23: Scalar and Vector Field Surface Integrals - Valuable Vector Calculus

so we are going to talk about surface integrals in scalar and vector fields starting out with a scalar field and i'm going to use this as an example of a surface now if you aren't already familiar with the concept of line integrals that's a very good thing to learn before moving on to surface integrals which are a more advanced version so if you aren't familiar with line integrals already i've left some links in the description to videos explaining that you can check those out and then come back here now the idea of line integrals is that we would integrate over a curve that had a range of movement in for example two dimensions i'll put some visuals on the screen now so you can see how this works basically because our input curve is already in two dimensions the idea of a line integral is that we compute the surface area of something that looks like a curtain as it moves through three dimensions when we're looking at a surface integral our surface is already a three-dimensional shape and so if we want to represent the function evaluated at some point on this shape that exists in three dimensions we would require a fourth dimension to represent that height so one way to think about the surface integral is that it's a three-dimensional volume where we take a height on top of this surface that exists in four dimensions now if it's difficult to think about a shape in four dimensions one way you can think about the surface integral is that we first take our surface in three dimensions and flatten it so that it exists only in two dimensions from there evaluating the function at every point on the surface corresponds to giving this a height so we're just looking at an ordinary three-dimensional function on top of this surface and we take the volume that would be our surface integral remember that when we were doing ordinary double integrals we would consider a two dimensional input region and the way that we computed the integral was by first splitting that region into a bunch of rectangles we use those rectangles as the base and then we would take the height to be the function evaluated at each point so this becomes a three-dimensional idea then that one small part of the integral is the volume of this rectangular prism the base times the height when we're doing a surface integral we're looking at the same idea but instead of looking at a two-dimensional input region our input region is a surface in three dimensions and therefore instead of just looking at a change in area along this region we're looking at an area on this surface and we call the area of a small region on this surface d s so this expression here is saying that we take the function evaluated at some point on the surface that's the height of this rectangular prism and then we multiply by the base which is the area of a small part of this surface d s so now that we have this expression for a scalar field surface integral we need to figure out a way to actually compute it and the first step to doing that is understanding how to describe our surface the way that we accomplish that is through a parametrization now in order to parametrize a surface we need to have two different variables and the reason for that is that with a surface there are two different independent directions of movement so in order to describe every point on the surface we need to have a parametrization in terms of two variables so we have a vector-valued function r of two variables u and v that takes in these two variables and outputs a point on the surface then we can describe the range of values u and v that gives every point on the surface we're interested in now that we have this parametrization r of u and v we can start to think about how to calculate d s remember that d s is an area of a small section on this surface so we need to think about how we can represent that in terms of u and v remember that r of u and v is going to give us a point on the surface if we change the value of u by some small amount so that we're looking at u plus d u instead that's going to create a particular movement on the surface a movement in some direction similarly if we do a small change in the value of v so that we're looking at v plus dv that's going to create a movement along this surface in a different direction we can complete this picture by connecting to the final point u plus d u comma v plus dv so this area represents a small range of u and v values and how they map onto the surface we want to find the area of this part of the surface one important element of this calculation is that as we make d u and dv very very small we can approximate the area on this surface as a parallelogram and that's because as we get smaller and smaller ranges on the surface we can start to approximate it as flat in that small region and therefore we're just looking at a parallelogram if we're looking at a parallelogram with side lengths a and b and an angle of theta between those two sides the area is going to be the base times the height but we can write the height in terms of b sine theta by looking at a right triangle just like this where we have the side b as the hypotenuse then b sine theta becomes this side we can use this formula to find the area of our parallelogram over here to use this formula the first step here is to find the side lengths of these parallelograms remember that each of these sides represents a movement purely due to a change in one variable along this side we aren't changing v at all we're only moving in the direction of increasing u from u to u plus d u if we're looking at a vector valued function r of u v the movement caused by a small change in u is going