Surface Integrals

hello my name is Joel Schneider and in this lesson I'd like to talk with you about surface integrals and how they relate to double integrals which we already know we'll start with looking at surface integrals and understanding the concepts behind them I we'll go through a couple of examples of surface integral calculations okay so let's start with looking at what we already know about double integrals double integrals if we look in three dimensions we're integrating over a region in the XY plane represented here by this blue shape since we're only in the XY plane it's a two dimensional shape we only have two degrees of freedom in addition we only have two variables changing Z does not change whatsoever inside our region it is completely irrelevant X and Y are the only variables that change this means that we can only move freely in two directions our position along the vertical direction or the Z axis is completely determined by our locations in X and Y in other words the shape that we are confined to is a two-dimensional shape now let's compare this to surface integrals in surface integrals instead of integrating over a flat region in the XY plane we're integrating over some generic surface that can be anywhere in three space like this section of blue surface here in this situation ation we also have two degrees of freedom because as soon as our X and Y locations are determined our Z location is fixed the shape that we're integrating over is still a two-dimensional surface it's simply in three space this time however unlike with double integrals now we have three variables changing X Y and Z all change depend on where we are on the surface this complicates our situation somewhat we can't calculate this surface integral straightforwardly like we did with double integrals with double integrals we only have two integral symbols how can we use that for surface integrals if three variables are changing instead what we'll have to do is do a change of variables in order to convert our surface integral into a double integral we can visualize this by taking our surface and using a projection to move it down into the XY plane so we can use a double integral like we are already accustomed to doing so now that we know the background on Surface integrals let's move on to looking at how we can actually calculate them surface integrals are usually one of two forms there's the flux form which looks like the double integral over a surface of some Vector field dotted with the normal Vector times the surface element what this essentially does is it looks at every surface element DS and finds the component of the Vector field that is in the same direction as the normal vector and then we add this up over the entire surface this calculates the total flux of a given Vector field through a surface the other form of a surface integral is a much more generic form it is simply a scalar function of three variables times an element of surface area we usually use this type of surface integral when we're trying to integrate some scalar quantity over a surface although these two forms look very different we'll find out that the way we evaluate them is actually very similar because of this we'll simply start with the flux integral as it is the one that we usually use most commonly and along the way we'll discover how we can solve the scalar surface integral as a starting point let's begin by looking at a normal Vector well if we're trying to find the normal Vector to a surface we already know a method that we can solve this problem we can use the gradient we know that the gradient of a function is going to produce a vector that is perpendicular or normal to the level curve so we can use the partial derivatives in the gradient operator on our function that is our surface in order to be able to find the normal Vector if we have a function of our surface that is in terms of X Y and Z the normal Vector is going to be f subx f suby f subz we also looked at a special case where we already have the formula for our surface explicitly given in terms of Z equals a function of X and Y in this special case the partial derivative with respect to Z ends up being -1 either of these vectors gives us the normal vector and if we divide by its magnitude then we can find the unit normal once we have our unit normal we can take the dotproduct of our desired Vector field with this normal vector this will give us a scalar function of X Y and Z just like we have in our generic form of the surface integral the only step that we have left in converting our surface integral into a double integral is to rewrite our surface in terms of area to do this we need to find DS in terms of Da we need to perform a change of variables that represents the projection we're trying to do if you look here at these two different elements of surface area DS when we project them into the XY plane we'll get exactly the same da but you can see that because the slope of the surface is much different the area in the surface DS of the right surface element is much greater than that of the left so our change of variables is going to have to take into account the angle of our surface to do this we can find the Jacobian if you recall the Jacobian is a scaling factor that we add into our expression for our integral when we're trying to convert from one variable system to another so we can rewrite DS as our Jacobian time da if you recall from