6.2 Examples on Disk Washer Part 1

All right. So let's continue with the discussion of uh you know with some examples. So the so we're going to give you some um examples on finding the volume of the solid obtained by rotating the given curves for the specific line.

Now we need to sketch the region the solid and a typical disc or or washer. So we don't quite need this to be um beautiful but at least we know what we're doing. Okay.

So first we get the region. Okay. So this region well before we sketch the region we need to know how to sketch the curve of course.

So first I have y = to 1 /x and then I have y = to 0. So let me let me use different colors. Okay.

So we got a x axis. We have a yaxis and y = 0 actually it is the x axis. Okay.

And then x = 1 and x = 4. So y = 0 x = 1 x = 4. So basically this is the region.

Okay. And We're going to rotate this region about the x-axxis. Okay.

So when we rotate the region about the x axis. Okay. So now I'm going to only cut the region.

Okay. Just like this. So this is the region and we're rotating about the x axis.

So the way to do that we're trying to get a sort of like a mirror. Okay. And then we're going to draw some circles.

So this is the solid. Okay. And now I'm going to use this to get a um disc or washer.

So I'm going to get this uh well, I'll say the crosssection. So just get the cross-section. Okay.

I'm going to get a cross-section. Okay. So, namely the crosssection is a circle.

Okay. So, this is basically a circle. Okay.

Now this is the radius R. Okay. So area is of course right R square.

Now I want to emphasize that this is our R. Okay. So this is what I call connection.

Okay. made a connection. All right.

So, let's get the uh let's um get the volume. Okay. So, first thing first, I have a cross-section like this.

So cross sections. So I have this cross-section and I'm going to have something like this. Okay.

So those are the cross-sections. And as you can see that the cross-section is sum up sum up in this direction. Okay.

So in this direction therefore v is a b area oops sorry so area dx okay now where area is pi r² because this is the um a circle. Okay, the cross-section is a circle. Now this r actually comes from here.

So with the with the connection with the relation. So this is r and this is r. So the orientation of r is top minus bottom.

Okay? Top minus bottom. Keep that in mind.

Okay? because cuz this r is on top here and that's the bottom which means we're going to have x top sorry it's y okay so this is the y direction so this is yt minus yb okay so yt this is our yt 1 /x minus yb so y bottom so that's the bottom, right? So, this is the bottom zero.

Okay? So, that's how we analyze it. And therefore, we have pi 1 /x², which is um * 1 /x².

So, that's the area. And pretty much we're done, right? Okay.

So, that's our pi * 1x^2 dx. Now we have the boundary. Okay, the boundary.

So this boundary comes up here comes here. Okay, because we are doing it in terms of x, right? So in this direction.

So the boundary is from here to here which is x = 1 and x = to 4. So one to four and that's how we set it up. And after we set it up, simply solve it.

So anti-derivative pi is a constant. Anti-derivative of um 1 /x² is - 1 /x. Okay.

And then plug it in. Evaluate this 1 I mean 1 /x from 1 to 4. So pi take it out.

-1 over4 minus negative be careful on this negatives. Okay. And once we do that we have 3 over 4 pi.

Easy as that. Okay. So that complete the first examples.

This complete the first example.