower bound. The upper bound is fixed, and equal to y. In the y-direction our lower bound starts at the intersection of x and y x. This means our lower bound in the y-direction is, while the upper bound is always just y. Altogether, we have ( y ) dv dxdy d. bounds Finally, to find the bounds on the -axis we look at the lowest and highest values of the -component of points in. This is the top vertex p (,, ), and the bottom plane. So, we can fill in our bounds to get y dv dxdyd. 3
Figure. Left: the tetrahedron. Right: Slices of orthogonal to the -axis Computing this: y dxdyd (y ) dyd ( ) y y d ( ) + d ( ) + 3 xample.. This example will use cylindrical coordinates. We ll find the volume underneath a paraboloid +x +y and contained inside a cylinder x +y using triple integration. We ll proceed like before, separating the integral into a single integral of a double integral. The slices of Y in this picture are all circles, and they are completely determined by what the height of the slice is. Notice that, below the intersection of the cylinder and the paraboloid the radius of the circles of these slices is constant and equal to. Above the intersection of the cylinder and the paraboloid, the radius of these circles is. Also, equating the two defining equations of these objects shows this intersection occurs at x y or or 3. The above discussion shows our volume integral can be split up into a sum Y dv 3 r drdθd + 3 r drdθd.
Figure 3. The region Y, and horiontal slices of Y Computing the left summand is simple because it s a cylinder: 3 r drdθd 3. You don t even need to compute the integral if you remember the formula for the volume of a cylinder. For the latter summand, we can integrate: r drdθd r drd d ( ) 3 3 3 3 adding the two we find that the volume we were trying to find was equal to 3 +. xample.3. This example will use spherical coordinates. Instead of finding a volume, let s integrate a function whose domain is an object in R 3. For the function, we ll use and for the region, we ll use the region lying inside of the singular-conic x y and inside the sphere x + y +. Again we ll set up a triple integral dv W but, this time we will move directly to spherical coordinates. In Cartesian and cylindrical coordinates, it s often easier think of the integral as a single integral of a double integral, and consider slicing your domain in order to find appropriate bounds of your integral. In spherical coordinates, since we no longer use a linear coordinate in the integration (i.e. a term like dx, dy, d) it won t be easier to break up the integral. Instead we ll need to analye the region directly: Luckily, the function is symmetric about the r, θ-plane. This means we can just integrate the top half of the cone inside of the sphere of radius and then double this result: dv ρ sin(ϕ) dρdθdϕ. W bounds What s left is to change all Cartesian functions to spherical ones, and to find the lower and upper bounds for ϕ. Since we know ρ cos(ϕ), we only need to find bounds for ϕ. Since ϕ starts at the -axis, this just means we need to find the angle from the -axis to the cone x + y ; the lower bound will then be at the -axis, when ϕ, and the upperbound will be the angle we find at the cone. To find this angle, we can work in the y -plane and use some trigonometry, see Figure. below. Here one can find that is the height of intersection of the cone x + y and
the sphere of radius by substitution. Rewriting x + y and x + y, we find is the height of intersection which after solving for gives. Now, all we need to know is cos(ϕ) to find ϕ. Figure. In the plane y. Plugging everything into our integral: dv W ρ cos (ϕ) sin(ϕ) dρdθdϕ allows us to start computing. ρ cos (ϕ) sin(ϕ) dρdθdϕ cos (ϕ) sin(ϕ) dθ dϕ cos (ϕ) sin(ϕ) dϕ cos (ϕ) sin(ϕ) dϕ
substituting u cos(ϕ) this is cos (ϕ) sin(ϕ) dϕ u du bounds ( cos 3 ) (ϕ) 3 ( ) 3 ( ). 7