6.2 Examples on Disk Washer Part 2

Well, let's take a look at the second problem. So again, no matter what, we got to sketch the region and then the solid and then the cross-section. Okay, so what we have this time y = to x².

So probably this one, a long one, you know, taller one. Okay. And then I have this guy.

So y = x². And then I have y = 2x. So this one y = 2x.

And the region is right here. Okay. So that's the region.

Okay. Now we are going to rotate this about the y axis. break down.

So, I have this, I have this, and then I'm going to have a uh something like that. Okay. So, we rotate this region.

We rotate this region to this side. So, what we have actually is let me use this color purple. Okay.

So we have this solid. Okay. And it is not that hard to see that when we sum sum them up in the direction.

So this is the direction. Then we use dy. Okay, because we stack up, right?

We stack up the the solid. We stack up the solid. All right.

Now, this is the cross-section. Okay. And this cross-section has the shape like a ring.

Okay, not necessarily a beautiful graph, but as long as you can see it. Okay, now this area, the area of a ring equals pi big r² - pi little r². If you want to, you can do this.

It doesn't really matter. Okay. Now, this is little R and this is big R.

Okay. Now, we're going to take it out here. This Well, what color should I use?

Uh, let me get a yellow. This is my big R. Okay.

So that's the big R. So maybe zero to cover it. So this is big R.

Okay. Uh what color should I use next? Rainbow.

Okay. So this is Well, this rainbow sucks. Um light orange.

Let's use orange. So this is little r where this is my little r. Okay.

Now if we come back here so we're drawing the connection right. So connection. Okay.

So connection this is my little R and this is my big R and the orientation will be X will be right and left right X right and X left okay so that's how we analyze ize it. Therefore, the volume would be the integral of area dy in this case. Okay, dy in this case because the orientation right is going up and down.

Now the boundary the boundary for this one, right? Okay, where we set y = x² and y = 2x. Okay, so we set system of equation.

Basically, we're looking for the two intersection. Okay, set set this equation. Okay, try to find x and y value that satisfy both equations.

Okay, and then solving it, I have 2x = to x². Move everything to one side. Factor.

Don't cancel. Okay, don't cancel the x. And uh we know that our x = 0 or x = to 2.

Now because we're doing dy, we also need to figure it out. y would be 0 squar which is zero or y would be 2^ 2 which is 4. Okay.

So I'm just using one of this equation. Therefore from here we know this is from zero to four. And now the area.

Okay. For area of the cross-section equals pi r² - r. Now big R.

Okay. So big R will be this curve. So XR - XL.

Now that curve come from y equals so this one comes from y = x² and I want to figure out what's x right so that means my x is plus minus radical y in this case we take the positive because it's on the right hand side and this one would be um y = to 2x well no not even that it is the um yaxis. So the yaxis is basically x equal to zero. Therefore I have radical y - 0.

Okay. And similarly I can have you know those yellow color actually is my scratch paper. And then for little r.

So x r - x l. And this guy is the line. And this line is y = 2x.

Which means x is 12 y. And this again would be the um yaxis which is x= 0. Okay.

So finally this is just 12 y minus 0. Okay. All right.

So from the analysis I know that this is radical y^² - y / 2². Okay. Because my r equals uh do it this way because x² minus uh yaxis in this case it will be this.

Okay, namely this one and little r would be y = to 2x. So that line with this y axis and when we solve it solve it for for x would be 12 y - 0. So that's how we got that.

So finally I have y - y^2 over 4. Plug it in and done. So pi y - y^² over 4 d y and solve it anti-derivative 12 y^ 2 - 1 / 4 anti-derivative will be 1/3 y cub from 0 to 4 and solve it pi 12 4^ 2 - 1 12 4 cub - 0 because it's just zero right so pi time I think it's 16 um how can I do that over six 16 * 1 is 6 uh 16 * 1 / 6 which is 8 over 3 pi.

Okay. Uh well I can do it. So that's 8 minus uh 36 over 3, right?

And then you'll get that. Okay. Anyway, so we got the answer for this problem.

Now the textbook actually um the textbook actually called this. Okay. So the textbook called this disk method.

Okay. Because the cross-section is a disk. So no hole.

Okay. And we call this method washer. This is a washer method because there's a hole.

It come from the ring, right? The hole come from the ring. Now, in the next section, actually, we're using another method called cylindrical methods.

So, basically, we have two big methods. One is about a cross-section. The other one is the cylindrical shell shell method.

Within the within the cross-section method, we have washer and disk. So, please make sure that you can tell them apart. Okay?

So, uh I'm going to give you two more example. Actually the next two example come from the same region but um I'm going to to uh rotate that region about different axis which will result in um washer and disc separately.