Integrals

Vectors

Vector Functions

### Discussion

You must be signed in to discuss.

Lectures

Join Bootcamp

### Video Transcript

so for us to evaluate this line integral. What we conduce do is rewrite it, using this formula of top here. So let's go ahead and do that. So this is going to be, um X minus y plus three and then I'll just put see down here and then DS. And so this is going to translate into will go from 0 to 2 pi 0 to 2 pi of what we plug in x and y into f or in this case, X of T is going to be co sign T. Y T is going to be signed because of the I and J components. So we just plug those in. So we're plugging it into right here. So this would be cosign t minus sign t plus three. And then we need to multiply by the square root of the derivatives of each of these. Just like when we did, like with the arc length formula. So the derivative of coastline is going to be negative signs, hopefully square. That gives us a sign. Square T signs derivative is co sign square that we get cosign square tea and we'll signs where plus coastlines. Where does Pythagoras. That's one Route one. So this is just all going to be one. So I'm just ignore that right? And now we just go ahead and integrate this so it'd be 0 to 2 pi co sign T minus sign T plus three d t. And so co sign If we integrate that gives a sign of tea and then integrating sign of t gives us negative co science that this would be plus co sign and then they'll just be plus three t and we evaluate from 0 to 2 pi. So when we plug into pie, um, sign a zero co sign is one and then three t is going to be six pie, but only princess. So six pie and then we subtract off when we plug in zeros. The signs also zero co sign is one again, and then three t is going to be zero. So actually, the ones here counts out, and we'll just be left with six pie. So our line integral will be equivalent to six pie

University of North Texas

Integrals

Vectors

Vector Functions

Lectures

Join Bootcamp