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0:00

Felicia S.

If $f(x)=x+\sqrt{2-x}$ and $g(u)=u+\sqrt{2-u},$ is it true that $f=g ?$

00:33

Dungarsinh P.

00:51

Heather Z.

00:06

Jeffery W.

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alrighty. So again, what we're looking for is the flux out of this region or across the surface. That's the boundary of this region of the Specter field, which is technically a surface integral. But we don't want to do that. That's hard. So what we do want to dio is convert this to a triple integral of the divergence. And so the volume region will call are well described in a second is going to turn out toe be a little bit easier to describe in cylindrical coordinates, but we'll get to that. But we're just integrating the divergence. And so what is the divergence? Well, partial with respect to X of this zero partial with respect to this is to Why so that's two why and then we just have TV. And like I said, this region I think we want to describe in terms of cylindrical coordinates. So Z, first of all, is going from zero toe Y minus three now in polar coordinates. Why is our sign data? So this is really are signed data minus three. That's too. Why dizzy and then our goes from zero to three because the projection onto the X Y plane is just a circle of radius three, and then data goes from 0 to 2 pi. And don't forget that we actually have it. Are D R? These data in the two y we can actually look place but in polar coordinates as to our side data already. And so this is the integral is your two pi 0 to 3 of thio are signed. Data times are signed data minus three and then we have another are DRD data. All right, so here we have a r squared and then another are sexually r cubed. And here we'll have a r squared. So this will be this first term Will Havenaar cubes. So that will be art of the fourth two are the fourth over four Just part of this fourth over to then we'll have a sine squared data and then we'll have a are to the third divided by three times three. So just one side data and then that's evaluated from 0 to 3 data. Okay, so this is just going to be well, 81 over to this is actually going to be zero now. It's really going to be 81 time signed data. But when I integrate science data from 0 to 2 pi, that's going to give me zero. So I'm just gonna have the integral of 81/2 0 to 2 pi sine squared data teeth data. But now the integral of sine squared data from 0 to 2 pi is something I'm sure you've seen before so we can use the power reducing trick identity make sine squared equal toe one minus cosine, two theta over to. And when we when we evaluate that from 0 to 2 pi, we just get pie. So the final answer is 81 pie divided by two. And so that's the flux through this cylindrical surface and we found it by converting the surface integral to a triple integral, using the divergence through him.