to be in the direction of the partial derivative of r with respect to u that partial derivative represents how does r change as we change u and that's the direction we're looking for and if we want that change in terms of a small movement d u the length of this side is going to be r sub u times d u similarly the side length up here is going to be r sub v times dv and from here we can start plugging things into this formula the area is going to be the length of this side which will be the magnitude of r sub u times d u times the length of this side the magnitude of r sub v times dv times the sine of the angle between them this formula is actually something that we've studied before it is the formula for the magnitude of the cross product the cross product between these two vectors r sub u and r sub v here because d u and dv are both just scalars i'm going to factor them out of that cross product so because the area of this parallelogram is given by the magnitude of the first vector times the magnitude of the second vector times sine theta that expression is just the magnitude of the cross product so the area of this parallelogram d s is equal to the magnitude of r sub u cross r sub v times d u dv now we can plug this formula for d s into our original surface integral here and transform it into an ordinary double integral in terms of u and v in this case r is talking about the range of uv values that describe our entire surface and because r of u v our parametrization gives the x y and z values on the surface we can easily plug those into the function as well and so this integral is something that we can actually compute now we're going to go through one example of a scalar field surface integral evaluate the integral over s of x z plus y squared ds where s is the surface described by x squared plus y squared equals 16 for z between zero and three just like when we were deriving the formula the first step here is to get a parametrization of our surface so let's think about what it means to have x squared plus y squared equals 16. well we know that that's going to describe a circle in the x y plane where we have a constant distance from the origin of 4. in that case we could use some angle theta to represent every point on the circle so we would have x equals 4 cosine theta and y equals 4 sine theta so that lets us describe the circle x squared plus y squared equals 16. but we're not just looking at a circle we're looking at a surface where we take this circle for a range of z values but notice that this circle stays the same regardless of what z we plug in and that means that the surface we're looking at is actually going to be a cylinder in that case what we can use as our second variable for the parametrization is simply z so the two variables that we're going to use to parameterize our surface are theta and z and we can use the parametrization here r of theta z equals our x value four cosine theta our y value for sine theta and our z value just being the variable z we also need to think about the range of values of theta and z that give us this surface for the z values it's pretty clear that we're going to have z between 0 and 3. for theta remember that this surface has all of the values with x squared plus y squared equals 16 meaning we're going to go all the way around the circle and one full revolution of a circle is values of theta between 0 and 2 pi so that's going to be the range for our theta variable the last thing that we need to do is calculate the magnitude of this cross product in our case that's going to be r sub theta cross r sub z so we need to figure out those partial derivatives of this function first of all r sub theta that's going to be the partial derivative of each of these components with respect to theta for the first one we're going to have the derivative of 4 cosine theta which is negative 4 sine theta the derivative of 4 sine theta will be for cosine theta and then the derivative of z with respect to theta is going to be 0 because we're looking at a partial derivative if we want r sub z notice that the x and y components are just in terms of theta so if we differentiate with respect to z those are constants the derivative is zero and we have the partial derivative of z which is just one now we need to take the cross product r sub theta cross r sub z and we can use that determinant formula to actually compute this so if we take our two vectors r sub theta and r sub z plug them into the cross product formula and compute this determinant the result that we get is the vector 4 cosine theta 4 sine theta and 0. our goal is to get the magnitude of this vector so if we take the magnitude here what we're going to get is the square root of four squared times cosine squared theta plus sine squared theta and here cosine squared theta plus sine squared theta is just one so we have the square root of four squared which gives us four that's the magnitude of this cross product now we're ready to turn this into an actual integral we can plug everything into the formula that we have right here first of all we're going to take the double integral over the region r that describes the range of u and v values for our surface we know that theta is going to go between 0 and 2 pi on our surface and we know that z is going to go between 0 and 3. so that gives us the range for our double integral then we want f of x y z we know that x y z values are described by this vector over here so if we want x times z that's going to be 4 cosine theta times z and then we add y squared which is 4 sine theta squared from here we're going to multiply this entire expression which is just f of x y z times the magnitude of our cross product we got that the magnitude of