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Hovedindhold

# Stokes' teorem bevis del 5

## Video udskrift

Now that we have a parameterization for the boundary of our surface right up here, let's think a little bit about what the line integral-- and this was the left side of our original Stokes' theorem statement-- what the line integral over the path C of F, our original vector field F, dot dr is going to be. And now, once again, this R we're talking about the path of this boundary right over here, C. Not C1 that we did to kind of get some of our foundational material going. So F dot dr-- well, we remember what F is. F was all the way up here. Its i, j, and k components were just P, Q, and R. So that's kind of easy to remember. But let's think about what dr is equal to. And we're going to have to break out a little bit of our three-dimensional or our multivariable chain rule right over here. So dr is the same thing as dr dt times dt. So we really just have to figure out what the derivative of R with respect to t. And this one we're going to have to break out a little bit of the chain rule. So let me write this down. So dr dt is going to be equal to the derivative of x with respect to t times i plus the derivative of y with respect to t-- I'm just taking the derivative with respect to t-- j. Because z is a function of x, which is a function of t, and z is also a function of y, which is a function of t, we're going to have to break out our multivariable chain rule. So if we want to take the derivative of z with respect to t-- I'll do it separately here, and then I'll write it down down here. The derivative of z with respect to t-- the way I conceptualize is, what's all the different ways that z can change from a change in t? Well, it could change because x is changing due to a change in t. So z could change due to x, the partial of z with respect to x, when x changes with respect to t. But then that's not the only way that z can change. We have to add to that how z can change with respect to y, partial of z with respect to y times how fast or how y is changing with respect to t. And this is just our multivariable chain rule. And so this is dz dt. So I'll just write it right over here. And I'll use slightly different notation that's consistent with what we were doing before, and it'll help make things a little bit clearer. So it's going to be the partial of z with respect to x, dx dt, plus-- actually, let me write it this way-- plus the partial of z with respect to y dy dt. And then we're going to multiply everything times our k. So with this out of the way, if we wanted what dr is, dr is just going to be this whole thing times dt. So let's do that. So now we can rewrite our line integral right over here. And we're going to now go into the t domain. And so t is going to go between a and b. And F dot dr-- remember, F's components were just the functions P, Q, and R. And each of those were functions of x, y, and z. And z is a function of x and y, so we'll have to think about all of that in a little bit. Use a little more of our multivariable chain rule. But when we take the dot products, we're just going to take the corresponding components and multiply them. So it is going to be-- actually, let me just copy and paste it. Let me rewrite it down here. Our vector field F-- and I'm going to write it a little bit shorter. Our vector field F is P times i plus Q times j plus R times k. So when we take our dot product of F dot dr, we're essentially taking the dot product of this and this. And we have to throw a dt at the end. So we're going to get P times dx dt plus Q times dy dt plus R times all of this business over here, which is the partial of z with respect to x dx dt plus the partial of z with respect to y dy dt. And then we have to multiply all of this times dt. We can't forget that part right over there. So we're going to multiply all of this right here times dt. Now, I'm going to leave you there in this video just because I'm afraid of making careless mistakes. What we're going to do now is we're going to rearrange this whole thing, recognize that this is the same thing as this right over here. Then that gives it a form that we can apply Green's theorem using this boundary right over here. And then when we do a little bit more algebraic manipulation, we're going to see that this thing simplifies to this thing right over here and proves Stokes' theorem for our special case.