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JOEL LEWIS: Hi. Welcome back to recitation. In lecture, you've been learning about triple integration. And I have a problem here for you on computing a volume of a region using a triple integral. So let's look at this. So I have a volume and I'm describing it to you; it's the volume inside the paraboloid z equals x squared plus y squared and bounded by the plane z equals 2y. So I've drawn a little picture here for you. So this is the paraboloid here. And we're just taking a plane cut of it. And so this is going to slice off some chunk of that paraboloid, and what I want to know is, what's the volume of that piece that gets cut off by that plane there? So below the plane and above the paraboloid. So, why don't you pause the video, take some time, work out this problem, come back, and we can work on it together. I hope you had some luck with this problem. I think it's a bit of a tricky one, so let's start to work through it together. ...

JOEL LEWIS: Hi. Welcome back to recitation. You've been doing some work on trig integration. I have a nice example here of a problem that requires trig integration in order to solve. So what I'd like you to do is to compute the volume of the solid that you get when you take one hump of the curve y equals sine a*x and you revolve it around the x-axis. So you take the curve between two consecutive roots, and then you, you know, revolve that around the x-axis and that gives you some, I don't know, vaguely football-shaped thing. And so then the question is, what's the volume of that solid? So why don't you pause the video, take a little while to work that out, come back, and we can work it out together. Welcome back. In order to solve this problem, we just are going to apply our usual methods for computing a volume of a solid of rotation. So in order to do that, remember that one of the things you need is you need to know the region over which you're i...

JOEL LEWIS: Hi, welcome back to recitation. In lecture, Professor Miller did a bunch of examples of integrals involving trigonometric functions. So I thought I would give you a couple more, well I guess three more examples, some of them are a little different flavor than the ones he did, but some nice examples of some integrals you can compute now. One thing you might need for them is you're going to have to remember some of your trig identities. Just like you had to remember some of them last time. So in particular, one identity that you might need today that you didn't need in lecture, was the angle sum identity for cosine. So let me just remind you what that is, it says the cosine of a plus b is equal to cosine a cosine b minus sine a sine b. So you're going to need that formula to compute one of these three integrals. So why don't you pause the video, take some time to work these out. Come back you can check your answers against my work. Hopefully you ...

JOEL LEWIS: Hi. Welcome back to recitation. In lecture, you've been learning about using gradients to compute tangent planes to surfaces. So I have an example of a practice problem here for you. So what I'd like you to do in part a is to use gradients to find the tangent plane to the surface z equals x cubed plus 3x y squared at the point (1, 2, 13). And in part b, I'd like you to do something similar, which is to use gradients to find the tangent line to the curve x cubed plus 2xy plus y squared equals 9 at the point (1, 2). So why don't you pause the video, have a couple goes at those. Come back and we can work on them together. So hopefully, you had some good luck working on these problems. Let's just take a look at them. So for part a, you're given a function in the sort of usual form that we use to graph it, which is you're given z equals a function of x and y. But in order to apply this gradient method, what we really want is we want to l...

JOEL LEWIS: Hi. Welcome back to recitation. You've been learning in lecture about matrices and their various applications, and one of them is to solving systems of linear equations. So I have here a system of three linear equations for you. 2x plus c*z equals 4, x minus y plus 2z equals pi, and x minus 2y plus 2z equals minus 12. So what I'd like you to do is the following. Find the value of c-- or all values of c-- for which, first of all, there's a unique solution to this system. Second of all, for which the corresponding homogeneous system has a unique solution. So remember that the corresponding homogeneous system is the system where you just replace these constants on the right by 0. So it's a very similar-looking system. The left-hand sides are all the same, but the right-hand sides are replaced with 0. So you want to find the value of c for which this system has a unique solution, the value of c for which the corresponding homogeneous system has a u...

JOEL LEWIS: Hi. Welcome back to recitation. In lecture, you've been learning about Stokes' Theorem. And I have a nice question here for you that can put Stokes' Theorem to the test. So what I'd like you to do is I'd like you to consider this field F. So its components are 2z, x, and y. And the surface S that is the top half of the unit sphere. So it's the sphere of radius 1 centered at the origin, but only its top half. Only the part where z is greater than or equal to 0. So what I'd like you to do is to verify Stokes' Theorem for this surface. So that is, I'd like you to compute the surface integral that comes from Stokes' Theorem for this surface, and the line integral that comes from Stokes' Theorem for the surface, and check that they're really equal to each other. Now, before we start, we should just say one brief thing about compatible orientation. So I didn't give you any orientations, but of course, it doesn'...