the cross product was equal to four so we're just going to multiply a four out here d z d theta since we're focusing on surface integrals in this video i'm not going to go through all the computations of this double integral but i've written out one step in the process here and i've written our final answer if we evaluate this integral which is 192 times pi so that's all we need to do to evaluate a surface integral in a scalar field and now it's time to start looking at vector fields so the purpose of a vector field surface integral is to measure the component of some vector field f normal to a surface s across the entire surface in order to understand how this can be calculated we have to start by thinking about how we can measure the component of f normal to a surface s at a single point in order to do that we can think about taking our surface and constructing a unit normal vector a vector that by definition is normal to the surface pointing straight out of the surface at a particular point now if we want to measure the component of f normal to the surface that's the same thing as measuring the component of f in the direction of the normal vector and the way that we do that is with a dot product if we take the dot product of f with a unit normal vector n that gives the component of f in the direction of n because the dot product talks about how parallel the two vectors are now remember that f dot n is a numerical value it's just a number that represents the component of f in the direction of n and because it's just a number we can make a graph of it in order to graph f dot n we'll start by taking our surface s and flattening it so that it becomes a two-dimensional region so let's say we flattened our original surface s and got this two-dimensional region right here in that case we can take f dot n for every point along this surface and that's going to correspond to some height above our two dimensional input region at every single point on this two-dimensional region here corresponding to some point on the original surface s the height above that point represents f dot n at that single point if instead of considering just a single point we want to aggregate all that information and look at the entire surface instead of just taking the height at a point we can take the volume under the entire shape and the volume of this function f dot n above the two dimensional region representing s that is what we call the vector field surface integral which we denote by the integral over our surface s of f dot n times d s where d s is the area of a small part of that original surface we also sometimes write it as f dot d s where d s is itself a vector and that's because once we get to the computation it's often easier to group and ds together as just one thing from here turning this vector field surface integral into something we can actually compute follows a lot of the same steps as we did with scalar field surface integrals just like before our first step will be to parametrize the surface in terms of some function r of u and v then we need to figure out how to find the unit in normal vector n times d s we can start with the formula that we found when we were doing scalar field surface integrals that d s is equal to the magnitude of r sub u times the magnitude of r sub v times the sine of the angle between them times d u dv from here we don't just want ds we want n ds that unit normal vector but if we want the unit normal vector on the left side we can also multiply it on the right side over here notice what this part of the formula is we have the magnitude of the first vector times the magnitude of the second vector times sine theta times a unit normal vector and this unit normal vector is normal to the surface but remember that r sub u and r sub v are partial derivatives of our vector function which means that they're going to move along the surface both of these vectors represent movement in the surface so this unit normal vector here because it's normal to the surface it's also going to be normal to both of these vectors and therefore this entire formula here is the exact formula for r sub u cross r sub v and we don't even need the magnitude this time because this unit normal vector represents the direction of that cross product so when we have a parametrization r of u v that describes our surface we can write the unit normal vector times d s as r sub u cross r sub v d u dv and therefore we can turn our original surface integral into a double integral over a region r of uv values that gives our surface of the vector field f evaluated at the point on the surface r of u v dotted with n d s which is what we have right here r sub u cross r sub v d u d v and that's all we need to compute the surface integral now there's one more thing that we need to understand with vector field surface integrals and that is to pay attention to the orientation of the unit normal vector to see what that means let's say the surface that we're looking at for our integral is the surface of a sphere in that case we could have a unit normal vector that points off of the sphere just like this that would be normal to the sphere however there is another unit normal vector that we could have chosen instead and that is the unit normal vector that looks like this the vector pointing in the exact opposite direction so there are always two unit normal vectors to a surface there is the original vector n and there is negative n when we do a vector field surface integral we always have to recognize that we're choosing an orientation for the normal vector now when problems describe the normal vector they'll usually say that it's either upward versus downward or outward versus inward in the case of the