class we actually derived what this scaling factor is by using the angle of the surface we found that the unit normal time DS is equal to plus or minus the vector - F subx - F suby 1 this was for the case in which we had Z of our surface explicitly given as a function of X and Y in the case when our surface is given as an implicit function of X Y and Z we modify this slightly to get plus or minus Vector FX over FZ FY over FZ1 now since this was just a Jacobian scaling Factor Vector we have to make sure that we include our da so we have each of these vectors time da now that we have found the normal Vector time DS as a function of the variables given to us in our surface we can also calculate the scalar DS since the unit normal Vector has magnitude one we can take the magnitude of the vector we found above and will be left with Ds we can use this to solve the scalar form of the surface integral that we mentioned earlier sometimes the quantity n hat DS is abbreviated by DS with a vector sign over the S both of these two representations are completely equivalent now we need to deal with the plus or minus symbol that we had in our n hat DS above by convention the normal Vector on our surface is usually oriented such that Z is positive however it's usually good to delineate whether we're asking for upwards flux or downwards flux just to avoid confusion however sometimes we are asked to find inwards or outward flux these don't necessarily have some direct relationship between upwards and downwards for example if we look at this upper half of a hemisphere the inward flux points downwards and the outward flux points upwards so when problems are referring to inwards and outward fluxes we need to pay particular attention to which direction our Z is pointing so now we have every piece we need to be able to calculate our surface integrals so let's outline the general problem solving approach to solving a surface integral the first step is to sketch a picture it doesn't have to be terribly accurate it just needs to help you be able to visualize what the problem is and what areas we're trying to find integrals over the next step is to identify the surface you're interested in and figure out what its boundaries are often times these boundaries and their equations will be very helpful for us once we've projected into the XY plane the next step is to use the Expressions we found earlier to calculate n hat DS you also need to make sure that after you've calculated this you identify which direction is the correct Z Direction based on the context the problem next you'll project your surface that you're interested in into the XY plane and using the boundaries that you know figure out what the boundaries of your region are in the XY plane plug in your expression for n hat DS that we calculated earlier and then you should be left with a normal double integral in the XY plane that we can evaluate as usual so let's try and apply this problem solving approach to a couple practice problems the first problem is to find the upward flux of the vector field f = x i + y j hat plus z k hat through the surface s where s is z equal 4 - x^2 - y^ 2 only above the XY plane so our first step is to draw picture so let's start with that if we draw in our axes here we know that since we have four - x^2 - y^ 2 that as X and Y get very large our Z is going to decrease to negative infinity and that our maximum will be when X and Y are both zero because the X and the Y are each squared and we have no square roots anywhere in here we know that this is a paraboloid so we can sketch in our paraboloid here with maximum at 0 0 and our boundary is going to be when the surface intersects the XY plane this is when Z = 4 - x^2 - y^2 is equal to 0 this means that our boundary is going to be x^2 + y^2 = 4 which is a circle in addition we need to make sure that we note what direction of flux we're looking for we're trying to find the outward flux denoted by this orange arrow here since regardless of our location on the surface the outward flux always has a positive Z component that means that the normal Vector we're looking for has positive Z component as well so if we calculate our n hat DS in the next step we know that we have FX fy1 * da if we plug in our function of our surface for f then we can solve to be 2x 2 y1 * da now we need to look at the region that our surface makes when it's projected into the XY plane if we draw our diagram again the projected region is going to be this purple area at the bottom we already found the boundary was a circle so our projected region R is going to be x^2 + y sared is less than or equal to 4 so if we take our surface integral double integral over the surface of f do n hat DS we can plug in all of the values we have and turn this into a double integral over the region R of the vector XYZ which is our F dotted with our Vector 2x 2y 1 da if we multiply this out we'll get the double integral over the region of 2x^2 + 2 y^2 + z * da since we're in the XY plane we don't want a variable Z in our expression we only have variables X and Y so we will need to convert our Z into X and Y we use the function of our surface which we already found because everywhere in our region Z is always satisfying this relationship of Z = 4 - x^2 - y^2 if we plug in this expression for Z and simplify then we'll have the double integral over the region of 4 + x^2 + y^2 * da since the region we're integrating over as a