JOEL LEWIS: Hi. Welcome back to recitation. In lecture you've started learning about quadratic approximation. So today we're just going to do a quick example of it. So I have a question written here on the board: What is the quadratic approximation of the function f of x equals e to the x plus x squared-- so here x and plus x squared is the exponent, so it's e to the quantity x plus x squared-- near x equals 0. So why don't you take a minute or two, pause the video, work this out on your own and then come back and we can do it together. All right. Welcome back. So there are two different ways we can do this problem. Let's first just do it the totally straightforward way, which is that you have this formula for quadratic approximations in terms of the derivatives of your function. And so, so we can just apply that formula. So here, so the formula is that the quadratic approximation of the function f-- so here, near the point 0-- is equal to f of 0 plus ...

JOEL LEWIS: Hi. Welcome back to recitation. One of the things you've been learning about in lecture is how to solve some max-min problems. How to find the maximum or minimum of a given function given some constraints and figure out where it reaches its largest or smallest value. So here I have a particular example of such a problem. So we're going to build a cardboard box. And our cardboard box has to meet the following criteria. So its volume has to be 3 units. The front and back of the cardboard box are going to be made just of single-thickness cardboard. But the two sides, the left and right, are going to be made double-thick. And the bottom is going to be made triple-thick. We're not going to have a top, it's just going to be an open box. So it's going to have five sides: two of single thickness, two of double thickness, and with the bottom of triple thickness. So, the question is-- there are, you know, a lot of different shapes of box with total v...

JOEL LEWIS: Hi. Welcome back to recitation. In lecture, you've been learning about line integrals of vector fields. And I have a couple of nice questions on that subject for you right here. So I want F to be the vector field whose first coordinate is x*y and whose second coordinate is x squared plus y squared. And so what I'd like you do is compute the line integral of F around two different curves C. So both curves start at the point (1, 1) and they end at the point (2, 4). So in part a, the curve is just the straight line that connects the point (1, 1) to (2, 4). And in part b, the curve is this sort of piecewise-- it's two sides of a rectangle, right? It goes straight up until it gets to the point (1, 4), and then it goes across to the point (2, 4). So it's a piecewise smooth curve, path, that connects those two points. So I'd like you to compute the integral over each of these curves of F dot dr. So why don't you pause the video, have a go at t...

JOEL LEWIS: Hi. Welcome back to recitation. In lecture, you've been learning about line integrals of vector fields. And I have one problem here that requires you to do two line integrals of vector fields on that subject. So OK. So in both problems, F is going to be the vector field whose coordinates are x*y and x squared plus y squared. And C is going to be the arc of the parabola y equals x squared that starts at the point (1, 1) and ends at the point (2, 4). All right? So what I'd like you to do is to compute the integral over this curve C of F dot dr in two different ways. So the first time, I'd like you to use the natural parametrization x equals t, y equals t squared. For me, at least, that's the first parametrization that I think of when I think about this curve. But I'd also like you to do it again using a different parametrization, using the parametrization x equals e to the t, y equals e to the 2t. So this still parametrizes the curve y equals...

JOEL LEWIS: Hi. Welcome back to recitation. In lecture, you've been learning about line integrals and computing them around curves and closed curves and in various different ways. So here I have some problems on line integrals for you. So in all cases I want C to be the circle of radius b. So b is some constant, some positive constant. It's the circle of radius b centered at the origin, and I want to orient it counterclockwise. And then what I'd like you to do is for each of the following vector fields F, I'd like you to compute the line integral around C of F dot dr. So in the first case, where F is x*i plus y*j. In the second, where F is g of x, y times x*i plus y*j. So here g of x, y is some scalar function. But you don't know a formula for this function. So your answer might be in terms of g, for example. You can assume it's a continuous, differentiable nice function. And then the third one, F is minus y*i plus x*j. Now before you start, I want...