sphere here it would make sense to describe the normal vector as either pointing outward away from the sphere or inward towards the center of the sphere when the normal vector is described as upward versus downward that's talking about the z component of the normal vector so if our normal vector for example were 0 2 negative 1 we could tell that this normal vector would be oriented downward because the z component is negative if we wanted it to be oriented upward we would take the negative of this vector so that the z component was positive now when the normal vector is described as inward versus outward it's not always as clear which direction our normal vector is pointing so in those cases if we can't tell whether our normal vector is pointing in the right direction what we can do is choose some values for u and v just plug in some values there determine the point on the surface and the normal vector and from there we can see if it's pointing in the right direction for example let's say that our surface is this sphere here if we took our parametrization and plugged in a certain value for u and v we would get a specific point on the sphere which say is one zero zero then we can plug in those same values for u and v to get our unit normal vector and say that our unit normal vector at this point was equal to negative one zero zero from there it's pretty easy to see that our normal vector is going to point back towards the center of the sphere and therefore it is oriented inward from there if the problem said the normal vector is oriented outward we would need to take the negative of our normal vector so that we got the proper orientation so now we'll go through an example of a vector field surface integral evaluate the integral over s of f dot d s where f is z over x x squared e to the x squared plus z squared and s is the surface described by x squared plus y squared equals 16 for z between zero and three with d s oriented outward remember that in this case d s this vector is just a shorthand for the unit normal vector times d s in this case the surface s is the exact same surface as we were looking at with the scalar field surface integral which means that we can use the same parametrization r of theta and z as four cosine theta for sine theta and z and from there we can also use the same cross product in this case we're not going to take the magnitude we just want the actual vector from our same computations with the scalar field example we got that r sub theta cross r sub z was 4 cosine theta 4 sine theta and 0. now is this vector oriented outward like we want it to be well let's think about what would happen if we plugged in a particular point say we plugged in theta equals zero and z equals zero in that case we're going to have r of zero and zero which will be four times the cosine of zero is one so that's just four four times the sine of zero is zero so that's zero and z equals zero so this is our point and then our normal vector is going to be given by this cross product evaluated at 0 0 that's going to be 4 cosine 0 is 4 4 sine 0 is 0 and then we have a 0. so this is the normal vector to the surface remember that this surface is a cylinder so we're looking at a surface that looks something like this the point 4 0 0 is going to be at the bottom here in the positive x direction in the normal vector is going to point outwards in the positive x direction which means that this normal vector is indeed oriented outward so this normal vector is the one that we need we don't need to take the negative now before we plug everything into the integral let's compute the dot product of f with r sub theta cross r sub z at the beginning that dot product is going to be first this vector field f if we plug in the point here now we dot that with the cross product that we have right here and when we do this dot product out notice first that we're going to have the x components multiplied together the 4 cosine theta on the top and bottom will cancel so we'll just have a z then we add the product of these two which is going to be 4 squared times 4 is 64 cosine squared theta sine theta and then we're going to add e to the four cosine theta squared plus z squared but just times zero because the z component of our normal vector is zero it doesn't matter what the z component of our vector field is that's not going to contribute to our answer at all so we don't need to worry about it that means that our final integral is going to be the double integral over the same ranges as in the scalar field case so the integral from 0 to 2 pi of the integral from 0 to 3 of the dot product that we just calculated so that's z plus 64 cosine squared theta sine theta times dz d theta from there it's just an ordinary double integral so i'll skip through the computations and we'll get to the final answer which is 9 pi so that is the value for our vector field surface integral so that's how we do surface integrals in scalar and vector fields instead of a double integral that has a flat two dimensional input range or a line integral that integrates over a curve in space a surface integral integrates over a surface in three dimensions when we're looking at a vector field we need to find the unit normal vector times d s once we parametrize our curve we can do that using the cross product between the partial derivatives and for a scalar field we don't need the normal vector at all so we can just take the magnitude of that cross product once we plug in our parametrization it just becomes an ordinary double integral and we can solve it to get to our solution [Music]

Channel: Mu Prime Math

Views: 49,754

Keywords: math

Id: hVBoEEJlNuI

Length: 27min 50sec (1670 seconds)

Published: Fri Jul 17 2020