circle it seems natural to convert convert this to Polar coordinates so that we can evaluate it more easily if we're converting it to Polar coordinates then x^2 + Y 2 is equal to R 2 so our integrant will be 4 + r 2 Da in polar coordinates is R Dr D Theta for a circle since we have x^2 + y^2 = 4 is our boundary that means the radius is 2 so our radius ranges from 0 to 2 and our Theta ranges from 0 to 2 2 pi if we integrate this with respect to R and evaluate then we'll be left with the integral from 0 to 2 pi of 8 + 4 d Theta this simplifies to a final answer of 24 Pi okay let's try another practice problem let's say we want to find the flux of the same Vector field F = XYZ out of the surface x^2 + y^2 + z^2 = 4 we can graph this in three dimensions we always want to start with a picture and we get a sphere of radius 2 now since we're trying to find the flux outward that means the normal Vector we're looking for are these two orange arrows here however because our shape is a sphere we notice that we don't have a consistently upwards pointing or downwards pointing normal Vector depending on where we are on our surface sometimes outwards means up sometimes out words means down in order to simplify our problem let's try using symmetry if we look at the top half of our sphere you can see that the vector field has a positive Z component as does the normal Vector if we look in the bottom half of the sphere our Vector field is exactly the same as it is in the top half of the sphere however the z-coordinate is negative so it's been reflected over the XY plane the same thing is true of the normal Vector because it's a sphere the noral Vector is exactly the same in the bottom half as it is in the top half just with a negated z-coordinate so the flux or the top half of our sphere is going to be exactly the same as the flux over the bottom half of the sphere so if we want to calculate the overall flux what we can do instead is calculate just the flux over2 and then multiply by two at the end since we can choose whichever one we want to figure out the flux let's just say we choose the top half in this case our normal Vector points up it also points radially outwards since this is a sphere that means that our normal Vector is going to be equal to XYZ however we need to make sure that we have a unit normal Vector so in order to do this we have to divide by the magnitude of our normal Vector since we're on the surface of a sphere with radius 2 that means the magnitude of our Vector is always going to be a constant two so now we're going to work with actually calculating our surface integral to find the flux instead of converting into a double integral let's see if we can continue to use symmetry to simplify our calculations instead of figuring out n hat DS in terms of Da what if we can just explicitly find it we've already calculated n hat by itself is there a way that we can calculate DS by itself well we have a sphere and it seems natural to explore our sphere in spherical coordinates let's draw out what a different itial unit of volume looks like in spherical coordinates here is our small unit of volume and we know that DS that we are looking at since we're on a surface of a sphere is this curved surface here we also know that the perpendicular length to the surface is going to be D row since that's the radial Direction so we have something that is roughly cubically shaped with one surface of surface area DS and a height of D row we already know that this differential element of volume has volume Row 2 sin f d r d d Theta we also know that the volume of a cube is the surface area of the base times the height so if we want to find the surface area of the base or DS we can take DV / d r that means that DS is equal to r^ 2 sin 5 d d Theta now let's see if we can move forward with actually calculating our surface integral we have the double integral of f do n hat DS over the surface well we know that XYZ is equal to our Vector field and we found that our normal Vector is equal to 12 XYZ so if we take the dotproduct here we're going to get x^2 + y^2 + z sared since we know that we're on the surface of a sphere we know that this quantity is equal to the radius squar so 2^ 2 so we have the 1/2 still out front then the 2^ squar from product of our two vectors and now we can substitute the expression we found for DS however since we're on the surface of a sphere we know that row equals 2 so we can plug that in as well if we're looking at a constant Theta what are the ranges for fi since we're only looking at the upper half of a sphere we know that fi ranges from zero at the Z axis to Pi / 2 on the X Y plane and then Theta ranges from 0 to 2 pi if we evaluate this integral we'll get 16 pi and since our final answer is twice that we get a final total outward flux of 32 Pi I hope that this video helped to summarize the concepts behind surface integrals as well as how to calculate surface integrals we started by looking at surface integrals and examining how we can use projections to to convert them into double integrals which are easier to calculate we also looked at using symmetry to solve surface integrals simply by directly calculating the normal vector and the element of surface area if we can take advantage of symmetry thanks for watching

Channel: MIT's Experimental Study Group

Views: 256,860

Keywords: Education

Id: 8V_G75YlV9o

Length: 19min 45sec (1185 seconds)

Published: Tue Apr 29 2014